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\begin{document}

\title{Examples}


\maketitle

\phantomsection
\label{section-phantom}

\tableofcontents

\section{Introduction}
\label{section-introduction}

\noindent
This chapter will contain examples which illuminate the theory.






\section{A nonintegral connected scheme whose local rings are domains}
\label{section-connected-locally-integral-not-integral}

\noindent
We give an example of an affine scheme $X = \Spec(A)$ which is
connected, all of whose local rings are domains, but which is not integral.
Connectedness of $X$ means $A$ has no nontrivial idempotents, see
Algebra, Lemma \ref{algebra-lemma-disjoint-decomposition}.
The local rings of $X$ are domains if, whenever $fg = 0$ in $A$, every
point of $X$ has a neighborhood where either $f$ or $g$ vanishes.
As long as $A$ is not a domain, then $X$ is not integral
(Properties, Definition \ref{properties-definition-integral}).

\medskip\noindent
Roughly speaking, the construction is as follows: let $X_0$ be the cross
(the union of coordinate axes) on the affine plane. Then let $X_1$ be
the (reduced) full preimage of $X_0$ on the blow-up of the plane ($X_1$
has three rational components forming a chain).  Then blow up the
resulting surface at the two singularities of $X_1$, and let $X_2$ be
the reduced preimage of $X_1$ (which has five rational components), etc.
Take $X$ to be the inverse limit. The only problem with this construction
is that blow-ups glue in a projective line, so $X_1$ is not affine. Let us
correct this by glueing in an affine line instead (so our scheme will be an
open subset in what was described above).

\medskip\noindent
Here is a completely algebraic construction: For every $k \ge 0$, let $A_k$
be the following ring: its elements are collections of
polynomials $p_i \in \mathbf{C}[x]$ where $i = 0, \dots, 2^k$ such that
$p_i(1) = p_{i + 1}(0)$. Set $X_k = \Spec(A_k)$. Observe that $X_k$ is
a union of $2^k + 1$ affine lines that meet transversally in a chain.
Define a ring homomorphism $A_k \to A_{k + 1}$ by
$$
(p_0, \dots, p_{2^k})
\longmapsto
(p_0, p_0(1), p_1, p_1(1), \dots, p_{2^k}),
$$
in other words, every other polynomial is constant. This identifies
$A_k$ with a subring of $A_{k + 1}$. Let $A$ be the direct limit of $A_k$
(basically, their union). Set $X = \Spec(A)$. For every $k$, we have
a natural embedding $A_k \to A$, that is, a map $X\to X_k$.
Each $A_k$ is connected but not integral; this implies that $A$ is
connected but not integral. It remains to show that the local rings of
$A$ are domains.

\medskip\noindent
Take $f, g \in A$ with $fg = 0$ and $x \in X$. Let us construct a
neighborhood of $x$ on which one of $f$ and $g$ vanishes. Choose $k$
such that $f, g \in A_{k - 1}$ (note the $k - 1$ index).
Let $y$ be the image of $x$ in $X_k$. It suffices to prove that $y$ has
a neighborhood on which either $f$ or $g$ viewed as sections of
$\mathcal{O}_{X_k}$ vanishes.
If $y$ is a smooth point of $X_k$, that is, it lies on only one of the
$2^k + 1$ lines, this is obvious. We can therefore assume that $y$ is one
of the $2^k$ singular points, so two components of $X_k$ pass through
$y$. However, on one of these two components (the one with odd index),
both $f$ and $g$ are constant, since they are pullbacks of functions on
$X_{k - 1}$. Since $fg = 0$ everywhere, either $f$ or $g$ (say, $f$)
vanishes on the other component.
This implies that $f$ vanishes on both components, as required.






\section{Noncomplete completion}
\label{section-noncomplete-completion}

\noindent
Let $R$ be a ring and let $\mathfrak m$ be a maximal ideal. Consider the
completion
$$
R^\wedge = \lim R/\mathfrak m^n.
$$
Note that $R^\wedge$ is a local ring with maximal ideal
$\mathfrak m' = \text{Ker}(R^\wedge \to R/\mathfrak m)$.
Namely, if $x = (x_n) \in R^\wedge$ is not in $\mathfrak m'$, then
$y = (x_n^{-1}) \in R^\wedge$ satisfies $xy = 1$, whence $R^\wedge$ is local by
Algebra, Lemma \ref{algebra-lemma-characterize-local-ring}. Now it is
always true that $R^\wedge$ complete in its limit topology (see the
discussion in
More on Algebra, Section \ref{more-algebra-section-topological-ring}).
But beyond that, we have the following questions:
\begin{enumerate}
\item Is it true that $\mathfrak m R^\wedge = \mathfrak m'$?
\item Is $R^\wedge$ viewed as an $R^\wedge$-module $\mathfrak m'$-adically
complete?
\item Is $R^\wedge$ viewed as an $R$-module $\mathfrak m$-adically complete?
\end{enumerate}
It turns out that of these questions all have a negative answer.
The example below was taken from an unpublished note of
Bart de Smit and Hendrik Lenstra. See also
\cite[Exercise III.2.12]{Bourbaki-CA} and
\cite[Example 1.8]{Yekutieli}

\medskip\noindent
Let $k$ be a field, $R = k[x_1, x_2, x_3, \ldots]$, and
$\mathfrak m = (x_1, x_2, x_3, \ldots)$.
We will think of an element $f$ of $R^\wedge$ as a (possibly) infinite sum
$$
f = \sum a_I x^I
$$
(using multi-index notation) such that for each $d \geq 0$ there
are only finitely many nonzero $a_I$ for $|I| = d$. The maximal
ideal $\mathfrak m' \subset R^\wedge$ is the collection of $f$ with
zero constant term. In particular, the element
$$
f = x_1 + x_2^2 + x_3^3 + \ldots
$$
is in $\mathfrak m'$ but not in $\mathfrak m R^\wedge$ which
shows that (1) is false in this example. Note that we do have
$\mathfrak m R^\wedge \subset \mathfrak m'$. Hence, $R^\wedge$
is not $\mathfrak m$-adically complete as an $R$-module, then it is also not
$\mathfrak m'$-adically complete. To show that $R^\wedge$ is not
$\mathfrak m$-adically complete (as an $R$-module)
it suffices to show that $K_2 = \text{Ker}(R^\wedge \to R/\mathfrak m^2)$
is not equal to $\mathfrak m^2R^\wedge$, see
Algebra, Lemma \ref{algebra-lemma-hathat}.
Note that an element of $\mathfrak m^2R^\wedge \subset (\mathfrak m')^2$
can be written as a finite sum
\begin{equation}
\label{equation-sum}
\sum\nolimits_{i = 1, \ldots, t} f_i g_i
\end{equation}
with $f_i, g_i \in R^\wedge$ having vanishing constant terms.
To get an example we are going to choose an $z \in K_2$
of the form
$$
z = z_1 + z_2 + z_3 + \ldots
$$
with the following properties
\begin{enumerate}
\item there exist sequences $1 < d_1 < d_2 < d_3 < \ldots $ and
$0 < n_1 < n_2 < n_3 < \ldots$ such that
$z_i \in k[x_{n_i}, x_{n_i + 1}, \ldots, x_{n_{i + 1} - 1}]$
homogeneous of degree $d_i$, and
\item in the ring $k[[x_{n_i}, x_{n_i + 1}, \ldots, x_{n_{i + 1} - 1}]]$
the element $z_i$ cannot be written as a sum (\ref{equation-sum})
with $t \leq i$.
\end{enumerate}
Clearly this implies that $z$ is not in $(\mathfrak m')^2$
because the image of the relation (\ref{equation-sum}) in the
ring $k[[x_{n_i}, x_{n_i + 1}, \ldots, x_{n_{i + 1} - 1}]]$
for $i$ large enough would produce a contradiction. Hence it suffices
to prove that for all $t > 0$ there exists a $d \gg 0$ and an integer
$n$ such that we can find an homogeneous element
$z \in k[x_1, \ldots, x_n]$ of degree $d$ which cannot be written as
a sum (\ref{equation-sum}) for the given $t$ in $k[[x_1, \ldots, x_n]]$.
Take $n > 2t$ and any $d > 1$ prime to the characteristic of $p$ and
set $z = \sum_{i = 1, \ldots, n} x_i^d$. Then the vanishing locus
of the ideal
$$
(\frac{\partial z}{\partial x_1}, \ldots, \frac{\partial z}{\partial x_n})
=
(dx_1^{d - 1}, \ldots, dx_n^{d - 1})
$$
consists of one point. On the other hand,
$$
\frac{\partial ( \sum\nolimits_{i = 1, \ldots, t} f_i g_i ) }{\partial x_j}
\in (f_1, \ldots, f_t, g_1, \ldots, g_t)
$$
by the Leibniz rule and hence the vanishing locus of these derivatives
contains at least
$$
V(f_1, \ldots, f_t, g_1, \ldots, g_t) \subset
\Spec(k[[x_1, \ldots, x_n]]).
$$
Hence this is a contradiction as the dimension of
$V(f_1, \ldots, f_t, g_1, \ldots, g_t)$ is at least $n - 2t \geq 1$.

\begin{lemma}
\label{lemma-noncomplete-completion}
There exists a local ring $R$ and a maximal ideal $\mathfrak m$ such that
the completion $R^\wedge$ of $R$ with respect to $\mathfrak m$ has the
following properties
\begin{enumerate}
\item $R^\wedge$ is local, but its maximal ideal is not equal to
$\mathfrak m R^\wedge$,
\item $R^\wedge$ is not a complete local ring, and
\item $R^\wedge$ is not $\mathfrak m$-adically complete as an $R$-module.
\end{enumerate}
\end{lemma}

\begin{proof}
This follows from the discussion above as (with $R = k[x_1, x_2, x_3, \ldots]$)
the completion of the localization $R_{\mathfrak m}$ is equal to the
completion of $R$.
\end{proof}





\section{Noncomplete quotient}
\label{section-noncomplete-quotient}

\noindent
Let $k$ be a field. Let
$$
R = k[t, z_1, z_2, z_3, \ldots, w_1, w_2, w_3, \ldots, x]/
(z_it - x^iw_i, z_i w_j)
$$
Note that in particular $z_iz_jt = 0$ in this ring. Any element $f$ of $R$
can be uniquely written as a finite sum
$$
f = \sum\nolimits_{i = 0, \ldots, d} f_i x^i
$$
where each $f_i \in k[t, z_i, w_j]$ has no terms involving the products
$z_it$ or $z_iw_j$. Moreover, if $f$ is written in this way, then
$f \in (x^n)$ if and only if $f_i = 0$ for $i < n$.
So $x$ is a nonzerodivisor and $\bigcap (x^n) = 0$.
Let $R^\wedge$ be the completion of $R$ with respect to the ideal $(x)$.
Note that $R^\wedge$ is $(x)$-adically complete, see
Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}.
By the above we see that an element of $R^\wedge$ can be uniquely written
as an infinite sum
$$
f = \sum\nolimits_{i = 0}^\infty f_i x^i
$$
where each $f_i \in k[t, z_i, w_j]$ has no terms involving the products
$z_it$ or $z_iw_j$. Consider the element
$$
f = \sum\nolimits_{i = 1}^\infty x^{i - 1} w_i =
xw_1 + x^2w_2 + x^3w_3 + \ldots
$$
i.e., we have $f_n = w_n$. Note that $f \in (t , x^n)$ for every $n$
because $x^mw_m \in (t)$ for all $m$.
We claim that $f \not \in (t)$. To prove this assume that
$tg = f$ where $g = \sum g_lx^l$ in canonical form as above.
Since $tz_iz_j = 0$ we may as well assume that none of the $g_l$ have
terms involving the products $z_iz_j$. Examining the process to
get $tg$ in canonical form we see the following:
Given any term $c m$ of $g_l$ where $c \in k$ and $m$ is a
monomial in $t, z_i, w_j$ and we make the following replacement
\begin{enumerate}
\item if the monomial $m$ does not involve any $z_i$, then $ctm$ is
a term of $f_l$, and
\item if the monomial $m$ does involve a $z_i$ then it is equal to
$m = z_i$ and we see that $cw_i$ is term of $f_{l + i}$.
\end{enumerate}
Since $g_0$ is a polynomial only finitely many of the variables $z_i$
occur in it. Pick $n$ such that $z_n$ does not occur in $g_0$.
Then the rules above show that $w_n$ does not occur in $f_n$ which is
a contradiction. It follows that $R^\wedge/(t)$ is not complete, see
Algebra, Lemma \ref{algebra-lemma-quotient-complete}.

\begin{lemma}
\label{lemma-noncomplete-quotient}
There exists a ring $R$ complete with respect to a principal ideal
$I$ and a principal ideal $J$ such that $R/J$ is not $I$-adically
complete.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{Completion is not exact}
\label{section-completion-not-exact}

\noindent
A quick example is the following. Suppose that $R = k[t]$. Let
$P = K = \bigoplus_{n \in \mathbf{N}} R$ and
$M = \bigoplus_{n \in \mathbf{N}} R/(t^n)$. Then there is a short exact
sequence $0 \to K \to P \to M \to 0$ where the first map is given by
multiplication by $t^n$ on the $n$th summand. We claim that
$0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0$ is not exact in the middle.
Namely, $\xi = (t^2, t^3, t^4, \ldots) \in P^\wedge$ maps to zero in
$M^\wedge$ but is not in the image of $K^\wedge \to P^\wedge$, because
it would be the image of $(t, t, t, \ldots)$ which is not an element of
$K^\wedge$.

\medskip\noindent
A ``smaller'' example is the following. In the situation of
Lemma \ref{lemma-noncomplete-quotient}
the short exact sequence $0 \to J \to R \to R/J \to 0$ does not remain
exact after completion. Namely, if $f \in J$ is a generator, then
$f : R \to J$ is surjective, hence $R \to J^\wedge$ is surjective, hence
the image of $J^\wedge \to R$ is $(f) = J$ but the fact that
$R/J$ is noncomplete means that the kernel of the surjection
$R \to (R/J)^\wedge$ is strictly bigger than $J$, see
Algebra, Lemmas \ref{algebra-lemma-completion-generalities} and
\ref{algebra-lemma-quotient-complete}.
By the same token the sequence
$R \to R \to R/(f) \to 0$ does not remain exact on completion.

\begin{lemma}
\label{lemma-completion-not-exact}
Completion is not an exact functor in general; it is not even
right exact in general. This holds even when $I$ is finitely
generated on the category of finitely presented modules.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{The category of complete modules is not abelian}
\label{section-non-abelian}


\noindent
Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal.
Consider the category $\mathcal{A}$ of $I$-adically complete
$R$-modules, see
Algebra, Definition \ref{algebra-definition-complete}.
Let $\varphi : M \to N$ be a morphism of $\mathcal{A}$.
The cokernel of $\varphi$ in $\mathcal{A}$ is the completion
$(\text{Coker}(\varphi))^\wedge$ of the usual cokernel
(as $I$ is finitely generated this completion is complete, see
Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}).
Let $K = \text{Ker}(\varphi)$. We claim that $K$ is complete and
hence is the kernel of $\varphi$ in $\mathcal{A}$. Namely, let
$K^\wedge$ be the completion. As $M$ is complete we obtain a factorization
$$
K \to K^\wedge \to M \xrightarrow{\varphi} N
$$
Since $\varphi$ is continuous for the $I$-adic topology, $K \to K^\wedge$
has dense image, and $K = \text{Ker}(\varphi)$ we conclude that $K^\wedge$
maps into $K$. Thus $K^\wedge = K \oplus C$ and $K$ is a direct sum
of a complete module, hence complete.

\medskip\noindent
We will give an example that shows that $\text{Im} \not = \text{Coim}$
in general. We take $R = \mathbf{Z}_p = \lim_n \mathbf{Z}/p^n\mathbf{Z}$
to be the ring of $p$-adic integers and we take $I = (p)$.
Consider the map
$$
\text{diag}(1, p, p^2, \ldots) :
\left(\bigoplus\nolimits_{n \geq 1} \mathbf{Z}_p\right)^\wedge
\longrightarrow
\prod\nolimits_{n \geq 1} \mathbf{Z}_p
$$
where the left hand side is the $p$-adic completion of the direct sum.
Hence an element of the left hand side is a vector $(x_1, x_2, x_3, \ldots)$
with $x_i \in \mathbf{Z}_p$ with $p$-adic valuation $v_p(x_i) \to \infty$ as
$i \to \infty$. This maps to $(x_1, px_2, p^2x_3, \ldots)$. Hence we see
that $(1, p, p^2, \ldots)$ is in the closure of the image but not in
the image. By our description of kernels and cokernels above it is
clear that $\text{Im} \not = \text{Coim}$ for this map.

\begin{lemma}
\label{lemma-complete-modules-not-abelian}
Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal.
The category of $I$-adically complete $R$-modules has kernels and
cokernels but is not abelian in general.
\end{lemma}

\begin{proof}
See above.
\end{proof}


\section{Regular sequences and base change}
\label{section-regular-base-change}

\noindent
We are going to construct a ring $R$ with a regular sequence
$(x, y, z)$ such that there exists a nonzero element $\delta \in R/zR$
with $x\delta = y\delta = 0$.

\medskip\noindent
To construct our example we first
construct a peculiar module $E$ over the ring $k[x, y, z]$
where $k$ is any field. Namely, $E$ will be a push-out as
in the following diagram
$$
\xymatrix{
\frac{xk[x, y, z, y^{-1}]}{xyk[x, y, z]} \ar[r] \ar[d]^{z/x} &
\frac{k[x, y, z, x^{-1}, y^{-1}]}{yk[x, y, z, x^{-1}]} \ar[r] \ar[d] &
\frac{k[x, y, z, x^{-1}, y^{-1}]}{yk[x, y, z, x^{-1}] + xk[x, y, z, y^{-1}]}
\ar[d] \\
\frac{k[x, y, z, y^{-1}]}{yzk[x, y, z]} \ar[r] &
E \ar[r] &
\frac{k[x, y, z, x^{-1}, y^{-1}]}{yk[x, y, z, x^{-1}] + xk[x, y, z, y^{-1}]}
}
$$
where the rows are short exact sequences (we dropped the outer zeros due
to typesetting problems). Another way to describe $E$ is as
$$
E = \{(f, g) \mid f \in k[x, y, z, x^{-1}, y^{-1}],
g \in k[x, y, z, y^{-1}] \}/\sim
$$
where $(f, g) \sim (f', g')$ if and only if there exists a
$h \in k[x, y, z, y^{-1}]$ such that
$$
f = f' + xh \bmod yk[x, y, z, x^{-1}], \quad
g = g' - zh \bmod yzk[x, y, z]
$$
We claim: (a) $x : E \to E$ is injective, (b)
$y : E/xE \to E/xE$ is injective, (c) $E/(x, y)E = 0$, (d) there
exists a nonzero element $\delta \in E/zE$ such that
$x\delta = y\delta = 0$.

\medskip\noindent
To prove (a) suppose that $(f, g)$ is a pair that gives rise to an
element of $E$ and that $(xf, xg) \sim 0$. Then there exists a
$h \in k[x, y, z, y^{-1}]$ such that $xf + xh \in yk[x, y, z, x^{-1}]$
and $xg - zh \in yzk[x, y, z]$. We may assume that
$h = \sum a_{i, j, k}x^iy^jz^k$ is a sum of monomials where only
$j \leq 0$ occurs. Then $xg - zh \in yzk[x, y, z]$ implies that
only $i > 0$ occurs, i.e., $h = xh'$ for some $h' \in k[x, y, z, y^{-1}]$.
Then $(f, g) \sim (f + xh', g - zh')$ and we see that we may assume
that $g = 0$ and $h = 0$. In this case $xf \in yk[x, y, z, x^{-1}]$
implies $f \in yk[x, y, z, x^{-1}]$ and we see that $(f, g) \sim 0$.
Thus $x : E \to E$ is injective.

\medskip\noindent
Since multiplication by $x$ is an isomorphism on
$\frac{k[x, y, z, x^{-1}, y^{-1}]}{yk[x, y, z, x^{-1}]}$ we see that
$E/xE$ is isomorphic to
$$
\frac{k[x, y, z, y^{-1}]}{
yzk[x, y, z] + xk[x, y, z, y^{-1}] + zk[x, y, z, y^{-1}]}
=
\frac{k[x, y, z, y^{-1}]}{xk[x, y, z, y^{-1}] + zk[x, y, z, y^{-1}]}
$$
and hence multiplication by $y$ is an isomorphism on $E/xE$. This clearly
implies (b) and (c).

\medskip\noindent
Let $e \in E$ be the equivalence class of $(1, 0)$.
Suppose that $e \in zE$. Then there exist $f \in k[x, y, z, x^{-1}, y^{-1}]$,
$g \in k[x, y, z, y^{-1}]$, and $h \in k[x, y, z, y^{-1}]$ such that
$$
1 + zf + xh \in yk[x, y, z, x^{-1}], \quad
0 + zg - zh \in yzk[x, y, z].
$$
This is impossible: the monomial $1$ cannot occur in
$zf$, nor in $xh$. On the other hand, we have $ye = 0$ and
$xe = (x, 0) \sim (0, -z) = z(0, -1)$. Hence setting $\delta$
equal to the congruence class of $e$ in $E/zE$ we obtain (d).

\begin{lemma}
\label{lemma-strange-regular-sequence}
There exists a local ring $R$ and a regular sequence $x, y, z$
(in the maximal ideal) such that there exists a nonzero element
$\delta \in R/zR$ with $x\delta = y\delta = 0$.
\end{lemma}

\begin{proof}
Let $R = k[x, y, z] \oplus E$ where $E$ is the module above considered
as a square zero ideal. Then it is clear that $x, y, z$ is a regular
sequence in $R$, and that the element $\delta \in E/zE \subset R/zR$
gives an element with the desired properties. To get a local example
we may localize $R$ at the maximal ideal $\mathfrak m = (x, y, z, E)$.
The sequence $x, y, z$ remains a regular sequence (as localization is
exact), and the element $\delta$ remains nonzero as it is supported
at $\mathfrak m$.
\end{proof}

\begin{lemma}
\label{lemma-base-change-regular-sequence}
There exists a local homomorphism of local rings $A \to B$
and a regular sequence $x, y$ in the maximal ideal of $B$ such that
$B/(x, y)$ is flat over $A$, but such that the images
$\overline{x}, \overline{y}$ of $x, y$ in $B/\mathfrak m_AB$ do not
form a regular sequence, nor even a Koszul-regular sequence.
\end{lemma}

\begin{proof}
Set $A = k[z]_{(z)}$ and let $B = (k[x, y, z] \oplus E)_{(x, y, z, E)}$.
Since $x, y, z$ is a regular sequence in $B$, see proof of
Lemma \ref{lemma-strange-regular-sequence},
we see that $x, y$ is a regular sequence in $B$ and that
$B/(x, y)$ is a torsion free $A$-module, hence flat.
On the other hand, there exists a nonzero element
$\delta \in B/\mathfrak m_AB = B/zB$ which is annihilated
by $\overline{x}, \overline{y}$. Hence
$H_2(K_\bullet(B/\mathfrak m_AB, \overline{x}, \overline{y})) \not = 0$.
Thus $\overline{x}, \overline{y}$ is not Koszul-regular, in particular
it is not a regular sequence, see
More on Algebra, Lemma \ref{more-algebra-lemma-regular-koszul-regular}.
\end{proof}





\section{A Noetherian ring of infinite dimension}
\label{section-Noetherian-infinite-dimension}

\noindent
A Noetherian local ring has finite dimension as we saw in
Algebra, Proposition \ref{algebra-proposition-dimension}.
But there exist Noetherian rings of infinite dimension.
See \cite[Appendix, Example 1]{Nagata}.

\medskip\noindent
Namely, let $k$ be a field, and consider the ring
$$
R = k[x_1, x_2, x_3, \ldots ].
$$
Let $\mathfrak p_i = (x_{2^{i - 1}}, x_{2^{i - 1} + 1}, \ldots, x_{2^i - 1})$
for $i = 1, 2, \ldots$ which are prime ideals of $R$. Let $S$
be the multiplicative subset
$$
S = \bigcap\nolimits_{i \geq 1} (R \setminus \mathfrak p_i).
$$
Consider the ring $A = S^{-1}R$.
We claim that
\begin{enumerate}
\item The maximal ideals of the ring $A$ are the ideals
$\mathfrak m_i = \mathfrak p_iA$.
\item We have $A_{\mathfrak m_i} = R_{\mathfrak p_i}$ which is
a Noetherian local ring of dimension $2^i$.
\item The ring $A$ is Noetherian.
\end{enumerate}
Hence it is clear that this is the example we are looking for.
Details omitted.




\section{Local rings with nonreduced completion}
\label{section-local-completion-nonreduced}

\noindent
In Algebra, Example \ref{algebra-example-bad-dvr-char-p} we gave an example
of a characteristic $p$ Noetherian local domain $R$ of dimension $1$
whose completion is nonreduced. In this section we present the example
of \cite[Proposition 3.1]{Ferrand-Raynaud} which gives a similar
ring in characteristic zero.

\medskip\noindent
Let $\mathbf{C}\{x\}$ be the ring of convergent power series over
the field $\mathbf{C}$ of complex numbers. The ring of all power series
$\mathbf{C}[[x]]$ is its completion. Let $K = \mathbf{C}\{x\}[1/x] = f.f.(B)$
be the field of convergent Laurent series. The $K$-module
$\Omega_{K/\mathbf{C}}$ of algebraic differentials
of $K$ over $\mathbf{C}$ is an infinite dimensional $K$-vector space
(proof omitted). We may choose $f_n \in x\mathbf{C}\{x\}$,
$n \geq 1$ such that
$
\text{d}x, \text{d}f_1, \text{d}f_2, \ldots
$
are part of a basis of $\Omega_{K/\mathbf{C}}$. Thus we can
find a $\mathbf{C}$-derivation
$$
D : \mathbf{C}\{x\} \longrightarrow \mathbf{C}((x))
$$
such that $D(x) = 0$ and $D(f_i) = x^{-n}$. Let
$$
A = \{f \in \mathbf{C}\{x\} \mid D(f) \in \mathbf{C}[[x]]\}
$$
We claim that
\begin{enumerate}
\item $\mathbf{C}\{x\}$ is integral over $A$,
\item $A$ is a local domain,
\item $\dim(A) = 1$,
\item the maximal ideal of $A$ is generated by $x$ and $xf_1$,
\item $A$ is Noetherian, and
\item the completion of $A$ is equal to the ring of dual numbers
over $\mathbf{C}[[x]]$.
\end{enumerate}
Since the dual numbers are nonreduced the ring $A$ gives the example.

\medskip\noindent
Note that if $0 \not = f \in x\mathbf{C}\{x\}$ then
we may write $D(f) = h/f^n$ for some $n \geq 0$ and $h \in \mathbf{C}[[x]]$.
Hence $D(f^{n + 1}/(n + 1)) \in \mathbf{C}[[x]]$
and $D(f^{n + 2}/(n + 2)) \in \mathbf{C}[[x]]$. Thus we
see $f^{n + 1}, f^{n + 2} \in A$!
In particular we see (1) holds. We also conclude that
the fraction field of $A$ is equal to the fraction field of
$\mathbf{C}\{x\}$. It also follows immediately that
$A \cap x\mathbf{C}\{x\}$ is the set of nonunits of $A$, hence
$A$ is a local domain of dimension $1$. If we can show (4)
then it will follow that $A$ is Noetherian (proof omitted).
Suppose that $f \in A \cap x\mathbf{C}\{x\}$. Write
$D(f) = h$, $h \in \mathbf{C}[[x]]$. Write $h = c + xh'$
with $c \in \mathbf{C}$, $h' \in \mathbf{C}[[x]]$. Then
$D(f - cxf_1) = c + xh' - c = xh'$. On the other hand
$f - cxf_1 = xg$ with $g \in \mathbf{C}\{x\}$, but by the
computation above we have $D(g) = h' \in \mathbf{C}[[x]]$
and hence $g \in A$. Thus $f = cxf_1 + xg \in (x, xf_1)$ as desired.

\medskip\noindent
Finally, why is the completion of $A$ nonreduced? Denote $\hat A$ the
completion of $A$. Of course this maps surjectively to the completion
$\mathbf{C}[[x]]$ of $\mathbf{C}\{x\}$ because $x \in A$. Denote
this map $\psi : \hat A \to \mathbf{C}[[x]]$.
Above we saw that $\mathfrak m_A = (x, xf_1)$
and hence $D(\mathfrak m_A^n) \subset (x^{n - 1})$ by an easy
computation. Thus $D : A \to \mathbf{C}[[x]]$ is continuous and
gives rise to a continuous derivation $\hat D : \hat A \to \mathbf{C}[[x]]$
over $\psi$. Hence we get a ring map
$$
\psi + \epsilon \hat D :
\hat A
\longrightarrow
\mathbf{C}[[x]][\epsilon].
$$
Since $\hat A$ is a one dimensional Noetherian complete local ring, if we
can show this arrow is surjective then it will follow that $\hat A$
is nonreduced. Actually the map is an isomorphism but we omit the
verification of this. The subring $\mathbf{C}[x]_{(x)} \subset A$
gives rise to a map $i : \mathbf{C}[[x]] \to \hat A$ on completions such
that $i \circ \psi = \text{id}$ and such that $D \circ i = 0$
(as $D(x) = 0$ by construction). Consider the elements $x^nf_n \in A$.
We have
$$
(\psi + \epsilon D)(x^nf_n) = x^n f_n + \epsilon
$$
for all $n \geq 1$. Surjectivity easily follows from these remarks.




\section{A non catenary Noetherian local ring}
\label{section-non-catenary-Noetherian-local}

\noindent
Even though there is a succesful dimension theory of Noetherian local rings
there are non-catenary Noetherian local rings. An example may be found in
\cite[Appendix, Example 2]{Nagata}. In fact, we will present this example
in the simplest case. Namely, we will construct a local Noetherian domain $A$
of dimension $2$ which is not universally catenary. (Note that $A$ is
automatically catenary, see
Exercises, Exercise
\ref{exercises-exercise-Noetherian-local-domain-dim-2-catenary}.)
The existence of a Noetherian local ring which is not universally
catenary implies the existence of a Noetherian local ring which
is not catenary -- and we spell this out at the end of this section
in the particular example at hand.

\medskip\noindent
Let $k$ be a field, and consider the formal power series ring
$k[[x]]$ in one variable over $k$. Let
$$
z = \sum\nolimits_{i = 1}^\infty a_i x^i
$$
be a formal power series. We assume $z$ as an element of the Laurent
series field $k((x)) = f.f.(k[[x]])$ is transcendental over $k(x)$.
Put
$$
z_j
=
x^{-j}(z - \sum\nolimits_{i = 1, \ldots, j - 1} a_i x^i)
=
\sum\nolimits_{i = j}^\infty a_i x^{i - j}
\in k[[x]].
$$
Note that $Z = z_1$.
Let $R$ be the subring of $k[[x]]$ generated by $x$, $z$ and all of the
$z_j$, in other words
$$
R = k[x, z_1, z_2, z_3, \ldots ] \subset k[[x]].
$$
Consider the ideals $\mathfrak m = (x)$ and
$\mathfrak n = (x - 1, z_1, z_2, \ldots)$ of $R$.

\medskip\noindent
We have $x(z_{j + 1} + a_j) = z_j$. Hence $R/\mathfrak m = k$
and $\mathfrak m$ is a maximal ideal. Moreover, any element of $R$
not in $\mathfrak m$ maps to a unit in $k[[x]]$ and hence
$R_{\mathfrak m} \subset k[[x]]$. In fact it is easy to deduce
that $R_{\mathfrak m}$ is a discrete valuation ring and residue
field $k$.

\medskip\noindent
We claim that
$$
R/(x - 1) =
k[x, z_1, z_2, z_3, \ldots ]/(x - 1)
\cong
k[z].
$$
Namely, the relation above implies that
$(x - 1)(z_{j + 1} + a_j) = -z_{j + 1} - a_j + z_j$, and hence
we may express the class of $z_{j + 1}$ in terms of $z_j$ in
the quotient $R/(x - 1)$. Since the fraction field of $R$
has transcendence degree $2$ over $k$ by construction we see that $z$ is
transcendental over $k$ in $R/(x - 1)$, whence the desired isomorphism.
Hence $\mathfrak n = (x - 1, z)$ and is a maximal ideal. In fact the
map
$$
k[x, x^{-1}, z]_{(x - 1, z)} \longrightarrow R_{\mathfrak n}
$$
is an isomorphism (since $x^{-1}$ is invertible in $R_{\mathfrak n}$
and since $z_{j + 1} = x^{-1}z_j - a_j = \ldots = f_j(x, x^{-1}, z)$).
This shows that $R_{\mathfrak n}$ is a regular local ring
of dimension $2$ and residue field $k$.

\medskip\noindent
Let $S$ be the multiplicative subset
$$
S =
(R \setminus \mathfrak m) \cap (R \setminus \mathfrak n) =
R \setminus (\mathfrak m \cup \mathfrak n)
$$
and set $B = S^{-1}R$. We claim that
\begin{enumerate}
\item The ring $B$ is a $k$-algebra.
\item The maximal ideals of the ring $B$ are the two ideals
$\mathfrak mB$ and $\mathfrak nB$.
\item The residue fields at these maximal ideals is $k$.
\item We have $B_{\mathfrak mB} = R_{\mathfrak m}$
and $B_{\mathfrak nB} = R_{\mathfrak n}$
which are Noetherian regular local rings of dimensions $1$ and $2$.
\item The ring $B$ is Noetherian.
\end{enumerate}
We omit the details of the verifications.

\medskip\noindent
Whenever given a $k$-algebra $B$ with the properties listed above we
get an example as follows. Take $A = k + \text{rad}(B) \subset B$,
in our case $\text{rad}(B) = \mathfrak mB + \mathfrak nB$.
It is easy to see that $B$ is finite over $A$ and hence $A$ is
Noetherian by Eakin's theorem (see \cite{Eakin}, or
\cite[Appendix A1]{Nagata}, or insert future reference here).
Also $A$ is a local domain with the same fraction field as $B$ and
residue field $k$. Since the dimension of $B$ is $2$ we see that $A$
has dimension $2$ as well, by
Algebra, Lemma \ref{algebra-lemma-integral-sub-dim-equal}.

\medskip\noindent
If $A$ were universally catenary then the dimension formula,
Algebra, Lemma \ref{algebra-lemma-dimension-formula}
would give $\dim(B_{\mathfrak mB}) = 2$ contradiction.

\medskip\noindent
Note that $B$ is generated by one element over $A$.
Hence $B = A[x]/\mathfrak p$ for some prime
$\mathfrak p$ of $A[x]$. Let $\mathfrak m' \subset A[x]$ be
the maximal ideal corresponding to $\mathfrak mB$. Then on
the one hand $\dim(A[x]_{\mathfrak m'}) = 3$ and on the
other hand
$$
(0)
\subset \mathfrak pA[x]_{\mathfrak m'}
\subset \mathfrak m'A[x]_{\mathfrak m'}
$$
is a maximal chain of primes. Hence $A[x]_{\mathfrak m'}$ is
an example of a non catenary Noetherian local ring.




\section{Non-quasi-affine variety with quasi-affine normalization}
\label{section-nonquasi-affine}

\noindent
The existence of an example of this kind is mentioned in
\cite[II Remark 6.6.13]{EGA}. They refer to the fifth volume of
EGA for such an example, but the fifth volume did not appear.

\medskip\noindent
Let $k$ be a field.
Let $Y = \mathbf{A}^2_k \setminus \{(0, 0)\}$.
We are going to construct a finite surjective birational morphism
$\pi : Y \longrightarrow X$
with $X$ a variety over $k$ such that $X$ is not quasi-affine.
Namely, consider the following curves in $Y$:
$$
\begin{matrix}
C_1 & : & x = 0 \\
C_2 & : & y = 0
\end{matrix}
$$
Note that $C_1 \cap C_2 = \emptyset$. We choose the isomorphism
$\varphi : C_1 \to C_2$, $(0, y) \mapsto (y^{-1}, 0)$.
We claim there is a unique morphism $\pi : Y \to X$ as above
such that
$$
\xymatrix{
C_1
\ar@<1ex>[rr]^{\text{id}} \ar@<-1ex>[rr]_{\varphi}
& &
Y \ar[r]^\pi & X
}
$$
is a coequalizer diagram in the category of varieties (and even in
the category of schemes). Accepting this for the moment let us
show that such an $X$ cannot be quasi-affine. Namely, it is clear
that we would get
$$
\Gamma(X, \mathcal{O}_X) =
\{ f \in k[x, y] \mid f(0, y) = f(y^{-1}, 0)\} =
k \oplus (xy) \subset k[x, y].
$$
In particular these functions do not separate the points $(1, 0)$
and $(-1, 0)$ whose images in $X$ (we will see below) are distinct
(if the characteristic of $k$ is not $2$).

\medskip\noindent
To show that $X$ exists consider the Zariski open
$D(x + y) \subset Y$ of $Y$. This is the spectrum
of the ring
$k[x, y, 1/(x + y)]$
and the curves $C_1$, $C_2$ are completely contained in
$D(x + y)$. Moreover the morphism
$$
C_1 \coprod C_2
\longrightarrow
D(x + y) \cap Y = \Spec(k[x, y, 1/(x + y)])
$$
is a closed immersion. It follows from
More on Algebra, Lemma \ref{more-algebra-lemma-fibre-product-finite-type}
that the ring
$$
A =
\{f \in k[x, y, 1/(x + y)] \mid f(0, y) = f(y^{-1}, 0)\}
$$
is of finite type over $k$. On the other hand we have the open
$D(xy) \subset Y$ of $Y$ which is disjoint from the curves $C_1$
and $C_2$. It is the spectrum of the ring
$$
B = k[x, y, 1/xy].
$$
Note that we have $A_{xy} \cong B_{x + y}$ (since $A$ clearly contains
the elements $xyP(x, y)$ any polynomial $P$ and the element $xy/(x + y)$).
The scheme $X$ is obtained by glueing the affine schemes
$\Spec(A)$ and $\Spec(B)$ using the isomorphism
$A_{xy} \cong B_{x + y}$ and hence is clearly of finite type over
$k$. To see that it is separated one has to show that the
ring map $A \otimes_k B \to B_{x + y}$ is surjective. To see
this use that $A \otimes_k B$ contains the element
$xy/(x + y) \otimes 1/xy$ which maps to $1/(x + y)$.
The morphism $X \to Y$ is given by the natural maps
$D(x + y) \to \Spec(A)$ and $D(xy) \to \Spec(B)$.
Since these are both finite we deduce that $X \to Y$ is finite
as desired. We omit the verification that $X$ is indeed the
coequalizer of the displayed diagram above, however, see
(insert future reference for pushouts in the category of schemes
here). Note that the morphism $\pi : Y \to X$ does
map the points  $(1, 0)$
and $(-1, 0)$ to distinct points in $X$ because the
function $(x + y^3)/(x + y)^2 \in A$ has value
$1/1$, resp.\ $-1/(-1)^2 = -1$ which are always distinct
(unless the characteristic is $2$ -- please find your own points
for characteristic $2$). We summarize this discussion in the
form of a lemma.

\begin{lemma}
\label{lemma-quasi-affine-normalization-not-quasi-affine}
Let $k$ be a field.
There exists a variety $X$ whose normalization is quasi-affine but
which is itself not quasi-affine.
\end{lemma}

\begin{proof}
See discussion above and (insert future reference on normalization here).
\end{proof}






\section{A locally closed subscheme which is not open in closed}
\label{section-strange-immersion}

\noindent
This is a copy of
Morphisms, Example \ref{morphisms-example-thibaut}.
Here is an example of an immersion which is not a composition of an
open immersion followed by a closed immersion.
Let $k$ be a field.
Let $X = \Spec(k[x_1, x_2, x_3, \ldots])$.
Let $U = \bigcup_{n = 1}^{\infty} D(x_n)$.
Then $U \to X$ is an open immersion.
Consider the ideals
$$
I_n =
(x_1^n, x_2^n, \ldots, x_{n - 1}^n, x_n - 1, x_{n + 1}, x_{n + 2}, \ldots)
\subset
k[x_1, x_2, x_3, \ldots][1/x_n].
$$
Note that $I_n k[x_1, x_2, x_3, \ldots][1/x_nx_m] = (1)$
for any $m \not = n$. Hence the quasi-coherent ideals
$\widetilde I_n$ on $D(x_n)$ agree on $D(x_nx_m)$, namely
$\widetilde I_n|_{D(x_nx_m)} = \mathcal{O}_{D(x_n x_m)}$ if
$n \not = m$. Hence these ideals glue to a quasi-coherent sheaf of ideals
$\mathcal{I} \subset \mathcal{O}_U$.
Let $Z \subset U$ be the closed subscheme corresponding to
$\mathcal{I}$. Thus $Z \to X$ is an immersion.

\medskip\noindent
We claim that we cannot factor $Z \to X$ as
$Z \to \overline{Z} \to X$, where $\overline{Z} \to X$ is closed
and $Z \to \overline{Z}$ is open. Namely, $\overline{Z}$ would
have to be defined by an ideal $I \subset k[x_1, x_2, x_3, \ldots]$
such that $I_n = I k[x_1, x_2, x_3, \ldots][1/x_n]$.
But the only element $f \in k[x_1, x_2, x_3, \ldots]$
which ends up in all $I_n$ is $0$! Hence $I$ does not exist.




\section{Nonexistence of suitable opens}
\label{section-nonexistence-opens}

\noindent
This section complements the results of
Properties, Section \ref{properties-section-finding-affine-opens}.

\medskip\noindent
Let $k$ be a field and let $A = k[z_1, z_2, z_3, \ldots]/I$ where $I$
is the ideal generated by all pairwise products
$z_iz_j$, $i \not = j$, $i, j \in \mathbf{N}$. Set $S = \Spec(A)$.
Let $s \in S$ be the closed point corresponding to the maximal
ideal $(z_i)$. We claim there is no
quasi-compact open $V \subset S \setminus \{s\}$ which is
dense in $S \setminus \{s\}$. Note that $S \setminus \{s\} = \bigcup D(z_i)$.
Each $D(z_i)$ is open and irreducible with generic point
$\eta_i$. We conclude that $\eta_i \in V$ for all $i$.
However, a principal affine open of $S \setminus \{s\}$ is
of the form $D(f)$ where $f \in (z_1, z_2, \ldots)$. Then
$f \in (z_1, \ldots, z_n)$ for some $n$ and we see that $D(f)$ contains
only finitely many of the points $\eta_i$. Thus $V$ cannot be quasi-compact.

\medskip\noindent
Let $k$ be a field and let $B = k[x, z_1, z_2, z_3, \ldots]/J$ where $J$
is the ideal generated by the products $xz_i$, $i \in \mathbf{N}$ and by
all pairwise products $z_iz_j$, $i \not = j$, $i, j \in \mathbf{N}$.
Set $T = \Spec(B)$. Consider the principal open $U = D(x)$.
We claim there is no quasi-compact open $V \subset S$ such that
$V \cap U = \emptyset$ and $V \cup U$ is dense in $S$.
Let $t \in T$ be the closed point corresponding to the maximal
ideal $(x, z_i)$. The closure of $U$ in $T$ is
$\overline{U} = U \cup \{t\}$. Hence $V \subset \bigcup_i D(z_i)$
is a quasi-compact open. By the arguments of the previous paragraph
we see that $V$ cannot be dense in $\bigcup D(z_i)$.

\begin{lemma}
\label{lemma-complement-of-affine-does-not-contain-qc-dense-open}
Nonexistence quasi-compact opens of affines:
\begin{enumerate}
\item There exist an affine scheme $S$ and affine open $U \subset S$
such that there is no quasi-compact open $V \subset S$ with
$U \cap V = \emptyset$ and $U \cup V$ dense in $S$.
\item There exists an affine scheme $S$ and a closed point $s \in S$ such that
$S \setminus \{s\}$ does not contain a quasi-compact dense open.
\end{enumerate}
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}

\noindent
Let $X$ be the glueing of two copies of the affine scheme $T$ (see above)
along the affine open $U$. Thus there is a morphism $\pi : X \to T$ and
$X = U_1 \cup U_2$ such that $\pi$ maps $U_i$ isomorphically to $T$ and
$U_1 \cap U_2$ isomorphically to $U$. Note that $X$ is quasi-separated
(by Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated})
and quasi-compact. We claim there does not exist a separated, dense,
quasi-compact open $W \subset X$. Namely, consider the two closed
points $x_1 \in U_1$, $x_2 \in U_2$ mapping to the closed point $t \in T$
introduced above. Let $\tilde \eta \in U_1 \cap U_2$ be the generic point
mapping to the (unique) generic point $\eta$ of $U$.
Note that $\tilde\eta \leadsto x_1$ and $\tilde\eta \leadsto x_2$
lying over the specialization $\eta \leadsto s$.
Since $\pi|_W : W \to T$ is separated we conclude that we cannot have both
$x_1$ and $x_2 \in W$ (by the valuative criterion of separatedness
Schemes, Lemma \ref{schemes-lemma-valuative-criterion-separatedness}).
Say $x_1 \not \in W$. Then $W \cap U_1$ is a quasi-compact (as $X$
is quasi-separated) dense open of $U_1$ which does not contain $x_1$.
Now observe that there exists an isomorphism $(T, t) \cong (S, s)$
of schemes (by sending $x$ to $z_1$ and $z_i$ to $z_{i + 1}$).
Hence by the first paragraph of this section we arrive at a contradiction.

\begin{lemma}
\label{lemma-no-dense-separated-quasi-compact-open-in-qcqs}
There exists a quasi-compact and quasi-separated scheme $X$ which does
not contain a separated quasi-compact dense open.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}







\section{Nonexistence of quasi-compact dense open subscheme}
\label{section-nonexistence-qc-dense-open-subscheme}

\noindent
Let $X$ be a quasi-compact and quasi-separated algebraic space over a field
$k$. We know that the schematic locus $X' \subset X$ is a dense open
subspace, see
Properties of Spaces, Proposition
\ref{spaces-properties-proposition-locally-quasi-separated-open-dense-scheme}.
In fact, this result holds when $X$ is reasonable, see
Decent Spaces, Proposition
\ref{decent-spaces-proposition-reasonable-open-dense-scheme}.
A natural question is whether one can find a quasi-compact dense
open subscheme of $X$. It turns out this is not possible in general.

\medskip\noindent
Assume the characteristic of $k$ is not 2.
Let $B = k[x, z_1, z_2, z_3, \ldots]/J$ where $J$ is the ideal generated by
the products $xz_i$, $i \in \mathbf{N}$ and by all pairwise products
$z_iz_j$, $i \not = j$, $i, j \in \mathbf{N}$. Set $U = \Spec(B)$.
Denote $0 \in U$ the closed point all of whose coordinates are zero.
Set
$$
j : R = \Delta \coprod \Gamma \longrightarrow U \times_k U
$$
where $\Delta$ is the image of the diagonal morphism of $U$ over $k$ and
$$
\Gamma = \{((x, 0, 0, 0, \ldots), (-x, 0, 0, 0, \ldots))
\mid x \in \mathbf{A}^1_k, x \not = 0\}.
$$
It is clear that $s, t : R \to U$ are \'etale, and hence
$j$ is an \'etale equivalence relation. The quotient $X = U/R$
is an algebraic space (Spaces, Theorem \ref{spaces-theorem-presentation}).
Note that $j$ is not an immersion because
$(0, 0) \in \Delta$ is in the closure of $\Gamma$.
Hence $X$ is not a scheme. On the other hand, $X$ is quasi-separated
as $R$ is quasi-compact. Denote $0_X$ the image of the point $0 \in U$.
We claim that $X \setminus \{0_X\}$ is a scheme, namely
$$
X \setminus \{0_X\} =
\Spec\left(k[x^2, x^{-2}]\right) \coprod
\Spec\left(k[z_1, z_2, z_3, \ldots]/(z_iz_j)\right) \setminus \{0\}
$$
(details omitted). On the other hand, we have seen in
Section \ref{section-nonexistence-opens} that the scheme
on the right hand side does not contain
a quasi-compact dense open.

\begin{lemma}
\label{lemma-nonexistence-qc-dense-open-subscheme}
There exists a quasi-compact and quasi-separated algebraic space
which does not contain a quasi-compact dense open subscheme.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}

\noindent
Using the construction of 
Spaces, Example \ref{spaces-example-non-representable-descent}
in the same manner as we used the construction of
Spaces, Example \ref{spaces-example-affine-line-involution}
above, one obtains an example of a quasi-compact, quasi-separated, and
locally separated algebraic space which does not contain a quasi-compact
dense open subscheme.


\section{Affines over algebraic spaces}
\label{section-embedding-affines}

\medskip\noindent
Suppose that $f : Y \to X$ is a morphism of schemes with $f$
locally of finite type and $Y$ affine. Then there exists an immersion
$Y \to \mathbf{A}^n_X$ of $Y$ into affine $n$-space over $X$.
See the slightly more general
Morphisms, Lemma \ref{morphisms-lemma-quasi-affine-finite-type-over-S}.

\medskip\noindent
Now suppose that $f : Y \to X$ is a morphism of algebraic spaces with
$f$ locally of finite type and $Y$ an affine scheme. Then it is not
true in general that we can find an immersion of $Y$ into affine
$n$-space over $X$.

\medskip\noindent
A first (nasty) counter example is $Y = \Spec(k)$ and
$X = [\mathbf{A}^1_k/\mathbf{Z}]$ where $k$ is a field of characteristic zero
and $\mathbf{Z}$ acts on $\mathbf{A}^1_k$ by translation $(n, t) \mapsto t + n$.
Namely, for any morphism $Y \to \mathbf{A}^n_X$ over $X$ we can pullback to
the covering $\mathbf{A}^1_k$ of $X$ and we get an infinite disjoint union of
$\mathbf{A}^1_k$'s mapping into $\mathbf{A}^{n + 1}_k$ which is not an
immersion.

\medskip\noindent
A second counter example is $Y = \mathbf{A}^1_k \to X = \mathbf{A}^1_k/R$
with $R = \{(t, t)\} \coprod \{(t, -t), t \not = 0\}$. Namely, in
this case the morphism $Y \to \mathbf{A}^n_X$ would be given by some
regular functions $f_1, \ldots, f_n$ on $Y$ and hence the
fibre product of $Y$ with the covering
$\mathbf{A}^{n + 1}_k \to \mathbf{A}^n_X$
would be the scheme
$$
\{(f_1(t), \ldots, f_n(t), t)\} \coprod
\{(f_1(t), \ldots, f_n(t), -t), t \not = 0\}
$$
with obvious morphism to $\mathbf{A}^{n + 1}_k$ which is not an immersion.
Note that this gives a counter example with $X$ quasi-separated.

\begin{lemma}
\label{lemma-cannot-embed-into-affine}
There exists a finite type morphism of algebraic spaces $Y \to X$
with $Y$ affine and $X$ quasi-separated, such that there does not exist
an immersion $Y \to \mathbf{A}^n_X$ over $X$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}










\section{Pushforward of quasi-coherent modules}
\label{section-push-quasi-coherent}

\noindent
In Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}
we proved that $f_*$ transforms quasi-coherent modules into quasi-coherent
modules when $f$ is quasi-compact and quasi-separated. Here are some
examples to show that these conditions are both necessary.

\medskip\noindent
Suppose that $Y = \Spec(A)$ is an affine scheme and that
$X = \coprod_{n \in \mathbf{N}} Y$. We claim that $f_*\mathcal{O}_X$
is not quasi-coherent where $f : X \to Y$ is the obvious morphism.
Namely, for $a \in A$ we have
$$
f_*\mathcal{O}_X(D(a)) = \prod\nolimits_{n \in \mathbf{N}} A_a
$$
Hence, in order for $f_*\mathcal{O}_X$ to be quasi-coherent we would need
$$
\prod\nolimits_{n \in \mathbf{N}} A_a
=
\left(\prod\nolimits_{n \in \mathbf{N}} A\right)_a
$$
for all $a \in A$. This isn't true in general, for example if
$A = \mathbf{Z}$ and $a = 2$, then $(1, 1/2, 1/4, 1/8, \ldots)$
is an element of the left hand side which is not in the right hand side.
Note that $f$ is a non-quasi-compact separated morphism.

\medskip\noindent
Let $k$ be a field. Set
$$
A = k[t, z, x_1, x_2, x_3, \ldots]/(tx_1z, t^2x_2^2z, t^3x_3^3z, \ldots)
$$
Let $Y = \Spec(A)$. Let $V \subset Y$ be the open subscheme
$V = D(x_1) \cup D(x_2) \cup \ldots$. Let $X$ be two copies of $Y$
glued along $V$. Let $f : X \to Y$ be the obvious morphism. Then we
have an exact sequence
$$
0 \to f_*\mathcal{O}_X \to
\mathcal{O}_Y \oplus \mathcal{O}_Y \xrightarrow{(1, -1)} j_*\mathcal{O}_V
$$
where $j : V \to Y$ is the inclusion morphism. Since
$$
A \longrightarrow \prod A_{x_n}
$$
is injective (details omitted) we see that $\Gamma(Y, f_*\mathcal{O}_X) = A$.
On the other hand, the kernel of the map
$$
A_t \longrightarrow \prod A_{tx_n}
$$
is nonzero because it contains the element $z$. Hence
$\Gamma(D(t), f_*\mathcal{O}_X)$ is strictly bigger than
$A_t$ because it contains $(z, 0)$. Thus we see that $f_*\mathcal{O}_X$
is not quasi-coherent. Note that $f$ is quasi-compact but
non-quasi-separated.

\begin{lemma}
\label{lemma-pushforward-quasi-coherent}
Schemes, Lemma \ref{schemes-lemma-push-forward-quasi-coherent}
is sharp in the sense that one can neither drop the assumption
of quasi-compactness nor the assumption of quasi-separatedness.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}








\section{A nonfinite module with finite free rank 1 stalks}
\label{section-nonfree}

\noindent
Let $R = \mathbf{Q}[x]$. Set
$M = \sum_{n \in \mathbf{N}} \frac{1}{x - n}R$ as a submodule of
the fraction field of $R$. Then $M$ is not finitely generated, but
for every prime $\mathfrak p$ of $R$ we have
$M_{\mathfrak p} \cong R_{\mathfrak p}$ as an
$R_{\mathfrak p}$-module.




\section{A finite flat module which is not projective}
\label{section-finite-flat-not-projective}

\noindent
This is a copy of
Algebra, Remark \ref{algebra-remark-warning}.
It is not true that a finite $R$-module which is
$R$-flat is automatically projective. A counter
example is where $R = \mathcal{C}^\infty(\mathbf{R})$
is the ring of infinitely differentiable functions on
$\mathbf{R}$, and $M = R_{\mathfrak m} = R/I$ where
$\mathfrak m = \{f \in R \mid f(0) = 0\}$ and
$I = \{f \in R \mid \exists \epsilon, \epsilon > 0 :
f(x) = 0\ \forall x, |x| < \epsilon\}$.

\medskip\noindent
The morphism $\Spec(R/I) \to \Spec(R)$ is also
an example of a flat closed immersion which is not open.

\begin{lemma}
\label{lemma-finite-flat-non-projective}
Strange flat modules.
\begin{enumerate}
\item There exists a ring $R$ and a finite flat $R$-module $M$ which is
not projective.
\item There exists a closed immersion which is flat but not open.
\end{enumerate}
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{A projective module which is not locally free}
\label{section-projective-not-locally-free}

\noindent
We give two examples. One where the rank is between $0$ and $1$
and one where the rank is $\aleph_0$.

\begin{lemma}
\label{lemma-ideal-generated-by-idempotents-projective}
Let $R$ be a ring. Let $I \subset R$ be an ideal generated by
a countable collection of idempotents. Then $I$ is projective
as an $R$-module.
\end{lemma}

\begin{proof}
Say $I = (e_1, e_2, e_3, \ldots)$ with $e_n$ an idempotent of $R$.
After inductively replacing $e_{n + 1}$ by $e_n + (1 - e_n)e_{n + 1}$
we may assume that $(e_1) \subset (e_2) \subset (e_3) \subset \ldots$
and hence $I = \bigcup_{n \geq 1} (e_n) = \colim_n e_nR$.
In this case
$$
\Hom_R(I, M) = \Hom_R(\colim_n e_nR, M)
= \lim_n \Hom_R(e_nR, M) = \lim_n e_nM
$$
Note that the transition maps $e_{n + 1}M \to e_nM$ are given
by multiplication by $e_n$ are are surjective. Hence by
Algebra, Lemma \ref{algebra-lemma-ML-exact-sequence}
the functor $\Hom_R(I, M)$ is exact, i.e., $I$ is a projective
$R$-module.
\end{proof}

\begin{lemma}
\label{lemma-map-cannot-be-injective}
Let $R$ be a ring. Let $n \geq 1$. Let $M$ be an $R$-module generated
by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a
nonzero kernel.
\end{lemma}

\begin{proof}
Choose a surjection $R^{\oplus n - 1} \to M$.
We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$.
It suffices to prove $f'$ has a nonzero kernel.
The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a
matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say
$a = a_{ij}$ is not, then we can replace $A$ by the localization $A_a$
and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel
after localization then there was a nonzero kernel to start with as
localization is exact, see
Algebra, Proposition \ref{algebra-proposition-localization-exact}.
In this case we can do a base change on both $R^{\oplus n}$
and $R^{\oplus n - 1}$ and reduce to the case where
$$
A =
\left(
\begin{matrix}
1 & 0 & 0 \ldots \\
0 & a_{22} & a_{23} & \ldots \\
0 & a_{32} & \ldots \\
\ldots & \ldots
\end{matrix}
\right)
$$
Hence in this case we win by induction on $n$. If not then each
$a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that
$I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer
such that $I^m \not = 0$. Then we see that $(I^m)^{\oplus n}$ is
contained in the kernel of the map and we win.
\end{proof}

\noindent
Suppose that $P \subset Q$ is an inclusion of $R$-modules with $Q$ a
finite $R$-module and $P$ locally free, see
Algebra, Definition \ref{algebra-definition-locally-free}.
Suppose that $Q$ can be generated by $N$ elements as an $R$-module.
Then it follows from
Lemma \ref{lemma-map-cannot-be-injective}
that $P$ is finite locally free (with the free parts having rank
at most $N$). And in this case $P$ is a finite $R$-module, see
Algebra, Lemma \ref{algebra-lemma-finite-projective}.

\medskip\noindent
Combining this with the above we see that a non-finitely-generated
ideal which is generated by a countable collection of idempotents
is projective but not locally free. An explicit example is
$R = \prod_{n \in \mathbf{N}} \mathbf{F}_2$ and
$I$ the ideal generated by the idempotents
$$
e_n = (1, 1, \ldots, 1, 0, \ldots )
$$
where the sequence of $1$'s has length $n$.

\begin{lemma}
\label{lemma-ideal-projective-not-locally-free}
There exists a ring $R$ and an ideal $I$ such that $I$ is projective as
an $R$-module but not locally free as an $R$-module.
\end{lemma}

\begin{proof}
See above.
\end{proof}

\begin{lemma}
\label{lemma-chow-group-product}
Let $K$ be a field.
Let $C_i$, $i = 1, \ldots, n$ be smooth, projective, geometrically irreducible
curves over $K$. Let $P_i \in C_i(K)$ be a rational point and
let $Q_i \in C_i$ be a point such that $[\kappa(Q_i) : K] = 2$.
Then $[P_1 \times \ldots \times P_n]$ is nonzero in
$A_0(U_1 \times_K \ldots \times_K U_n)$ where $U_i = C_i \setminus \{Q_i\}$.
\end{lemma}

\begin{proof}
There is a degree map
$\deg : A_0(C_1 \times_K \ldots \times_K C_n) \to \mathbf{Z}$
Because each $Q_i$ has degree $2$ over $K$ we see that
any zero cycle supported on the ``boundary''
$$
C_1 \times_K \ldots \times_K C_n
\setminus
U_1 \times_K \ldots \times_K U_n
$$
has degree divisible by $2$.
\end{proof}

\noindent
We can construct another example of a projective but not locally free
module using the lemma above as follows. Let
$C_n$, $n = 1, 2, 3, \ldots$ be smooth, projective, geometrically irreducible
curves over $\mathbf{Q}$ each with a pair of points
$P_n, Q_n \in C_n$ such that $\kappa(P_n) = \mathbf{Q}$ and
$\kappa(Q_n)$ is a quadratic extension of $\mathbf{Q}$.
Set $U_n = C_n \setminus \{Q_n\}$; this is an affine curve.
Let $\mathcal{L}_n$ be the inverse of the ideal sheaf of $P_n$
on $U_n$. Note that $c_1(\mathcal{L}_n) = [P_n]$ in the group of
zero cycles $A_0(U_n)$. Set $A_n = \Gamma(U_n, \mathcal{O}_{U_n})$.
Let $L_n = \Gamma(U_n, \mathcal{L}_n)$ which is a locally
free module of rank $1$ over $A_n$. Set
$$
B_n = A_1 \otimes_{\mathbf{Q}} A_2 \otimes_{\mathbf{Q}} \ldots
\otimes_{\mathbf{Q}} A_n
$$
so that $\Spec(B_n) = U_1 \times \ldots \times U_n$ all products
over $\Spec(\mathbf{Q})$. For $i \leq n$ we set
$$
L_{n, i} =
A_1 \otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} M_i
\otimes_{\mathbf{Q}} \ldots \otimes_{\mathbf{Q}} A_n
$$
which is a locally free $B_n$-module of rank $1$. Note that this is
also the global sections of $\text{pr}_i^*\mathcal{L}_n$. Set
$$
B_\infty = \colim_n B_n
\quad\text{and}\quad
L_{\infty, i} = \colim_n L_{n, i}
$$
Finally, set
$$
M = \bigoplus\nolimits_{i \geq 1} L_{\infty, i}.
$$
This is a direct sum of finite locally free modules, hence projective.
We claim that $M$ is not locally free. Namely, suppose that
$f \in B_\infty$ is a nonzero function such that $M_f$ is free
over $(B_\infty)_f$. Let $e_1, e_2, \ldots$ be a basis. Choose
$n \geq 1$ such that $f \in B_n$.
Choose $m \geq n + 1$ such that $e_1, \ldots, e_{n + 1}$ are in
$$
\bigoplus\nolimits_{1 \leq i \leq m} L_{m, i}.
$$
Because the elements $e_1, \ldots, e_{n + 1}$ are part of a basis
after a faithfully flat base change we conclude that
the chern classes
$$
c_i(\text{pr}_1^*\mathcal{L}_1 \oplus \ldots \oplus
\text{pr}_m^*\mathcal{L}_m), \quad i = m, m - 1, \ldots, m - n
$$
are zero in the chow group of
$$
D(f) \subset U_1 \times \ldots \times U_m
$$
Since $f$ is the pullback of a function on $U_1 \times \ldots \times U_n$
this implies in particular that
$$
c_{m - n}(\mathcal{O}_W^{\oplus n} \oplus
\text{pr}_1^*\mathcal{L}_{n + 1} \oplus \ldots
\oplus \text{pr}_{m - n}^*\mathcal{L}_m) = 0.
$$
on the variety
$$
W = (C_{n + 1} \times \ldots \times C_m)_K
$$
over the field $K = \mathbf{Q}(C_1 \times \ldots \times C_n)$.
In other words the cycle
$$
[(P_{n + 1} \times \ldots \times P_m)_K]
$$
is zero in the chow group of zero cycles on $W$. This contradicts
Lemma \ref{lemma-chow-group-product}
above because the points $Q_i$, $n + 1 \leq i \leq m$
induce corresponding points $Q_i'$ on $(C_n)_K$ and as $K/\mathbf{Q}$
is geometrically irreducible we have $[\kappa(Q_i') : K] = 2$.

\begin{lemma}
\label{lemma-projective-not-locally-free}
There exists a countable ring $R$ and a projective module $M$
which is a direct sum of countably many locally free rank $1$
modules such that $M$ is not locally free.
\end{lemma}

\begin{proof}
See above.
\end{proof}







\section{Zero dimensional local ring with nonzero flat ideal}
\label{section-zero-dimensional-flat-ideal}

\noindent
In \cite{Lazard} there is an example of a zero dimensional local ring with a
nonzero flat ideal. Here is the construction. Let $k$ be a field.
Let $X_i, Y_i$, $i \geq 1$ be variables. Take
$R = k[X_i, Y_i]/(X_i - Y_i X_{i + 1}, Y_i^2)$. Denote $x_i$, resp.\ $y_i$
the image of $X_i$, resp.\ $Y_i$ in this ring. Note that
$$
x_i = y_i x_{i + 1} = y_i y_{i +1} x_{i + 2} =
y_i y_{i + 1} y_{i + 2} x_{i + 3} = \ldots
$$
in this ring. The ring $R$ has only
one prime ideal, namely $\mathfrak m = (x_i, y_i)$. We claim that
the ideal $I = (x_i)$ is flat as an $R$-module.

\medskip\noindent
Note that the annihilator of $x_i$ in $R$ is the ideal
$(x_1, x_2, x_3, \ldots, y_i, y_{i + 1}, y_{i + 2}, \ldots)$.
Consider the $R$-module $M$ generated by elements $e_i$, $i \geq 1$ and
relations $e_i = y_i e_{i + 1}$. Then $M$ is flat as it is the
colimit $\colim_i R$ of copies of $R$ with transition maps
$$
R \xrightarrow{y_1} R \xrightarrow{y_2} R \xrightarrow{y_3} \ldots
$$
Note that the annihilator of $e_i$ in $M$ is the ideal
$(x_1, x_2, x_3, \ldots, y_i, y_{i + 1}, y_{i + 2}, \ldots)$.
Since every element of $M$, resp.\ $I$ can be written as
$f e_i$, resp.\ $h x_i$ for some $f, h \in R$ we see that the
map $M \to I$, $e_i \to x_i$ is an isomorphism and $I$ is flat.

\begin{lemma}
\label{lemma-zero-dimensional-flat-ideal}
There exists a local ring $R$ with a unique prime ideal
and a nonzero ideal $I \subset R$ which is a flat $R$-module
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}








\section{An epimorphism of zero-dimensional rings which is not surjective}
\label{section-epimorphism-not-surjective}

\noindent
In \cite{Autour} one can find the following example.
Let $k$ be a field. Consider the ring homomorphism
$$
k[x_1, x_2, \ldots, z_1, z_2, \ldots]/
(x_i^{4^i}, z_i^{4^i})
\longrightarrow
k[x_1, x_2, \ldots, y_1, y_2, \ldots]/
(x_i^{4^i}, y_i - x_{i + 1}y_{i + 1}^2)
$$
which maps $x_i$ to $x_i$ and $z_i$ to $x_iy_i$. Note that $y_i^{4^{i + 1}}$
is zero in the right hand side but that $y_1$ is not zero (details omitted).
This map is not surjective: we can think of the above as a map of
$\mathbf{Z}$-graded algebras by setting $\deg(x_i) = -1$, $\deg(z_i) = 0$,
and $\deg(y_i) = 1$ and then it is clear that $y_1$ is not in the image.
Finally, the map is an epimorphism because
$$
y_{i - 1} \otimes 1 = x_i y_i^2 \otimes 1 = y_i \otimes x_i y_i =
x_i y_i \otimes y_i = 1 \otimes x_i y_i^2.
$$
hence the tensor product of the target over the source is isomorphic
to the target.

\begin{lemma}
\label{lemma-epi-not-surjective}
There exists an epimorphism of local rings of dimension $0$
which is not a surjection.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{Finite type, not finitely presented, flat at prime}
\label{section-ft-not-fp-flat-at-prime}

\noindent
Let $k$ be a field. Consider the local ring $A_0 = k[x, y]_{(x, y)}$.
Denote $\mathfrak p_{0, n} = (y + x^n + x^{n + 1})$. This is a prime ideal.
Set
$$
A = A_0[z_1, z_2, z_3, \ldots]/(z_n z_m, z_n(y + x^n + x^{2n + 1}))
$$
Note that $A \to A_0$ is a surjection whose kernel is an ideal of
square zero. Hence $A$ is also a local ring and the prime ideals of $A$
are in one-to-one correspondence with the prime ideals of $A_0$.
Denote $\mathfrak p_n$ the prime ideal of $A$ corresponding to
$\mathfrak p_{0, n}$. Observe that $\mathfrak p_n$ is the annihilator
of $z_n$ in $A$. Let
$$
C = A[z]/(xz^2 + z + y)[\frac{1}{2zx + 1}].
$$
Note that $A \to C$ is an \'etale ring map, see
Algebra, Example \ref{algebra-example-make-standard-smooth}.
Let $\mathfrak q \subset C$ be the maximal ideal generated by
$x$, $y$, $z$ and all $z_n$. As $A \to C$ is flat we see that the
annihilator of $z_n$ in $C$ is $\mathfrak p_nC$. We compute
\begin{align*}
C/\mathfrak p_n C
& =
A_0/(y + x^n + x^{2n + 1}) \\
& =
k[x]_{(x)}[z]/(xz^2 + z - x^n - x^{2n + 1}) \\
& =
k[x]_{(x)}[z]/(z - x^n) \times k[x]_{(x)}[z]/(xz + x^{n + 1} + 1) \\
& =
k[x]_{(x)} \times k(x)
\end{align*}
because $(z - x^n)(xz + x^{n + 1} + 1) = xz^2 + z - x^n - x^{2n + 1}$.
Hence we see that $\mathfrak p_nC = \mathfrak r_n \cap \mathfrak q_n$
with $\mathfrak r_n = \mathfrak p_nC + (z - x^n)C$ and
$\mathfrak q_n = \mathfrak p_nC + (xz + x^{n + 1} + 1)C$.
Since $\mathfrak q_n + \mathfrak r_n = C$ we also get
$\mathfrak p_nC = \mathfrak r_n \mathfrak q_n$,
It follows that $\mathfrak q_n$ is the annihilator of $\xi_n = (z - x^n)z_n$.
Observe that on the one hand $\mathfrak r_n \subset \mathfrak q$, and
on the other hand $\mathfrak q_n + \mathfrak q = C$. This follows for example
because $\mathfrak q_n$ is a maximal ideal of $C$ distinct from $\mathfrak q$.
Similarly we have $\mathfrak q_n + \mathfrak q_m = C$.
At this point we let
$$
B = \text{Im}(C \longrightarrow C_{\mathfrak q})
$$
We observe that the elements $\xi_n$ map to zero in $B$ as $xz + x^{n + 1} + 1$
is not in $\mathfrak q$. Denote $\mathfrak q' \subset B$ the image of
$\mathfrak q$. By construction $B$ is a finite type $A$-algebra, with
$B_{\mathfrak q'} \cong C_{\mathfrak q}$. In particular we see that
$B_{\mathfrak q'}$ is flat over $A$.

\medskip\noindent
We claim there does not exist an element $g' \in B$, $g' \not \in \mathfrak q'$
such that $B_{g'}$ is of finite presentation over $A$. We sketch a proof of
this claim. Choose an element $g \in C$ which maps to $g' \in B$.
Consider the map $C_g \to B_{g'}$. By
Algebra, Lemma \ref{algebra-lemma-finite-presentation-independent}
we see that $B_g$ is finitely presented over $A$ if and only if the kernel
of $C_g \to B_{g'}$ is finitely generated. But the element $g \in C$ is
not contained in $\mathfrak q$, hence maps to a nonzero element of
$A_0[z]/(xz^2 + z + y)$. Hence $g$ can only be contained in finitely
many of the prime ideals $\mathfrak q_n$, because the primes
$(y + x^n + x^{2n + 1}, xz + x^{n + 1} + 1)$ are an infinite collection
of codimension 1 points of the 2-dimensional irreducible Noetherian space
$\Spec(k[x, y, z]/(xz^2 + z + y))$. The map
$$
\bigoplus\nolimits_{g \not \in \mathfrak q_n} C/\mathfrak q_n
\longrightarrow
C_g, \quad
(c_n) \longrightarrow \sum c_n \xi_n
$$
is injective and its image is the kernel of $C_g \to B_{g'}$. We omit the
proof of this statement. (Hint: Write $A = A_0 \oplus I$ as an $A_0$-module
where $I$ is the kernel of $A \to A_0$. Similarly, write $C = C_0 \oplus IC$.
Write
$IC = \bigoplus Cz_n \cong \bigoplus (C/\mathfrak r_n \oplus C/\mathfrak q_n)$
and study the effect of multiplication by $g$ on the summands.)
This concludes the sketch of the proof of the claim.
This also proves that $B_{g'}$ is not flat over $A$ for any $g'$ as above.
Namely, if it were flat, then the annihilator of the image of $z_n$ in
$B_{g'}$ would be $\mathfrak p_nB_{g'}$, and would not contain $z - x^n$.

\medskip\noindent
As a consequence we can answer (negatively) a question posed in
\cite[Part I, Remarques (3.4.7) (\romannumeral 5)]{GruRay}.
Here is a precise statement.

\begin{lemma}
\label{lemma-example-raynaud-gruson}
There exists a local ring $A$, a finite type ring map $A \to B$ and a prime
$\mathfrak q$ lying over $\mathfrak m_A$ such that $B_{\mathfrak q}$ is flat
over $A$, and for any element $g \in B$, $g \not \in \mathfrak q$
the ring $B_g$ is neither finitely presented over $A$ nor flat over $A$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}


\section{Finite type, flat and not of finite presentation}
\label{section-finite-type-flat-not-finite-presentation}

\noindent
In this section we give some examples of ring maps and morphisms
which are of finite type and flat but not of finite presentation.

\medskip\noindent
Let $R$ be a ring which has an ideal $I$ such that $R/I$ is a finite
flat module but not projective, see
Section \ref{section-finite-flat-not-projective}
for an explicit example. Note that this means that $I$ is not
finitely generated, see
Algebra, Lemma \ref{algebra-lemma-finitely-generated-pure-ideal}.
Note that $I = I^2$, see
Algebra, Lemma \ref{algebra-lemma-pure}.
The base ring in our examples will be
$R$ and correspondingly the base scheme $S = \Spec(R)$.

\medskip\noindent
Consider the ring map $R \to R \oplus R/I\epsilon$ where
$\epsilon^2 = 0$ by convention. This is a finite, flat
ring map which is not of finite presentation. All the fibre rings
are complete intersections and geometrically irreducible.

\medskip\noindent
Let $A = R[x, y]/(xy, ay; a \in I)$. Note that as an $R$-module
we have $A = \bigoplus_{i \geq 0} Ry^i \oplus \bigoplus_{j > 0} R/Ix^j$.
Hence $R \to A$ is a flat finite type ring map which is not of finite
presentation. Each fibre ring
is isomorphic to either $\kappa(\mathfrak p)[x, y]/(xy)$
or $\kappa(\mathfrak p)[x]$.

\medskip\noindent
We can turn the previous example into a projective morphism by
taking $B = R[X_0, X_1, X_2]/(X_1X_2, aX_2; a \in I)$.
In this case $X = \text{Proj}(B) \to S$ is a proper flat morphism
which is not of finite presentation such that for each $s \in S$
the fibre $X_s$ is isomorphic either to $\mathbf{P}^1_s$ or to
the closed subscheme of $\mathbf{P}^2_s$ defined by the vanishing of
$X_1X_2$ (this is a projective nodal curve of arithmetic genus $0$).

\medskip\noindent
Let $M = R \oplus R \oplus R/I$. Set $B = \text{Sym}_R(M)$ the
symmetric algebra on $M$. Set $X = \text{Proj}(B)$.
Then $X \to S$ is a proper flat morphism, not of finite presentation
such that for $s \in S$ the geometric fibre is isomorphic to either
$\mathbf{P}^1_s$ or $\mathbf{P}^2_s$. In particular these fibres are
smooth and geometrically irreducible.

\begin{lemma}
\label{lemma-finite-type-flat-not-finitely-presented}
There exist examples of
\begin{enumerate}
\item a flat finite type ring map with geometrically irreducible
complete intersection fibre rings which is not of finite presentation,
\item a flat finite type ring map with geometrically connected,
geometrically reduced, dimension 1, complete intersection fibre rings
which is not of finite presentation,
\item a proper flat morphism of schemes $X \to S$ each of whose fibres
is isomorphic to either $\mathbf{P}^1_s$ or to the vanishing locus of
$X_1X_2$ in $\mathbf{P}^2_s$ which is not of finite presentation, and
\item a proper flat morphism of schemes $X \to S$ each of whose
fibres is isomorphic to either $\mathbf{P}^1_s$ or $\mathbf{P}^2_s$
which is not of finite presentation.
\end{enumerate}
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{Topology of a finite type ring map}
\label{section-topology-finite-type}

\noindent
Let $A \to B$ be a local map of local domains.
If $A$ is Noetherian, $A \to B$ is essentially of finite type,
and $A/\mathfrak m_A \subset B/\mathfrak m_B$ is finite then there
exists a prime $\mathfrak q \subset B$, $\mathfrak q \not = \mathfrak m_B$
such that $A \to B/\mathfrak q$ is the localization of a quasi-finite ring
map. See
More on Morphisms, Lemma
\ref{more-morphisms-lemma-quasi-finite-quasi-section-meeting-nearby-open}.

\medskip\noindent
In this section we give an example that shows this result is false $A$
is no longer Noetherian. Namely, let $k$ be a field and set
$$
A = \{a_0 + a_1 x + a_2 x^2 + \ldots \mid a_0 \in k, a_i \in k((y))
\text{ for }i\geq 1\}
$$
and
$$
C = \{a_0 + a_1 x + a_2 x^2 + \ldots \mid a_0 \in k[y], a_i \in k((y))
\text{ for }i\geq 1\}.
$$
The inclusion $A \to C$ is of finite type as $C$ is generated by $y$
over $A$. We claim that $A$ is a local ring with maximal ideal
$\mathfrak m = \{a_1 x + a_2 x^2 + \ldots \in A\}$ and no prime
ideals besides $(0)$ and $\mathfrak m$. Namely, an element
$f = a_0 + a_1 x + a_2 x^2 + \ldots$ of $A$ is invertible as soon as
$a_0 \not = 0$. If $\mathfrak q \subset A$ is a nonzero prime ideal,
and $f = a_i x^i + \ldots \in \mathfrak q$, then using properties of
power series one sees that for any $g \in k((y))$ the element
$g^{i + 1} x^{i + 1} \in \mathfrak q$, i.e., $gx \in \mathfrak q$.
This proves that $\mathfrak q = \mathfrak m$.

\medskip\noindent
As to the spectrum of the ring $C$, arguing in the same way as above
we see that any nonzero prime ideal contains the prime
$\mathfrak p = \{a_1 x + a_2 x^2 + \ldots \in C\}$ which lies
over $\mathfrak m$. Thus the only prime of $C$ which lies over
$(0)$ is $(0)$. Set $\mathfrak m_C = yC + \mathfrak p$ and
$B = C_{\mathfrak m_C}$. Then $A \to B$ is the desired example.

\begin{lemma}
\label{lemma-topology-finite-type}
There exists a local homomorphism $A \to B$ of local domains which is
essentially of finite type and such that $A/\mathfrak m_A \to B/\mathfrak m_B$
is finite such that for every prime
$\mathfrak q \not = \mathfrak m_B$ of $B$ the ring map
$A \to B/\mathfrak q$ is not the localization of a quasi-finite ring map.
\end{lemma}

\begin{proof}
See the discussion above.
\end{proof}



\section{Pure not universally pure}
\label{section-pure-not-universally}

\noindent
Let $k$ be a field. Let
$$
R = k[[x, xy, xy^2, \ldots]] \subset k[[x, y]].
$$
In other words, a power series $f \in k[[x, y]]$ is in $R$ if and only
if $f(0, y)$ is a constant. In particular $R[1/x] = k[[x, y]][1/x]$
and $R/xR$ is a local ring with a maximal ideal whose square
is zero. Denote $R[y] \subset k[[x, y]]$ the set of power series
$f \in k[[x, y]]$ such that $f(0, y)$ is a polynomial in $y$. Then
$R \to R[y]$ is a finite type but not finitely presented ring map
which induces an isomorphism after inverting $x$. Also there is a
surjection $R[y]/xR[y] \to k[y]$ whose kernel has square zero. Consider
the finitely presented ring map $R \to S = R[t]/(xt - xy)$.
Again $R[1/x] \to S[1/x]$ is an isomorphism and in this case
$S/xS \cong (R/xR)[t]/(xy)$ maps onto $k[t]$ with nilpotent kernel.
There is a surjection $S \to R[y]$, $t \longmapsto y$ which induces
an isomorphism on inverting $x$ and a surjection with nilpotent kernel
modulo $x$. Hence the kernel of $S \to R[y]$ is locally nilpotent.
In particular $S \to R[y]$ is a universal homeomorphism.

\medskip\noindent
First we claim that $S$ is an $S$-module which is relatively pure
over $R$. Since on inverting $x$ we obtain an isomorphism we only need
to check this at the maximal ideal $\mathfrak m \subset R$. Since
$R$ is complete with respect to its maximal ideal it is henselian
hence we need only check that every prime $\mathfrak p \subset R$,
$\mathfrak p \not = \mathfrak m$, the unique prime $\mathfrak q$
of $S$ lying over $\mathfrak p$ satisfies
$\mathfrak mS + \mathfrak q \not = S$. Since $\mathfrak p \not = \mathfrak m$
it corresponds to a unique prime ideal of $k[[x, y]][1/x]$. Hence
either $\mathfrak p = (0)$ or $\mathfrak p = (f)$ for some
irreducible element $f \in k[[x, y]]$ which is not associated to $x$
(here we use that $k[[x, y]]$ is a UFD -- insert future reference here).
In the first case $\mathfrak q = (0)$ and the result is clear. In the
second case we may multiply $f$ by a unit so that $f \in R[y]$
(Weierstrass preparation; details omitted). Then it is easy to see that
$R[y]/fR[y] \cong k[[x, y]]/(f)$ hence $f$ defines a prime ideal
of $R[y]$ and $\mathfrak mR[y] + fR[y] \not = R[y]$.
Since $S \to R[y]$ is a universal homeomorphism we deduce the
desired result for $S$ also.

\medskip\noindent
Second we claim that $S$ is not universally relatively pure over $R$.
Namely, to see this it sufffices to find a valuation ring
$\mathcal{O}$ and a local ring map $R \to \mathcal{O}$ such that
$\Spec(R[y] \otimes_R \mathcal{O}) \to \Spec(\mathcal{O})$
does not hit the closed point of $\Spec(\mathcal{O})$.
Equivalently, we have to find $\varphi : R \to \mathcal{O}$ such that
$\varphi(x) \not = 0$ and $v(\varphi(x)) > v(\varphi(xy))$ where $v$
is the valuation of $\mathcal{O}$.
(Because this means that the valuation of $y$ is negative.)
To do this consider the ring map
$$
R
\longrightarrow
\{a_0 + a_1 x + a_2 x^2 + \ldots \mid a_0 \in k[y^{-1}], a_i \in k((y))\}
$$
defined in the obvious way. We can find a valuation ring $\mathcal{O}$
dominating the localization of the right hand side at the maximal
ideal $(y^{-1}, x)$ and we win.

\begin{lemma}
\label{lemma-pure-not-universally-pure}
There exists a morphism of affine schemes of finite presentation
$X \to S$ and an $\mathcal{O}_X$-module $\mathcal{F}$ of finite presentation
such that $\mathcal{F}$ is pure relative to $S$, but not universally
pure relative to $S$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{A formally smooth non-flat ring map}
\label{section-formally-smooth-nonflat}

\noindent
Let $k$ be a field. Consider the $k$-algebra $k[\mathbf{Q}]$.
This is the $k$-algebra with basis $x_\alpha, \alpha \in \mathbf{Q}$
and multiplication determined by $x_\alpha x_\beta = x_{\alpha + \beta}$.
(In particular $x_0 = 1$.)
Consider the $k$-algebra homomorphism
$$
k[\mathbf{Q}] \longrightarrow k, \quad
x_\alpha \longmapsto 1.
$$
It is surjective with kernel $J$ generated by the elements $x_\alpha - 1$.
Let us compute $J/J^2$. Note that multiplication by $x_\alpha$ on $J/J^2$
is the identity map. Denote $z_\alpha$ the class of $x_\alpha - 1$ modulo
$J^2$. These classes generate $J/J^2$. Since
$$
(x_\alpha - 1)(x_\beta - 1) = x_{\alpha + \beta} - x_\alpha - x_\beta + 1 =
(x_{\alpha + \beta} - 1) - (x_\alpha - 1) - (x_\beta - 1)
$$
we see that $z_{\alpha + \beta} = z_\alpha + z_\beta$ in $J/J^2$.
A general element of $J/J^2$ is of the form $\sum \lambda_\alpha z_\alpha$
with $\lambda_\alpha \in k$ (only finitely many nonzero). Note that if the
characteristic of $k$ is $p > 0$ then
$$
0 = pz_{\alpha/p} = z_{\alpha/p} + \ldots + z_{\alpha/p} = z_\alpha
$$
and we see that $J/J^2 = 0$. If the characteristic of $k$ is zero, then
$$
J/J^2 = \mathbf{Q} \otimes_{\mathbf{Z}} k \cong k
$$
(details omitted) is not zero.

\medskip\noindent
We claim that $k[\mathbf{Q}] \to k$ is a formally smooth ring map
if the characteristic of $k$ is positive. Namely, suppose given a
solid commutative diagram
$$
\xymatrix{
k \ar[r] \ar@{..>}[rd] & A \\
k[\mathbf{Q}] \ar[u] \ar[r]^\varphi & A' \ar[u]
}
$$
with $A' \to A$ a surjection whose kernel $I$ has square zero.
To show that $k[\mathbf{Q}] \to k$ is formally smooth we have to prove
that $\varphi$ factors through $k$. Since $\varphi(x_\alpha - 1)$ maps
to zero in $A$ we see that $\varphi$ induces a map
$\overline{\varphi} : J/J^2 \to I$ whose vanishing is the obstruction
to the desired factorization. Since $J/J^2 = 0$ if the characteristic
is $p > 0$ we get the result we want, i.e., $k[\mathbf{Q}] \to k$ is
formally smooth in this case. Finally, this ring map is not flat, for
example as the nonzerodivisor $x_2 - 1$ is mapped to zero.

\begin{lemma}
\label{lemma-formally-smooth-nonflat}
There exists a formally smooth ring map which is not flat.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}


\section{A formally \'etale non-flat ring map}
\label{section-formally-etale-nonflat}

\noindent
In this section we give a counterexample to the final sentence in
\cite[0${}_{\rm{IV}}$, Example 19.10.3(i)]{EGA}
(this was not one of the items caught in their
later errata lists). Consider $A \to A/J$ for a local ring A and a nonzero
proper ideal $J$ such that $J^2 = J$ (so $J$ isn't finitely generated); the
valuation ring of an algebraically closed non-archimedean field with $J$
its maximal ideal is a source of such $(A, J)$. These non-flat quotient
maps are formally \'etale. Namely, suppose given a commutative diagram
$$
\xymatrix{
A/J \ar[r] & R/I \\
A \ar[u] \ar[r]^\varphi & R \ar[u]
}
$$
where $I$ is an ideal of the ring $R$ with $I^2 = 0$. Then $A \to R$
factors uniquely through $A/J$ because
$$
\varphi(J) = \varphi(J^2) \subset (\varphi(J)A)^2 \subset I^2 = 0.
$$
Hence this also provides a counterexample to the formally \'etale case
of the ``structure theorem'' for locally finite type and formally \'etale
morphisms in \cite[IV, Theorem 18.4.6(i)]{EGA}
(but not a counterexample to part (ii), which
is what people actually use in practice). The error in the proof of
the latter is that the very last step of the proof is to invoke the incorrect
\cite[0${}_{\rm{IV}}$, Example 19.3.10(i)]{EGA},
which is how the counterexample just mentioned creeps in.

\begin{lemma}
\label{lemma-formally-etale-not-presented}
There exist formally \'etale nonflat ring maps.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{A formally \'etale ring map with nontrivial cotangent complex}
\label{section-formally-etale-nontrivial-cotangent-complex}

\noindent
Let $k$ be a field. Consider the ring
$$
R = k[\{x_n\}_{n \geq 1}, \{y_n\}_{n \geq 1}]/(
x_1y_1, x_{nm}^m - x_n, y_{nm}^m - y_n)
$$
Let $A$ be the localization at the maximal ideal generated by
all $x_n, y_n$ and denote $J \subset A$ the maximal ideal. Set $B = A/J$.
By construction $J^2 = J$ and hence $A \to B$ is formally \'etale (see
Section \ref{section-formally-etale-nonflat}).
We claim that the element $x_1 \otimes y_1$ is a nonzero
element in the kernel of
$$
J \otimes_A J \longrightarrow J.
$$
Namely, $(A, J)$ is the colimit of the localizations
$(A_n, J_n)$ of the rings
$$
R_n = k[x_n, y_n]/(x_n^n y_n^n)
$$
at their corresponding maximal ideals. Then
$x_1 \otimes y_1$ corresponds to the element
$x_n^n \otimes y_n^n \in J_n \otimes_{A_n} J_n$ and is nonzero
(by an explicit computation which we omit). Since $\otimes$ commutes
with colimits we conclude. By
\cite[III Section 3.3]{cotangent}
we see that $J$ is not weakly regular. Hence by
\cite[III Proposition 3.3.3]{cotangent}
we see that the cotangent complex $L_{B/A}$ is not zero. In fact, we can
be more precise. We have $H_0(L_{B/A}) = \Omega_{B/A}$ and
$H_1(L_{B/A}) = 0$ because $J/J^2 = 0$. But from the five-term exact sequence
of Quillen's fundamental spectral sequence
(see Cotangent, Remark \ref{cotangent-remark-elucidate-ss} or
\cite[Corollary 8.2.6]{Reinhard})
and the nonvanishing of
$\text{Tor}_2^A(B, B) = \text{Ker}(J \otimes_A J \to J)$ we conclude that
$H_2(L_{B/A})$ is nonzero.

\begin{lemma}
\label{lemma-formally-etale-nontrivial-cotangent-complex}
There exists a formally \'etale surjective ring map $A \to B$
with $L_{B/A}$ not equal to zero.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}















\section{Ideals generated by sets of idempotents and localization}
\label{section-ideal-locally-idempotents}

\noindent
Let $R$ be a ring. Consider the ring
$$
B(R) = R[x_n; n \in \mathbf{Z}]/(x_n(x_n - 1), x_nx_m; n \not = m)
$$
It is easy to show that every prime $\mathfrak q \subset B(R)$
is either of the form
$$
\mathfrak q = \mathfrak pB(R) + (x_n; n \in \mathbf{Z})
$$
or of the form
$$
\mathfrak q =
\mathfrak pB(R) + (x_n - 1) + (x_m; n \not = m, m \in \mathbf{Z}).
$$
Hence we see that
$$
\Spec(B(R)) =
\Spec(R) \amalg \coprod\nolimits_{n \in \mathbf{Z}} \Spec(R)
$$
where the topology is not just the disjoint union topology. It has the
following properties: Each of the copies indexed by $n \in \mathbf{Z}$
is an open subscheme, namely it is the standard open $D(x_n)$.
The "central" copy of $\Spec(R)$ is in the closure of the union
of any infinitely many of the other copies of $\Spec(R)$.
Note that this last copy of $\Spec(R)$ is cut out by the ideal
$(x_n, n \in \mathbf{Z})$ which is generated by the idempotents $x_n$.
Hence we see that if $\Spec(R)$ is connected,
then the decomposition above is exactly the decomposition of
$\Spec(B(R))$ into connected components.

\medskip\noindent
Next, let $A = \mathbf{C}[x, y]/((y - x^2 + 1)(y + x^2 - 1))$.
The spectrum of $A$ consists of two irreducible components
$C_1 = \Spec(A_1)$, $C_2 = \Spec(A_2)$
with $A_1 = \mathbf{C}[x, y]/(y - x^2 + 1)$ and
$A_2 = \mathbf{C}[x, y]/(y + x^2 - 1)$. Note that these are
parametrized by $(x, y) = (t, t^2 - 1)$ and $(x, y) = (t, -t^2 + 1)$
which meet in $P = (-1, 0)$ and $Q = (1, 0)$. We can make a twisted
version of $B(A)$ where we glue $B(A_1)$ to $B(A_2)$ in the following
way: Above $P$ we let $x_n \in B(A_1) \otimes \kappa(P)$
correspond to $x_n \in B(A_2) \otimes \kappa(P)$, but above $Q$
we let $x_n \in B(A_1) \otimes \kappa(Q)$
correspond to $x_{n + 1} \in B(A_2) \otimes \kappa(Q)$.
Let $B^{twist}(A)$ denote the resulting $A$-algebra.
Details omitted. By construction
$B^{twist}(A)$ is Zariski locally over $A$ isomorphic to the untwisted
version. Namely, this happens over both the principal open
$\Spec(A) \setminus \{P\}$
and the principal open $\Spec(A) \setminus \{Q\}$.
However, our choice of glueing produces enough "monodromy" such that
$\Spec(B^{twist}(A))$ is connected (details omitted).
Finally, there is a central copy of
$\Spec(A) \to \Spec(B^{twist}(A))$
which gives a closed subscheme whose ideal is Zariski locally
on $B^{twist}(A)$ cut out by ideals generated by idempotents, but
not globally (as $B^{twist}(A)$ has no nontrivial idempotents).

\begin{lemma}
\label{lemma-not-generated-idempotents}
There exists an affine scheme $X = \Spec(A)$ and a
closed subscheme $T \subset X$ such that $T$ is Zariski locally
on $X$ cut out by ideals generated by idempotents, but
$T$ is not cut out by an ideal generated by idempotents.
\end{lemma}

\begin{proof}
See above.
\end{proof}



\section{A ring map which identifies local rings which is not ind-\'etale}
\label{section-not-ind-etale}

\noindent
Note that the ring map $R \to B(R)$ constructed in
Section \ref{section-ideal-locally-idempotents} is
a colimit of finite products of copies of $R$. Hence $R \to B(R)$
is ind-Zariski, see
Pro-\'etale Cohomology, Definition \ref{proetale-definition-ind-zariski}.
Next, consider the ring map $A \to B^{twist}(A)$
constructed in Section \ref{section-ideal-locally-idempotents}.
Since this ring map is Zariski locally on $\Spec(A)$ isomorphic to an
ind-Zariski ring map $R \to B(R)$ we conclude that it identifies local rings
(see Pro-\'etale Cohomology, Lemma \ref{proetale-lemma-ind-zariski-implies}).
The discussion in Section \ref{section-ideal-locally-idempotents}
shows there is a section
$B^{twist}(A) \to A$ whose kernel is not generated by idempotents.
Now, if $A \to B^{twist}(A)$ were ind-\'etale, i.e.,
$B^{twist}(A) = \colim A_i$ with $A \to A_i$ \'etale,
then the kernel of $A_i \to A$ would be generated by an idempotent
(Algebra, Lemmas \ref{algebra-lemma-map-between-etale} and
\ref{algebra-lemma-surjective-flat-finitely-presented}).
This would contradict the result mentioned above.

\begin{lemma}
\label{lemma-not-ind-etale}
There is a ring map $A \to B$ which identifies local rings but
which is not ind-\'etale. A fortiori it is not ind-Zariski.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}




\section{Non flasque quasi-coherent sheaf associated to injective module}
\label{section-nonflasque}

\noindent
For more examples of this type see \cite[Expos\'e II, Appendix I]{SGA6}
where Illusie explains some examples due to Verdier.

\medskip\noindent
Consider the affine scheme $X = \Spec(A)$
where
$$
A = k[f, g, x, y, \{a_n, b_n\}_{n \geq1}]/
(fy - gx, \{a_nf^n + b_ng^n\}_{n \geq 1})
$$
is the ring from
Properties, Example \ref{properties-example-does-not-work-in-general}.
Set $I = (f, g) \subset A$.
Consider the quasi-compact open $U = D(f) \cup D(g)$ of $X$.
We have seen in loc.\ cit.\ that there is a section
$s \in \mathcal{O}_X(U)$ which does not come from an $A$-module
map $I^n \to A$ for any $n \geq 0$.

\medskip\noindent
Let $\alpha : A \to J$ be the embedding of $A$ into an injective $A$-module.
Let $Q = J/\alpha(A)$ and denote $\beta : J \to Q$ the quotient map.
We claim that the map
$$
\Gamma(X, \widetilde{J})
\longrightarrow
\Gamma(U, \widetilde{J})
$$
is not surjective. Namely, we claim that $\alpha(s)$ is not in the image.
To see this, we argue by contradiction. So assume that $x \in J$ is an
element which restricts to $\alpha(s)$ over $U$. Then $\beta(x) \in Q$
is an element which restricts to $0$ over $U$. Hence we know that
$I^n\beta(x) = 0$ for some $n$, see
Properties,
Lemma \ref{properties-lemma-sections-over-quasi-compact-open-in-affine}.
This implies that we get a morphism
$\varphi : I^n \to A$, $h \mapsto \alpha^{-1}(hx)$. It is easy to see that
this morphism $\varphi$ gives rise to the section $s$ via the map of
Properties,
Lemma \ref{properties-lemma-sections-over-quasi-compact-open-in-affine}
which is a contradiction.

\begin{lemma}
\label{lemma-nonflasque}
There exists an affine scheme $X = \Spec(A)$ and an injective
$A$-module $J$ such that $\widetilde{J}$ is not a flasque sheaf on $X$.
Even the restriction $\Gamma(X, \widetilde{J}) \to \Gamma(U, \widetilde{J})$
with $U$ quasi-compact open need not be surjective.
\end{lemma}

\begin{proof}
See above.
\end{proof}





\section{A non-separated flat group scheme}
\label{section-non-separated-group-scheme}

\noindent
Every group scheme over a field is separated, see
Groupoids, Lemma \ref{groupoids-lemma-group-scheme-over-field-separated}.
This is not true for group schemes over a base.

\medskip\noindent
Let $k$ be a field. Let $S = \Spec(k[x]) = \mathbf{A}^1_k$.
Let $G$ be the affine line with $0$ doubled (see
Schemes, Example \ref{schemes-example-affine-space-zero-doubled})
seen as a scheme over $S$. Thus a fibre of $G \to S$ is either a
singleton or a set with two elements (one in $U$ and one in $V$).
Thus we can endow these fibres with the structure of a group (by
letting the element in $U$ be the zero of the group structure).
More precisely, $G$ has two opens $U, V$ which map isomorphically
to $S$ such that $U \cap V$ is mapped isomorphically to
$S \setminus \{0\}$. Then
$$
G \times_S G = U \times_S U \cup V \times_S U \cup
U \times_S V \cup V \times_S V
$$
where each piece is isomorphic to $S$. Hence we can define a multiplication
$m : G \times_S G \to G$ as the unique $S$-morphism which maps the first
and the last piece into $U$ and the two middle pieces into $V$. This matches
the pointwise description given above. We omit the
verification that this defines a group scheme structure.

\begin{lemma}
\label{lemma-non-separated-group-scheme}
There exists a flat group scheme of finite type over the affine line
which is not separated.
\end{lemma}

\begin{proof}
See the discussion above.
\end{proof}

\begin{lemma}
\label{lemma-non-quasi-separated-group-scheme}
There exists a flat group scheme of finite type over the infinite
dimensional affine space which is not quasi-separated.
\end{lemma}

\begin{proof}
The same construction as above can be carried out with the infinite dimensional
affine space $S = \mathbf{A}^\infty_k = \Spec k[x_1, x_2, \dotsc]$ as the base
and the origin $0 \in S$ corresponding to the maximal ideal
$(x_1, x_2, \ldots)$ as the closed point which is doubled in $G$.
The resulting group scheme $G \rightarrow S$ is 
not quasi-separated as explained in
Schemes, Example \ref{schemes-example-not-quasi-separated}. 
\end{proof}



\section{A non-flat group scheme with flat identity component}
\label{section-non-flat-group-scheme}

\noindent
Let $X \to S$ be a monomorphism of schemes. Let $G = S \amalg X$.
Let $m : G \times_S G \to G$ be the $S$-morphism
$$
G \times_S G = X \times_S X \amalg X \amalg X \amalg S
\longrightarrow G = X \amalg S
$$
which maps the summands $X \times_S X$ and $S$ into $S$ and
maps the summands $X$ into $X$ by the identity morphism.
This defines a group law. To see this we have to show that
$m \circ (m \times \text{id}_G) = m \circ (\text{id}_G \times m)$
as maps $G \times_S G \times_S G \to G$. Decomposing
$G \times_S G \times_S G$ into components as above, we see that
we need to verify this for the restriction to each of the $8$-pieces.
Each piece is isomorphic to either $S$, $X$, $X \times_S X$, or
$X \times_S X \times_S X$. Moreover, both maps map these pieces
to $S$, $X$, $S$, $X$ respectively. Having said this, the fact that
$X \to S$ is a monomorphism implies that $X \times_S X \cong X$
and $X \times_S X \times_S X \cong X$ and that there is in each case
exactly one $S$-morphism $S \to S$ or $X \to X$. Thus we see
that $m \circ (m \times \text{id}_G) = m \circ (\text{id}_G \times m)$.
Thus taking $X \to S$ to be any
nonflat monomorphism of schemes (e.g., a closed immersion)
we get an example of a group scheme
over a base $S$ whose identity component is $S$ (hence flat)
but which is not flat.

\begin{lemma}
\label{lemma-non-flat-group-scheme}
There exists a group scheme $G$ over a base $S$ whose identity
component is flat over $S$ but which is not flat over $S$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}




\section{A non-separated group algebraic space over a field}
\label{section-non-separated-group-space}

\noindent
Every group scheme over a field is separated, see
Groupoids, Lemma \ref{groupoids-lemma-group-scheme-over-field-separated}.
This is not true for group algebraic spaces over a field
(but see end of this section for positive results).

\medskip\noindent
Let $k$ be a field of characteristic zero.
Consider the algebraic space $G = \mathbf{A}^1_k/\mathbf{Z}$ from
Spaces, Example \ref{spaces-example-affine-line-translation}.
By construction $G$ is the fppf sheaf associated to the presheaf
$$
T \longmapsto \Gamma(T, \mathcal{O}_T) / \mathbf{Z}
$$
on the category of schemes over $k$. The obvious addition rule on the presheaf
induces an addition $m : G \times G \to G$ which turns $G$ into a group
algebraic space over $\Spec(k)$. Note that $G$ is not separated
(and not even quasi-separated or locally separated). On the other hand
$G \to \Spec(k)$ is of finite type!

\begin{lemma}
\label{lemma-non-separated-group-space}
There exists a group algebraic space of finite type over a field
which is not separated (and not even quasi-separated or locally separated).
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}

\noindent
Positive results: If the group algebraic space $G$ is either
quasi-separated, or locally separated, or more generally a
decent algebraic space, then $G$ is in fact separated, see
More on Groupoids in Spaces, Lemma
\ref{spaces-more-groupoids-lemma-group-scheme-over-field-separated}.
Moreover, a finite type, separated group algebraic space over a
field is in fact a scheme (insert future reference here). The idea
of the proof is that the schematic locus is open dense, see
Properties of Spaces, Proposition
\ref{spaces-properties-proposition-locally-quasi-separated-open-dense-scheme}
or
Decent Spaces, Theorem \ref{decent-spaces-theorem-decent-open-dense-scheme}.
By translating this open we see that
every point of $G$ has an open neighbourhood which is a scheme.




\section{Specializations between points in fibre \'etale morphism}
\label{section-specializations-fibre-etale}

\noindent
If $f : X \to Y$ is an \'etale, or more generally a locally quasi-finite
morphism of schemes, then there are no specializations between points of
fibres, see
Morphisms, Lemma \ref{morphisms-lemma-locally-quasi-finite-fibres}.
However, for morphisms of algebraic spaces this doesn't hold in general.

\medskip\noindent
To give an example, let $k$ be a field.
Set
$$
P = k[u, u^{-1}, y, \{x_n\}_{n \in \mathbf{Z}}].
$$
Consider the action
of $\mathbf{Z}$ on $P$ by $k$-algebra maps generated by the automorphism
$\tau$ given by the rules $\tau(u) = u$, $\tau(y) = uy$, and
$\tau(x_n) = x_{n + 1}$. For $d \geq 1$ set
$I_d = ((1 - u^d)y, x_n - x_{n + d}, n \in \mathbf{Z})$.
Then $V(I_d) \subset \Spec(P)$ is the fix point locus of $\tau^d$.
Let $S \subset P$ be the multiplicative subset generated by $y$ and
all $1 - u^d$, $d \in \mathbf{N}$. Then we
see that $\mathbf{Z}$ acts freely on $U = \Spec(S^{-1}P)$.
Let $X = U/\mathbf{Z}$ be the quotient algebraic space, see
Spaces, Definition \ref{spaces-definition-quotient}.

\medskip\noindent
Consider the prime ideals $\mathfrak p_n = (x_n, x_{n + 1}, \ldots)$ in
$S^{-1}P$. Note that $\tau(\mathfrak p_n) = \mathfrak p_{n + 1}$.
Hence each of these define point $\xi_n \in U$ whose image in $X$ is
the same point $x$ of $X$. Moreover we have the specializations
$$
\ldots \leadsto \xi_n \leadsto \xi_{n - 1} \leadsto \ldots
$$
We conclude that $U \to X$ is an example of the promised type.

\begin{lemma}
\label{lemma-specializations-fibre-etale}
There exists an \'etale morphism of algebraic spaces $f : X \to Y$
and a nontrivial specialization of points $x \leadsto x'$ in $|X|$ with
$f(x) = f(x')$ in $|Y|$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}







\section{A torsor which is not an fppf torsor}
\label{section-torsor-not-fppf}

\noindent
In
Groupoids, Remark \ref{groupoids-remark-fun-with-torsors}
we raise the question whether any $G$-torsor is a $G$-torsor for the
fppf topology. In this section we show that this is not always the case.

\medskip\noindent
Let $k$ be a field. All schemes and stacks are over $k$ in what follows.
Let $G \to \Spec(k)$ be the group scheme
$$
G = (\mu_{2, k})^\infty =
\mu_{2, k} \times_k \mu_{2, k} \times_k \mu_{2, k} \times_k \ldots =
\lim_n (\mu_{2, k})^n
$$
where $\mu_{2, k}$ is the group scheme of second roots of unity over
$\Spec(k)$, see
Groupoids, Example \ref{groupoids-example-roots-of-unity}.
As an inverse limit of affine schemes we see that $G$ is an affine group
scheme. In fact it is the spectrum of the ring
$k[t_1, t_2, t_3, \ldots]/(t_i^2 - 1)$. The multiplication map
$m : G \times_k G \to G$ is on the algebra level given by
$t_i \mapsto t_i \otimes t_i$.

\medskip\noindent
We claim that any $G$-torsor over $k$ is of the form
$$
P = \Spec(k[x_1, x_2, x_3, \ldots]/(x_i^2 - a_i))
$$
for certain $a_i \in k^*$ and with $G$-action $G \times_k P \to P$
given by $x_i \to t_i \otimes x_i$ on the algebra level.
We omit the proof.
Actually for the example we only need that $P$ is a $G$-torsor
which is clear since over $k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots)$
the scheme $P$ becomes isomorphic to $G$ in a $G$-equivariant manner.
Note that $P$ is trivial if and only if $k' = k$ since if
$P$ has a $k$-rational point then all of the $a_i$ are squares.

\medskip\noindent
We claim that $P$ is an fppf torsor if and only if the field extension
$k \subset k' = k(\sqrt{a_1}, \sqrt{a_2}, \ldots)$ is finite.
If $k'$ is finite over $k$, then
$\{\Spec(k') \to \Spec(k)\}$
is an fppf covering which trivializes $P$ and we see that $P$ is indeed
an fppf torsor. Conversely, suppose that $P$ is a $G$-torsor for the
fppf topology. This means that there exists an fppf covering
$\{S_i \to \Spec(k)\}$ such that each $P_{S_i}$ is trivial.
Pick an $i$ such that $S_i$ is not empty. Let $s \in S_i$ be a closed
point. By
Varieties, Lemma \ref{varieties-lemma-locally-finite-type-Jacobson}
the field extension $k \subset \kappa(s)$ is finite, and by construction
$P_{\kappa(s)}$ has a $\kappa(s)$-rational point. Thus we see that
$k \subset k' \subset \kappa(s)$ and $k'$ is finite over $k$.

\medskip\noindent
To get an explicit example take $k = \mathbf{Q}$ and $a_i = i$
for example (or $a_i$ is the $i$th prime if you like).

\begin{lemma}
\label{lemma-torsors-principal-spaces-not-equal}
Let $S$ be a scheme. Let $G$ be a group scheme over $S$.
The stack $G\textit{-Principal}$ classifying principal homogeneous $G$-spaces
(see Examples of Stacks, Subsection
\ref{examples-stacks-subsection-principal-homogeneous-spaces})
and the stack $G\textit{-Torsors}$ classifying fppf $G$-torsors
(see Examples of Stacks, Subsection
\ref{examples-stacks-subsection-fppf-torsors})
are not equivalent in general.
\end{lemma}

\begin{proof}
The discussion above shows that the functor
$G\textit{-Torsors} \to G\textit{-Principal}$ isn't essentially
surjective in general.
\end{proof}









\section{Stack with quasi-compact flat covering which is not algebraic}
\label{section-not-algebraic-stack}

\noindent
In this section we briefly describe an example due to Brian Conrad.
You can find the example online at
\href{http://mathoverflow.net/questions/15082/fpqc-covers-of-stacks/15269#15269}{this location}.
Our example is slightly different.

\medskip\noindent
Let $k$ be an algebraically closed field.
All schemes and stacks are over $k$ in what follows.
Let $G \to \Spec(k)$ be an affine group scheme.
In
Examples of Stacks,
Proposition \ref{examples-stacks-proposition-equal-quotient-stacks}
we have seen that
$\mathcal{X} = [\Spec(k)/G]$ is a stack in groupoids over
$(\Sch/\Spec(k))_{fppf}$ which can be described
as follows. A $1$-morphism $T \to \mathcal{X}$
corresponds by definition to an fppf $G_T$-torsor $P$ over $T$.
The diagonal $1$-morphism
$$
\Delta :
\mathcal{X}
\longrightarrow
\mathcal{X} \times_{\Spec(k)} \mathcal{X}
$$
is representable and affine. The reason for this is that given any pair
of $G_T$-torsors $P_1, P_2$ in the fppf topology over a scheme $S/k$ the
scheme $\mathit{Isom}(P_1, P_2)$ is affine over $T$.
The trivial $G$-torsor over $\Spec(k)$ defines a $1$-morphism
$$
f : \Spec(k) \longrightarrow \mathcal{X}.
$$
We claim that this is a surjective $1$-morphism. The reason is simply
that by definition for any $1$-morphism $T \to \mathcal{X}$ there exists
a fppf covering $\{T_i \to T\}$ such that $P_{T_i}$ is isomorphic
to the trivial $G_{T_i}$-torsor. Hence the compositions
$T_i \to T \to \mathcal{X}$ factor through $f$. Thus it is clear that
the projection $T \times_\mathcal{X} \Spec(k) \to \mathcal{X}$
is surjective (which is how we define the property that $f$ is surjective, see
Algebraic Stacks,
Definition \ref{algebraic-definition-relative-representable-property}).
In a similar way you show that $f$ is quasi-compact and flat (details omitted).
We also record here the observation that
$$
\Spec(k) \times_\mathcal{X} \Spec(k) \cong G
$$
as schemes over $k$.

\medskip\noindent
Suppose there exists a surjective smooth morphism
$p : U \to \mathcal{X}$ where $U$ is a scheme.
Consider the fibre product
$$
\xymatrix{
W \ar[d] \ar[r] & U \ar[d] \\
\Spec(k) \ar[r] & \mathcal{X}
}
$$
Then we see that $W$ is a nonempty smooth scheme over $k$ which
hence has a $k$-point. This means that we can factor $f$ through $U$.
Hence we obtain
$$
G \cong
\Spec(k) \times_\mathcal{X} \Spec(k) \cong
(\Spec(k) \times_k \Spec(k))
\times_{(U \times_k U)}
(U \times_\mathcal{X} U)
$$
and since the projections $U \times_\mathcal{X} U \to U$ were
assumed smooth we conclude that
$U \times_\mathcal{X} U \to U \times_k U$ is
locally of finite type, see
Morphisms, Lemma \ref{morphisms-lemma-permanence-finite-type}.
It follows that in this case $G$ is locally of finite type over $k$.
Altogether we have proved the following lemma (which can be
significantly generalized).

\begin{lemma}
\label{lemma-BG-algebraic}
Let $k$ be a field. Let $G$ be an affine group scheme over $k$.
If the stack $[\Spec(k)/G]$ has a smooth covering by a
scheme, then $G$ is of finite type over $k$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}

\noindent
To get an explicit example as in the title of this section, take for example
$G = (\mu_{2, k})^\infty$ the group scheme of
Section \ref{section-torsor-not-fppf}, which is not locally of finite type
over $k$. By the discussion above we see that
$\mathcal{X} = [\Spec(k)/G]$ has properties (1) and (2) of
Algebraic Stacks, Definition \ref{algebraic-definition-algebraic-stack},
but not property (3). Hence $\mathcal{X}$ is not an algebraic stack.
On the other hand, there does exists a scheme $U$ an a surjective,
flat, quasi-compact morphism $U \to \mathcal{X}$, namely the morphism
$f : \Spec(k) \to \mathcal{X}$ we studied above.






\section{Limit preserving on objects, not limit preserving}
\label{section-limit-preserving}

\noindent
Let $S$ be a nonempty scheme. Let $\mathcal{G}$ be an injective abelian sheaf
on $(\Sch/S)_{fppf}$. We obtain a stack in groupoids
$$
\mathcal{G}\textit{-Torsors} \longrightarrow (\Sch/S)_{fppf}
$$
over $S$, see Examples of Stacks, Lemma
\ref{examples-stacks-lemma-torsors-sheaf-stack-in-groupoids}.
This stack is limit preserving on objects over $(\Sch/S)_{fppf}$ (see
Criteria for Representability, Section \ref{criteria-section-limit-preserving})
because every $\mathcal{G}$-torsor is trivial. On the other hand,
$\mathcal{G}\textit{-Torsors}$ is in general not limit preserving (see
Artin's Axioms, Definition \ref{artin-definition-limit-preserving})
as $\mathcal{G}$ need not be limit preserving as a sheaf. For example,
take any nonzero injective sheaf $\mathcal{I}$ and set
$\mathcal{G} = \prod_{n \in \mathbf{Z}} \mathcal{I}$ to get an
example.

\begin{lemma}
\label{lemma-limit-preserving-on-objects-not-limit-preserving}
Let $S$ be a nonempty scheme. There exists a stack in groupoids
$p : \mathcal{X} \to (\Sch/S)_{fppf}$
such that $p$ is limit preserving on objects, but $\mathcal{X}$ is not
limit preserving.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{A non-algebraic classifying stack}
\label{section-non-algebraic}

\noindent
Let $S = \Spec(\mathbf{F}_p)$ and let $\mu_p$ denote the
group scheme of $p$th roots of unity over $S$. In
Groupoids in Spaces, Section \ref{spaces-groupoids-section-stacks}
we have introduced the quotient stack $[S/\mu_p]$ and in
Examples of Stacks, Section \ref{examples-stacks-section-group-quotient-stacks}
we have shown $[S/\mu_p]$ is the classifying stack for
fppf $\mu_p$-torsors: Given a scheme $T$ over $S$ the category
$\Mor_S(T, [S/\mu_p])$ is canonically equivalent to the
category of fppf $\mu_p$-torsors over $T$. Finally, in
Criteria for Representability, Theorem
\ref{criteria-theorem-flat-groupoid-gives-algebraic-stack}
we have seen that $[S/\mu_p]$ is an algebraic stack.

\medskip\noindent
Now we can ask the question: ``How about the category fibred in groupoids
$\mathcal{S}$ classifying \'etale $\mu_p$-torsors?'' (In other words
$\mathcal{S}$ is a category over $\Sch/S$ whose fibre category over a
scheme $T$ is the category of \'etale $\mu_p$-torsors over $T$.)

\medskip\noindent
The first objection is that this isn't a stack for the fppf topology,
because descent for objects isn't going to hold. For example the
$\mu_p$-torsor $\Spec(\mathbf{F}_p(t)[x]/(x^p - t))$ over
$T = \Spec(\mathbf{F}_p(T))$ is fppf locally trivial, but
not \'etale locally trivial.

\medskip\noindent
A fix for this first problem is to work with the \'etale topology and
in this case descent for objects does work. Indeed it is true that
$\mathcal{S}$ is a stack in groupoids over $(\Sch/S)_\etale$.
Moreover, it is also the case that the diagonal
$\Delta : \mathcal{S} \to \mathcal{S} \times \mathcal{S}$
is representable (by schemes). This is true because given two
$\mu_p$-torsors (whether they be \'etale locally trivial or not)
the sheaf of isomorphisms between them is representable by a scheme.

\medskip\noindent
Thus we can finally ask if there exists a scheme $U$ and a smooth
and surjective $1$-morphism $U \to \mathcal{S}$. We will show in two
ways that this is impossible: by a direct argument (which we advise
the reader to skip) and by an argument using a general result.

\medskip\noindent
Direct argument (sketch): Note that the $1$-morphism
$\mathcal{S} \to \Spec(\mathbf{F}_p)$ satisfies the
infinitesimal lifting criterion for formal smoothness.
This is true because given a first order infinitesimal thickening
of schemes $T \to T'$ the kernel of $\mu_p(T') \to \mu_p(T)$
is isomorphic to the sections of the ideal sheaf of $T$ in $T'$, and
hence $H^1_\etale(T, \mu_p) = H^1_\etale(T', \mu_p)$.
Moreover, $\mathcal{S}$ is a limit preserving stack. Hence
if $U \to \mathcal{S}$ is smooth, then $U \to \Spec(\mathbf{F}_p)$
is limit preserving and satisfies  the
infinitesimal lifting criterion for formal smoothness.
This implies that $U$ is smooth over $\mathbf{F}_p$.
In particular $U$ is reduced, hence $H^1_\etale(U, \mu_p) = 0$.
Thus $U \to \mathcal{S}$ factors as
$U \to \Spec(\mathbf{F}_p) \to \mathcal{S}$
and the first arrow is smooth. By descent of smoothness, we see
that $U \to \mathcal{S}$ being smooth would imply
$\Spec(\mathbf{F}_p) \to \mathcal{S}$ is smooth. However, this
is not the case as
$\Spec(\mathbf{F}_p) \times_\mathcal{S} \Spec(\mathbf{F}_p)$
is $\mu_p$ which is not smooth over $\Spec(\mathbf{F}_p)$.


\medskip\noindent
Structural argument: In
Criteria for Representability, Section \ref{criteria-section-stacks-etale}
we have seen that we can think of algebraic stacks as those
stacks in groupoids for the \'etale topology with diagonal
representable by algebraic spaces having a smooth covering.
Hence if a smooth surjective $U \to \mathcal{S}$ exists then
$\mathcal{S}$ is an algebraic stack, and in particular satisfies
descent in the fppf topology. But we've seen above that $\mathcal{S}$
does not satisfies descent in the fppf topology.

\medskip\noindent
Loosely speaking the arguments above show that the classifying
stack in the \'etale topology for \'etale locally trivial torsors
for a group scheme $G$ over a base $B$ is algebraic if and only
if $G$ is smooth over $B$. One of the advantages of working with
the fppf topology is that it suffices to assume that $G \to B$
is flat and locally of finite presentation. In fact the quotient
stack (for the fppf topology) $[B/G]$ is algebraic if and only
if $G \to B$ is flat and locally of finite presentation, see
Criteria for Representability, Lemma \ref{criteria-lemma-BG-algebraic}.







\section{Sheaf with quasi-compact flat covering which is not algebraic}
\label{section-not-algebraic}

\noindent
Consider the functor $F = (\mathbf{P}^1)^\infty$, i.e., for a scheme $T$
the value $F(T)$ is the set of $f = (f_1, f_2, f_3, \ldots)$ where each
$f_i : T \to \mathbf{P}^1$ is a morphism of schemes. Note that
$\mathbf{P}^1$ satisfies the sheaf property for fpqc coverings, see
Descent, Lemma \ref{descent-lemma-fpqc-universal-effective-epimorphisms}.
A product of sheaves is a sheaf, so $F$ also satisfies the sheaf property for
the fpqc topology. The diagonal of $F$ is representable: if $f : T \to F$
and $g : S \to F$ are morphisms, then $T \times_F S$ is the scheme theoretic
intersection of the closed subschemes $T \times_{f_i, \mathbf{P}^1, g_i} S$
inside the scheme $T \times S$. Consider the group scheme $\text{SL}_2$ which
comes with a surjective smooth affine morphism $\text{SL}_2 \to \mathbf{P}^1$.
Next, consider $U = (\text{SL}_2)^\infty$ with its canonical (product) morphism
$U \to F$. Note that $U$ is an affine scheme. We claim the morphism
$U \to F$ is flat, surjective, and universally open. Namely, suppose
$f : T \to F$ is a morphism. Then $Z = T \times_F U$ is the infinite
fibre product of the schemes $Z_i = T \times_{f_i, \mathbf{P}^1} \text{SL}_2$
over $T$. Each of the morphisms $Z_i \to T$ is surjective smooth and
affine which implies that
$$
Z = Z_1 \times_T Z_2 \times_T Z_3 \times_T \ldots
$$
is a scheme flat and affine over $Z$. A simple limit argument shows that
$Z \to T$ is open as well.

\medskip\noindent
On the other hand, we claim that $F$ isn't an algebraic space.
Namely, if $F$ where an algebraic space it would be a quasi-compact
and separated (by our description of fibre products over $F$) algebraic
space. Hence cohomology of quasi-coherent sheaves would vanish above a
certain cutoff (see
Cohomology of Spaces, Proposition \ref{spaces-cohomology-proposition-vanishing}
and remarks preceding it). But clearly by taking the pullback
of $\mathcal{O}(-2, -2, \ldots, -2)$ under the projection
$$
(\mathbf{P}^1)^\infty \longrightarrow (\mathbf{P}^1)^n
$$
(which has a section) we can obtain a quasi-coherent sheaf whose cohomology
is nonzero in degree $n$. Altogether we obtain an answer to a question asked
by Anton Geraschenko on mathoverflow.

\begin{lemma}
\label{lemma-not-algebraic}
There exists a functor $F : \Sch^{opp} \to \textit{Sets}$
which satisfies the sheaf condition for the fpqc topology, has representable
diagonal $\Delta : F \to F \times F$, and such that there exists a
surjective, flat, universally open, quasi-compact morphism
$U \to F$ where $U$ is a scheme, but such that $F$ is not an algebraic space.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}








\section{Sheaves and specializations}
\label{section-sheaves}

\noindent
In the following we fix a big \'etale site $\Sch_\etale$
as constructed in
Topologies, Definition \ref{topologies-definition-big-etale-site}.
Moreover, a scheme will be an object of this site.
Recall that if $x, x'$ are points of a scheme $X$ we say $x$ is a
{\it specialization} of $x'$ or we write $x' \leadsto x$ if
$x \in \overline{\{x'\}}$. This is true in particular if $x = x'$.

\medskip\noindent
Consider the functor $F : \Sch_\etale \to \textit{Ab}$
defined by the following rules:
$$
F(X) = \prod\nolimits_{x \in X}
\prod\nolimits_{x' \in X, x' \leadsto x, x' \not = x}
\mathbf{Z}/2\mathbf{Z}
$$
Given a scheme $X$ we denote $|X|$ the underlying set of points.
An element $a \in F(X)$ will be viewed as a map of sets
$|X| \times |X| \to \mathbf{Z}/2\mathbf{Z}$, $(x, x') \mapsto a(x, x')$
which is zero if $x = x'$ or if $x$ is not a specialization of $x'$.
Given a morphism of schemes $f : X \to Y$ we define
$$
F(f) : F(Y) \longrightarrow F(X)
$$
by the rule that for $b \in F(Y)$ we set
$$
F(f)(b)(x, x') =
\left\{
\begin{matrix}
0 & \text{if }x\text{ is not a specialization of }x' \\
b(f(x), f(x')) & \text{else.}
\end{matrix}
\right.
$$
Note that this really does define an element of $F(X)$. We claim that if
$f : X \to Y$ and $g : Y \to Z$ are composable morphisms then
$F(f) \circ F(g) = F(g \circ f)$. Namely, let $c \in F(Z)$ and let
$x' \leadsto x$ be a specialization of points in $X$, then
$$
F(g \circ f)(x, x') = c(g(f(x)), g(f(x'))) = F(g)(F(f)(c))(x, x')
$$
because $f(x') \leadsto f(x)$. (This also works if $f(x) = f(x')$.)

\medskip\noindent
Let $G$ be the sheafification of $F$ in the \'etale topology.

\medskip\noindent
I claim that if $X$ is a scheme and $x' \leadsto x$ is a specialization
and $x' \not = x$, then $G(X) \not = 0$. Namely, let $a \in F(X)$ be
an element such that when we think of $a$ as a function
$|X| \times |X| \to \mathbf{Z}/2\mathbf{Z}$ it is nonzero at $(x, x')$.
Let $\{f_i : U_i \to X\}$ be an \'etale covering of $X$. Then we can pick an
$i$ and a point $u_i \in U_i$ with $f_i(u_i) = x$. Since generalizations
lift along flat morphisms (see
Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat})
we can find a specialization $u'_i \leadsto u_i$
with $f_i(u'_i) = x'$. By our construction above we see that
$F(f_i)(a) \not = 0$. Hence $a$ determines a nonzero element of $G(X)$.

\medskip\noindent
Note that if $X = \Spec(k)$ where $k$ is a field (or more generally
a ring all of whose prime ideals are maximal), then $F(X) = 0$
and for every \'etale morphism $U \to X$ we have $F(U) = 0$ because there
are no specializations between distinct points in fibres of an \'etale
morphism. Hence $G(X) = 0$.

\medskip\noindent
Suppose that $X \subset X'$ is a thickening, see
More on Morphisms, Definition \ref{more-morphisms-definition-thickening}.
Then the category of schemes \'etale over $X'$ is equivalent to the
category of schemes \'etale over $X$ by the base change functor
$U' \mapsto U = U' \times_{X'} X$, see
\'Etale Cohomology,
Theorem \ref{etale-cohomology-theorem-topological-invariance}.
Since it is always the case that $F(U) = F(U')$ in this situation
we see that also $G(X) = G(X')$.

\medskip\noindent
As a variant we can consider the presheaf $F_n$ which associates
to a scheme $X$ the collection of maps
$a : |X|^{n + 1} \to \mathbf{Z}/2\mathbf{Z}$ where $a(x_0, \ldots, x_n)$
is nonzero only if $x_n \leadsto \ldots \leadsto x_0$ is a sequence of
specializations and $x_n \not = x_{n - 1} \not = \ldots \not = x_0$.
Let $G_n$ be the sheaf associated to $F_n$.
In exactly the same way as above one shows that $G_n$ is nonzero
if $\dim(X) \geq n$ and is zero if $\dim(X) < n$.

\begin{lemma}
\label{lemma-sheaf-zero-on-low-dimension}
There exists a sheaf of abelian groups $G$ on
$\Sch_\etale$ with the following properties
\begin{enumerate}
\item $G(X) = 0$ whenever $\dim(X) < n$,
\item $G(X)$ is not zero if $\dim(X) \geq n$, and
\item if $X \subset X'$ is a thickening, then $G(X) = G(X')$.
\end{enumerate}
\end{lemma}

\begin{proof}
See the discussion above.
\end{proof}

\begin{remark}
\label{remark-specialization}
Here are some remarks:
\begin{enumerate}
\item The presheaves $F$ and $F_n$ are separated presheaves.
\item It turns out that $F$, $F_n$ are not sheaves.
\item One can show that $G$, $G_n$ is actually a sheaf for the fppf topology.
\end{enumerate}
We will prove these results if we need them.
\end{remark}


\section{Sheaves and constructible functions}
\label{section-constructible-functions}

\noindent
In the following we fix a big \'etale site $\Sch_\etale$
as constructed in
Topologies, Definition \ref{topologies-definition-big-etale-site}.
Moreover, a scheme will be an object of this site.
A {\it constructible stratification}
of a scheme $X$ is a locally finite disjoint union decomposition
$X = \coprod_{i \in I} X_i$ such that each $X_i \subset X$ is a
locally constructible subset of $X$. Locally finite means that
for any quasi-compact open $U \subset X$ there are only finitely
many $i \in I$ such that $X_i \cap U$ is not empty.
Note that if $f : X \to Y$ is a morphism of schemes and
$Y = \coprod Y_j$ is a constructible stratification, then
$X = \coprod f^{-1}(Y_j)$ is a constructible stratification of $X$.
Given a set $S$ and a scheme $X$ a {\it constructible function}
$f : |X| \to S$ is a map such that $X = \coprod_{s \in S} f^{-1}(s)$
is a constructible stratification of $X$.
If $G$ is an (abstract group) and $a, b : |X| \to G$ are constructible
functions, then $ab : |X| \to G, x \mapsto a(x)b(x)$ is a constructible
function too. The reason is that given any two constructible stratifications
there is a third one refining both.

\medskip\noindent
Let $A$ be any abelian group. For any scheme $X$ we define
$$
F(X) =
\frac{\{a : |X| \to A \mid a \text{ is a constructible function}\}}{\{\text{locally constant functions }|X| \to A\}}
$$
We think of an element $a$ of $F(X)$ simply as a function well defined
up to adding a locally constant one. Given a morphism of schemes
$f : X \to Y$ and an element $b \in F(Y)$, then we define
$F(f)(b) = b \circ f$. Thus $F$ is a presheaf on $\Sch_\etale$.

\medskip\noindent
Note that if $\{f_i : U_i \to X\}$ is an fppf covering, and $a \in F(X)$
is such that $F(f_i)(a) = 0$ in $F(U_i)$, then $a \circ f_i$ is a locally
constant function for each $i$. This means in turn that $a$ is a locally
constant function as the morphisms $f_i$ are open. Hence $a = 0$ in $F(X)$.
Thus we see that $F$ is a separated presheaf (in the fppf topology
hence a fortiori in the \'etale topology).

\medskip\noindent
Let $G$ be the sheafification of $F$ in the \'etale topology.
Since $F$ is separated, and since $F(X) \not = 0$ for example when
$X$ is the spectrum of a discrete valuation ring, we see that $G$
is not zero.

\medskip\noindent
Let $X = \Spec(k)$ where $k$ is a field.
Then any \'etale covering of $X$ can be dominated
by a covering $\{\Spec(k') \to \Spec(k)\}$ with $k \subset k'$
a finite separable extension of fields. Since $F(\Spec(k')) = 0$
we see that $G(X) = 0$.

\medskip\noindent
Suppose that $X \subset X'$ is a thickening, see
More on Morphisms, Definition \ref{more-morphisms-definition-thickening}.
Then the category of schemes \'etale over $X'$ is equivalent to the
category of schemes \'etale over $X$ by the base change functor
$U' \mapsto U = U' \times_{X'} X$, see
\'Etale Cohomology,
Theorem \ref{etale-cohomology-theorem-topological-invariance}.
Since $F(U) = F(U')$ in this situation we see that also $G(X) = G(X')$.

\medskip\noindent
The sheaf $G$ is limit preserving, see
Limits of Spaces,
Definition \ref{spaces-limits-definition-locally-finite-presentation}.
Namely, let $R$ be a ring which is written as a directed colimit
$R = \colim_i R_i$ of rings. Set $X = \Spec(R)$ and
$X_i = \Spec(R_i)$, so that $X = \lim_i X_i$. Then
$G(X) = \colim_i G(X_i)$. To prove this one first proves that
a constructible stratification of $\Spec(R)$ comes from a constructible
stratifications of some $\Spec(R_i)$. Hence the result for $F$. To
get the result for the sheafification, use that any \'etale ring map
$R \to R'$ comes from an \'etale ring map $R_i \to R_i'$ for some $i$.
Details omitted.

\begin{lemma}
\label{lemma-weird-sheaf}
There exists a sheaf of abelian groups $G$ on
$\Sch_\etale$ with the following properties
\begin{enumerate}
\item $G(\Spec(k)) = 0$ whenever $k$ is a field,
\item $G$ is limit preserving,
\item if $X \subset X'$ is a thickening, then $G(X) = G(X')$, and
\item $G$ is not zero.
\end{enumerate}
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}




\section{The lisse-\'etale site is not functorial}
\label{section-lisse-etale-not-functorial}

\noindent
The {\it lisse-\'etale} site
$X_{lisse,\etale}$ of $X$ is the category of schemes smooth over $X$
endowed with (usual) \'etale coverings, see
Cohomology of Stacks, Section \ref{stacks-cohomology-section-lisse-etale}.
Let $f : X  \to Y$ be a morphism of schemes.
There is a functor
$$
u : Y_{lisse,\etale} \longrightarrow X_{lisse,\etale},\quad
V/Y \longmapsto V \times_Y X
$$
which is continuous. Hence we obtain an adjoint pair of functors
$$
u^s :
\Sh(X_{lisse,\etale})
\longrightarrow
\Sh(Y_{lisse,\etale}),
\quad
u_s :
\Sh(Y_{lisse,\etale})
\longrightarrow
\Sh(X_{lisse,\etale}),
$$
see Sites, Section \ref{sites-section-continuous-functors}.
We claim that, in general, $u$ does {\bf not} define a morphism of sites, see
Sites, Definition \ref{sites-definition-morphism-sites}.
In other words, we claim that $u_s$ is not left exact in general. Note that
representable presheaves are sheaves on lisse-\'etale sites. Hence, by
Sites, Lemma \ref{sites-lemma-pullback-representable-sheaf}
we see that $u_sh_V = h_{V \times_Y X}$. Now consider two morphisms
$$
\xymatrix{
V_1 \ar[rd] \ar@<1ex>[rr]^a \ar@<-1ex>[rr]_b & & V_2 \ar[ld] \\
& Y
}
$$
of schemes $V_1, V_2$ smooth over $Y$. Now if $u_s$ is left exact, then
we would have
$$
u_s \text{Equalizer}(h_a, h_b : h_{V_1} \to h_{V_2})
=
\text{Equalizer}(h_{a \times 1}, h_{b \times 1} :
h_{V_1 \times_Y X} \to h_{V_2 \times_Y X})
$$
We will take the morphisms $a, b : V_1 \to V_2$ such that there exists
no morphism from a scheme smooth over $Y$ into $(a = b) \subset V_1$, i.e.,
such that the left hand side is the empty sheaf, but such that after
base change to $X$ the equalizer is nonempty and smooth over $X$.
A silly example is to take $X = \Spec(\mathbf{F}_p)$,
$Y = \Spec(\mathbf{Z})$ and $V_1 = V_2 = \mathbf{A}^1_\mathbf{Z}$
with morphisms $a(x) = x$ and $b(x) = x + p$. Note that the equalizer
of $a$ and $b$ is the fibre of $\mathbf{A}^1_\mathbf{Z}$ over $(p)$.

\begin{lemma}
\label{lemma-lisse-etale-not-functorial}
The lisse-\'etale site is not functorial, even for morphisms of schemes.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}





\section{Derived pushforward of quasi-coherent modules}
\label{section-derived-push-quasi-coherent}

\noindent
Let $k$ be a field of characteristic $p > 0$. Let $S = \Spec(k[x])$.
Let $G = \mathbf{Z}/p\mathbf{Z}$ viewed either as an abstract group
or as a constant group scheme over $S$. Consider the algebraic stack
$\mathcal{X} = [S/G]$ where $G$ acts trivially on $S$, see
Examples of Stacks, Remark \ref{examples-stacks-remark-X-mod-G-group}
and
Criteria for Representability, Lemma \ref{criteria-lemma-BG-algebraic}.
Consider the structure morphism
$$
f : \mathcal{X} \longrightarrow S
$$
This morphism is quasi-compact and quasi-separated. Hence we get a functor
$$
Rf_{\textit{QCoh}, *} :
D^{+}_{\textit{QCoh}}(\mathcal{O}_\mathcal{X})
\longrightarrow
D^{+}_{\textit{QCoh}}(\mathcal{O}_S),
$$
see 
Derived Categories of Stacks, Proposition
\ref{stacks-perfect-proposition-derived-direct-image-quasi-coherent}.
Let's compute $Rf_{\textit{QCoh}, *}\mathcal{O}_\mathcal{X}$.
Since $D_{\textit{QCoh}}(\mathcal{O}_S)$ is equivalent to
the derived category of $k[x]$-modules (see
Derived Categories of Schemes, Lemma
\ref{perfect-lemma-affine-compare-bounded})
this is equivalent to computing
$R\Gamma(\mathcal{X}, \mathcal{O}_\mathcal{X})$.
For this we can use the covering $S \to \mathcal{X}$ and the spectral
sequence
$$
H^q(S \times_\mathcal{X} \ldots \times_\mathcal{X} S, O)
\Rightarrow H^{p + q}(\mathcal{X}, \mathcal{O}_\mathcal{X})
$$
see
Cohomology of Stacks, Proposition
\ref{stacks-cohomology-proposition-smooth-covering-compute-cohomology}.
Note that
$$
S \times_\mathcal{X} \ldots \times_\mathcal{X} S = S \times G^p
$$
which is affine. Thus the complex
$$
k[x] \to \text{Map}(G, k[x]) \to \text{Map}(G^2, k[x]) \to \ldots
$$
computes $R\Gamma(\mathcal{X}, \mathcal{O}_\mathcal{X})$.
Here for $\varphi \in \text{Map}(G^{p - 1}, k[x])$ its differential is
the map which sends $(g_1, \ldots, g_p)$ to
$$
\varphi(g_2, \ldots, g_p) +
\sum\nolimits_{i = 1}^{p - 1}
(-1)^i\varphi(g_1, \ldots, g_i + g_{i + 1}, \ldots, g_p)
+ (-1)^p\varphi(g_1, \ldots, g_{p - 1}).
$$
This is just the complex computing the group cohomology of $G$ acting
trivially on $k[x]$ (insert future reference here). The cohomology of
the cyclic group $G$ on $k[x]$ is exactly one copy of $k[x]$ in each
cohomological degree $\geq 0$ (insert future reference here). We conclude
that
$$
Rf_*\mathcal{O}_\mathcal{X} = \bigoplus\nolimits_{n \geq 0} \mathcal{O}_S[-n]
$$
Now, consider the complex
$$
E = \bigoplus\nolimits_{m \geq 0} \mathcal{O}_\mathcal{X}[m]
$$
This is an object of $D_{\textit{QCoh}}(\mathcal{X})$. We interrupt the
discussion for a general result.

\begin{lemma}
\label{lemma-is-limit}
Let $\mathcal{X}$ be an algebraic stack. Let $K$ be an object of
$D(\mathcal{O}_\mathcal{X})$ whose cohomology sheaves are locally
quasi-coherent (Sheaves on Stacks, Definition
\ref{stacks-sheaves-definition-locally-quasi-coherent})
and satisfy the flat base change property (Cohomology of Stacks,
Definition \ref{stacks-cohomology-definition-flat-base-change}).
Then there exists a distinguished triangle
$$
K \to
\prod\nolimits_{n \geq 0} \tau_{\geq -n} K \to
\prod\nolimits_{n \geq 0} \tau_{\geq -n} K \to K[1]
$$
in $D(\mathcal{O}_\mathcal{X})$. In other words, $K$ is the derived
limit of its canonical truncations.
\end{lemma}

\begin{proof}
Recall that we work on the ``big fppf site'' $\mathcal{X}_{fppf}$
of $\mathcal{X}$ (by our conventions
for sheaves of $\mathcal{O}_\mathcal{X}$-modules in the chapters
Sheaves on Stacks and Cohomology on Stacks). Let $\mathcal{B}$ be the set
of objects $x$ of $\mathcal{X}_{fppf}$ which lie over an affine scheme $U$. 
Combining
Sheaves on Stacks, Lemmas
\ref{stacks-sheaves-lemma-compare-fppf-etale},
\ref{stacks-sheaves-lemma-cohomology-restriction},
Descent, Lemma \ref{descent-lemma-quasi-coherent-and-flat-base-change},
and
Cohomology of Schemes, Lemma
\ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}
we see that $H^p(x, \mathcal{F}) = 0$ if $\mathcal{F}$ is
locally quasi-coherent and $x \in \mathcal{B}$.
Now the claim follows from
Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-is-limit}.
\end{proof}

\begin{lemma}
\label{lemma-sum-is-product}
Let $\mathcal{X}$ be an algebraic stack. If $\mathcal{F}_n$ is a collection
of locally quasi-coherent sheaves with the flat base change property on
$\mathcal{X}$, then $\oplus_n \mathcal{F}_n[n] \to \prod_n \mathcal{F}_n[n]$
is an isomorphism in $D(\mathcal{O}_\mathcal{X})$.
\end{lemma}

\begin{proof}
This is true because by Lemma \ref{lemma-is-limit} we see that the direct sum
is isomorphic to the product.
\end{proof}

\noindent
We continue our discussion. Since a quasi-coherent module is locally
quasi-coherent and satisfies the flat base change property
(Sheaves on Stacks, Lemma \ref{stacks-sheaves-lemma-quasi-coherent}) we get
$$
E = \prod\nolimits_{m \geq 0} \mathcal{O}_\mathcal{X}[m]
$$
Since cohomology commutes
with limits we see that
$$
Rf_*E = \prod\nolimits_{m \geq 0}
\left(\bigoplus\nolimits_{n \geq 0} \mathcal{O}_S[m - n]\right)
$$
Note that this complex is not an object of $D_{\textit{QCoh}}(\mathcal{O}_S)$
because the cohomology sheaf in degree $0$ is an infinite product of copies
of $\mathcal{O}_S$ which is not even a locally quasi-coherent
$\mathcal{O}_S$-module.

\begin{lemma}
\label{lemma-push-not-OK}
A quasi-compact and quasi-separated morphism
$f : \mathcal{X} \to \mathcal{Y}$ of algebraic stacks
need not induce a functor
$Rf_* : D_{\textit{QCoh}}(\mathcal{O}_\mathcal{X}) \to
D_{\textit{QCoh}}(\mathcal{O}_\mathcal{Y})$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



\section{A big abelian category}
\label{section-big}

\noindent
The purpose of this section is to give an example of a ``big'' abelian category
$\mathcal{A}$ and objects $M, N$ such that the collection of isomorphism
classes of extensions $\text{Ext}_\mathcal{A}(M, N)$ is not a set.
The example is due to Freyd, see \cite[page 131, Exercise A]{Freyd}.

\medskip\noindent
We define $\mathcal{A}$ as follows. An object of $\mathcal{A}$
consists of a triple $(M, \alpha, f)$ where $M$ is an abelian group and
$\alpha$ is an ordinal and $f : \alpha \to \text{End}(M)$ is a map.
A morphism $(M, \alpha, f) \to (M', \alpha', f')$ is given by a homomorphism
of abelian groups $\varphi : M \to M'$ such that for {\it any} ordinal $\beta$
we have
$$
\varphi \circ f(\beta) = f'(\beta) \circ \varphi
$$
Here the rule is that we set $f(\beta) = 0$ if $\beta$ is not in
$\alpha$ and similarly we set $f'(\beta)$ equal to zero if $\beta$ is not an
element of $\alpha'$. We omit the verification that the category so defined
is abelian.

\medskip\noindent
Consider the object $Z = (\mathbf{Z}, \emptyset, f)$, i.e., all the
operators are zero. The observation is that computed in
$\mathcal{A}$ the group $\text{Ext}^1_\mathcal{A}(Z, Z)$ is a proper class
and not a set. Namely, for each ordinal $\alpha$ we can find an extension
$(M, \alpha + 1, f)$ of $Z$ by $Z$ whose underlying group is
$M = \mathbf{Z} \oplus \mathbf{Z}$ and where the value of $f$ is
always zero except for
$$
f(\alpha) =
\left(
\begin{matrix}
0 & 1 \\
0 & 0
\end{matrix}
\right).
$$
This clearly produces a proper class of isomorphism classes of extensions.
In particular, the derived category of $\mathcal{A}$ has proper classes for
its collections of morphism, see
Derived Categories, Lemma \ref{derived-lemma-ext-1}. This means that some
care has to be exercised when defining Verdier quotients of triangulated
categories.

\begin{lemma}
\label{lemma-big-abelian-category}
There exists a ``big'' abelian category $\mathcal{A}$ whose
$\text{Ext}$-groups are proper classes.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}





\section{Weakly associated points and scheme theoretic density}
\label{section-weak-ass-dense}

\noindent
Let $k$ be a field. Let $R = k[z, x_i, y_i]/(z^2, zx_iy_i)$ where
$i$ runs over the elements of $\mathbf{N}$. Note that $R = R_0 \oplus M_0$ where
$R_0 = k[x_i, y_i]$ is a subring and $M_0$ is an ideal of square zero
with $M_0 \cong R_0/(x_iy_i)$ as $R_0$-module. The prime
$\mathfrak p = (z, x_i)$ is weakly associated to $R$ as an $R$-module
(Algebra, Definition \ref{algebra-definition-weakly-associated}).
Indeed, the element $z$ in $R_\mathfrak p$ is nonzero but annihilated
by $\mathfrak pR_\mathfrak p$. On the other hand, consider the open
subscheme
$$
U = \bigcup D(x_i) \subset \Spec(R) = S
$$
We claim that $U \subset S$ is scheme theoretically dense
(Morphisms, Definition \ref{morphisms-definition-scheme-theoretically-dense}).
To prove this it suffices to show that $\mathcal{O}_S \to j_*\mathcal{O}_U$
is injective where $j : U \to S$ is the inclusion morphism, see 
Morphisms, Lemma \ref{morphisms-lemma-characterize-scheme-theoretically-dense}.
Translated back into algebra, we have to show that for all $g \in R$
the map
$$
R_g \longrightarrow \prod R_{x_ig}
$$
is injective. Write $g = g_0 + m_0$ with $g_0 \in R_0$ and $m_0 \in M_0$.
Then $R_g = R_{g_0}$ (details omitted). Hence we may assume $g \in R_0$.
We may also assume $g$ is not zero. Now $R_g = (R_0)_g \oplus (M_0)_g$.
Since $R_0$ is a domain, the map $(R_0)_g \to \prod (R_0)_{x_ig}$ is
injective. If $g \in (x_iy_i)$ then $(M_0)_g = 0$ and there is nothing
to prove. If $g \not \in (x_iy_i)$ then, since
$(x_iy_i)$ is a radical ideal of $R_0$, we have to show that
$M_0 \to \prod (M_0)_{x_ig}$ is injective. The kernel of
$R_0 \to M_0 \to (M_0)_{x_n}$ is $(x_iy_i, y_n)$. Since
$(x_iy_i, y_n)$ is a radical ideal, if $g \not \in (x_iy_i, y_n)$
then the kernel of $R_0 \to M_0 \to (M_0)_{x_ng}$ is $(x_iy_i, y_n)$.
As $g \not \in (x_iy_i, y_n)$ for all $n \gg 0$ we conclude that
the kernel is contained in $\bigcap_{n \gg 0} (x_iy_i, y_n) = (x_iy_i)$
as desired.

\begin{lemma}
\label{lemma-example-schematically-dense-missing-weakly-associated-point}
There exists a scheme $X$ and a schematically dense open $U \subset X$
such that some weakly associated point $x \in X$ is not in $U$.
\end{lemma}

\begin{proof}
In the situation above we have $\mathfrak p \not \in U$.
\end{proof}



\section{Example of non-additivity of traces}
\label{section-non-additive}

\noindent
Let $k$ be a field and let $R = k[\epsilon]$ be the ring of dual numbers
over $k$. In other words, $R = k[x]/(x^2)$ and $\epsilon$ is the congruence
class of $x$ in $R$. Consider the short exact sequence of complexes
$$
\xymatrix{
0 \ar[d] \ar[r] &
R \ar[d]^\epsilon \ar[r]_1 &
R \ar[d] \\
R \ar[r]^1 & R \ar[r] & 0
}
$$
Here the columns are the complexes, the first row is placed in degree $0$, and
the second row in degree $1$. Denote the first complex (i.e., the
left column) by $A^\bullet$, the second by $B^\bullet$ and the third
$C^\bullet$. We claim that the diagram
\begin{equation}
\label{equation-commutes-up-to-homotopy}
\vcenter{
\xymatrix{
A^\bullet \ar[d]_{1 + \epsilon} \ar[r] &
B^\bullet \ar[r] \ar[d]_1 &
C^\bullet \ar[d]_1 \\
A^\bullet \ar[r] & B^\bullet \ar[r] & C^\bullet
}
}
\end{equation}
commutes in $K(R)$, i.e., is a diagram of complexes commuting up to homotopy.
Namely, the square on the right commutes and the one on the left is off by
the homotopy $1 : A^1 \to B^0$. On the other hand,
$$
\text{Tr}_{A^\bullet}(1 + \epsilon) + \text{Tr}_{C^\bullet}(1)
\not = \text{Tr}_{B^\bullet}(1).
$$

\begin{lemma}
\label{lemma-nonadditivity-of-trace}
There exists a ring $R$, a distringuished triangle
$(K, L, M, \alpha, \beta, \gamma)$ in the homotopy category $K(R)$,
and an endomorphism $(a, b, c)$ of this distinguished triangle, such that
$K$, $L$, $M$ are perfect complexes and
$\text{Tr}_K(a) + \text{Tr}_M(c) \not = \text{Tr}_L(b)$.
\end{lemma}

\begin{proof}
Consider the example above. The map $\gamma : C^\bullet \to A^\bullet[1]$
is given by multiplication by $\epsilon$ in degree $0$, see
Derived Categories, Definition \ref{derived-definition-distinguished-triangle}.
Hence it is also true that
$$
\xymatrix{
C^\bullet \ar[d] \ar[r]_\gamma & A^\bullet[1] \ar[d] \\
C^\bullet \ar[r]^\gamma & A^\bullet[1]
}
$$
commutes in $K(R)$ as $\epsilon(1 + \epsilon) = \epsilon$.
Thus we indeed have a morphism of distinguished triangles.
\end{proof}






\section{Being projective is not local on the base}
\label{section-non-descending-property-projective}

\noindent
In the chapter on descent we have seen that many properties of morphisms
are local on the base, even in the fpqc topology. See
Descent, Sections \ref{descent-section-descending-properties-morphisms},
\ref{descent-section-descending-properties-morphisms-fpqc}, and
\ref{descent-section-descending-properties-morphisms-fppf}.
This is not true for projectivity of morphisms.

\begin{lemma}
\label{lemma-non-descending-property-projective}
The properties
\begin{enumerate}
\item[] $\mathcal{P}(f) =$``$f$ is projective'', and
\item[] $\mathcal{P}(f) =$``$f$ is quasi-projective''
\end{enumerate}
are not Zariski local on the base. A fortiori, they are not fpqc local
on the base.
\end{lemma}

\begin{proof}
Following Hironaka \cite[Example B.3.4.1]{H},
we define a proper morphism of smooth complex 3-folds
$f:V_Y\to Y$
which is Zariski-locally projective, but not projective. Since $f$ is proper
and not projective, it is also not quasi-projective.

\medskip\noindent
Let $Y$ be projective 3-space over the complex numbers ${\mathbf C}$.
Let $C$ and $D$ be smooth conics in $Y$ such
that the closed subscheme $C\cap D$ is reduced and consists
of two complex points $P$ and $Q$. (For example,
let $C=\{ [x,y,z,w]: xy=z^2, w=0\}$, $D=\{ [x,y,z,w]:
xy=w^2, z=0\}$, $P=[1,0,0,0]$,
and $Q=[0,1,0,0]$.) 
On $Y-Q$, first blow up the curve $C$, and then blow
up the strict transform of the curve $D$ (Divisors, Definition 
\ref{divisors-definition-strict-transform}). On $Y-P$, first blow up
the curve $D$, and then blow up the strict transform of the curve
$C$. Over $Y-P-Q$, the two varieties we have constructed are canonically
isomorphic, and so we can glue them over $Y-P-Q$. The result
is a smooth proper 3-fold $V_Y$ over ${\mathbf C}$. The morphism
$f:V_Y\to Y$ is proper and Zariski-locally projective (since
it is a blow-up over $Y-P$ and over $Y-Q$), by Divisors,
Lemma \ref{divisors-lemma-blowing-up-projective}. We will show that
$V_Y$ is not projective over ${\mathbf C}$. That will imply that
$f$ is not projective.

\medskip\noindent
To do this, let $L$ be the inverse image in $V_Y$ of a complex point
of $C-P-Q$, and $M$ the inverse image of a complex point of $D-P-Q$.
Then $L$ and $M$ are isomorphic to the projective line
${\mathbf P}^1_{{\mathbf C}}$. 
Next, let $E$ be the inverse image in $V_Y$ of $C\cup D\subset Y$ in $V_Y$;
thus $E\rightarrow C\cup D$ is a proper morphism, with fibers
isomorphic to ${\mathbf P}^1$ over $(C\cup D)-\{P,Q\}$.
The inverse
image of $P$ in $E$ is a union of two lines $L_0$ and $M_0$, and we have
rational equivalences of cycles $L\sim L_0+M_0$ and $M\sim M_0$ on $E$
(using that $C$ and $D$ are isomorphic to ${\mathbf P}^1$).
Note the asymmetry resulting from the order in which we blew
up the two curves. Near $Q$, the opposite happens. So the inverse image
of $Q$ is the union of two lines $L_0'$ and $M_0'$, and we have
rational equivalences $L\sim L_0'$ and $M\sim L_0'+M_0'$ on $E$.
Combining these equivalences, we find that $L_0+M_0'\sim 0$
on $E$ and hence on $V_Y$. If $V_Y$ were projective over ${\mathbf C}$,
it would have
an ample line bundle $H$, which would have degree $>0$ on all curves
in $V_Y$. In particular $H$ would have positive degree on $L_0+M_0'$,
contradicting that the degree of a line bundle is well-defined
on 1-cycles modulo rational equivalence on a proper scheme
over a field (Chow Homology,
Lemma \ref{chow-lemma-proper-pushforward-rational-equivalence}
and Lemma \ref{chow-lemma-factors}).
So $V_Y$ is not projective over ${\mathbf C}$.
\end{proof}

\noindent
In different terminology, Hironaka's 3-fold $V_Y$ is a small
resolution of the blow-up $Y'$ of $Y$ along the reduced subscheme
$C\cup D$; here $Y'$ has two node singularities. If we define $Z$ by blowing
up $Y$ along $C$ and then along the strict transform
of $D$, then $Z$ is a smooth projective 3-fold, and the non-projective
3-fold $V_Y$ differs from $Z$ by a ``flop'' over $Y-P$.


\section{Descent data for schemes need not be effective, even for a projective morphism}
\label{section-non-effective-descent-projective}

\noindent
In the chapter on descent we have seen that descent data for schemes relative
to an fpqc morphism are effective for several classes
of morphisms. In particular, affine morphisms and more generally
quasi-affine morphisms satisfy descent for fpqc coverings
(Descent, Lemma \ref{descent-lemma-quasi-affine}).
This is not true for projective morphisms.

\begin{lemma}
\label{lemma-non-effective-descent-projective}
There is an etale covering $X\to S$ of schemes and a descent datum
$(V/X,\varphi)$ relative to $X\to S$ such that 
$V\to X$ is projective,
but the descent datum is not effective in the category of schemes.
\end{lemma}

\begin{proof}
We imitate Hironaka's example of a smooth separated complex
algebraic space of dimension 3
which is not a scheme \cite[Example B.3.4.2]{H}.

\medskip\noindent
Consider the action of the group $G = \mathbf{Z}/2 = \{1, g\}$
on projective 3-space
$\mathbf{P}^3$ over the complex numbers by
$$
g[x,y,z,w] = [y,x,w,z].
$$
The action is free outside the two disjoint lines
$L_1=\{ [x,x,z,z]\}$ and $L_2=\{ [x,-x,z,-z]\}$ in
${\mathbf P}^3$. Let $Y={\mathbf P}^3-(L_1\cup L_2)$. There is a
smooth quasi-projective scheme $S=Y/G$ over ${\mathbf C}$ such that
$Y\to S$ is a $G$-torsor (Groupoids,
Definition \ref{groupoids-definition-principal-homogeneous-space}).
Explicitly, we can define $S$ as the image of the open subset $Y$
in ${\mathbf P}^3$ under the morphism
\begin{align*}
{\mathbf P}^3 & \to \text{Proj } {\mathbf C}[x,y,z,w]^G\\
   & = \text{Proj } {\mathbf C}[u_0,u_1,v_0,v_1,v_2]/(v_0v_1=v_2^2),
\end{align*}
where $u_0=x+y$, $u_1=z+w$, $v_0=(x-y)^2$, $v_1=(z-w)^2$,
and $v_2=(x-y)(z-w)$, and the ring is graded with $u_0,u_1$
in degree 1 and $v_0,v_1,v_2$ in degree 2.

\medskip\noindent
Let $C=\{ [x,y,z,w]: xy=z^2, w=0\}$ and $D=\{ [x,y,z,w]:
xy=w^2, z=0\}$. These are smooth conic curves in ${\mathbf P}^3$, contained
in the $G$-invariant open subset $Y$, with $g(C)=D$. Also,
$C\cap D$ consists of the two points $P:=[1,0,0,0]$
and $Q:=[0,1,0,0]$, and these two points are switched by the action
of $G$. 

\medskip\noindent
Let $V_Y\to Y$ be the scheme which over $Y-P$
is defined by blowing up $D$ and then the strict transform
of $C$, and over $Y-Q$ is defined by blowing up $C$ and then
the strict transform of $D$. (This is the same construction
as in the proof of Lemma \ref{lemma-non-descending-property-projective},
except that $Y$ here denotes an open subset of ${\mathbf P}^3$
rather than all of ${\mathbf P}^3$.)
Then the action of $G$ on $Y$ lifts to an action of $G$ on $V_Y$,
which switches the inverse images of $Y-P$ and $Y-Q$. This action
of $G$ on $V_Y$ gives a descent datum $(V_Y/Y,\varphi_Y)$
on $V_Y$ relative to the $G$-torsor
$Y\to S$. The morphism $V_Y\to Y$
is proper but not projective, as shown in the proof
of Lemma \ref{lemma-non-descending-property-projective}.

\medskip\noindent
Let $X$ be the disjoint union of the open subsets $Y-P$ and $Y-Q$;
then we have surjective etale morphisms $X\to Y\to S$.
Let $V$ be the pullback of $V_Y\to Y$ to $X$; then the morphism
$V\to X$ is projective, since $V_Y\to Y$ is a blow-up over each of the open
subsets $Y-P$ and $Y-Q$. Moreover, the descent datum $(V_Y/Y,\varphi_Y)$
pulls back to a descent datum $(V/X,\varphi)$ relative to the
etale covering $X\to S$.

\medskip\noindent
Suppose that this descent datum is effective in the category
of schemes. That is, there is a scheme $U\to S$
which pulls back to the morphism $V\to X$ together
with its descent datum. Then $U$ would be the quotient
of $V_Y$ by its $G$-action.
$$
\xymatrix{
V \ar[r]\ar[d]&  X\ar[d] \\
V_Y \ar[r]\ar[d]& Y\ar[d] \\
U \ar[r]& S
}
$$

\medskip\noindent
Let $E$ be the inverse image of $C\cup D\subset Y$ in $V_Y$;
thus $E\rightarrow C\cup D$ is a proper morphism, with fibers
isomorphic to ${\mathbf P}^1$ over $(C\cup D)-\{P,Q\}$.
The inverse image of $P$ in $E$ is a union of two lines $L_0$
and $M_0$. It follows that the inverse image of $Q=g(P)$ in $E$
is the union of two lines $L_0'=g(M_0)$ and $M_0'=g(L_0)$.
As shown in the proof
of Lemma \ref{lemma-non-descending-property-projective},
we have a rational equivalence $L_0+M_0'=L_0+g(L_0)\sim 0$ on $E$.

\medskip\noindent
By descent of closed subschemes, there is a curve $L_1\subset U$
(isomorphic to ${\mathbf P}^1$)
whose inverse image in $V_Y$ is $L_0\cup g(L_0)$. (Use Descent, Lemma
\ref{descent-lemma-affine}, noting that a closed immersion is an affine
morphism.)
Let $R$ be a complex point of $L_1$. Since
we assumed that $U$ is a scheme, we can choose a function
$f$ in the local ring $O_{U,R}$ that vanishes at $R$ but not
on the whole curve $L_1$. Let $D_{\text{loc}}$ be an irreducible component
of the closed subset $\{ f=0\}$ in $\text{Spec }O_{U,R}$; then
$D_{\text{loc}}$ has codimension 1.
The closure of $D_{\text{loc}}$ in $U$ is an irreducible divisor $D_U$
in $U$ which contains the point $R$ but not the whole curve $L_1$.
The inverse image of $D_U$ in $V_Y$ is an effective divisor $D$
which intersects $L_0\cup g(L_0)$ but does not contain either
curve $L_0$ or $g(L_0)$.

\medskip\noindent
Since the complex 3-fold $V_Y$ is smooth, $O(D)$ is a line
bundle on $V_Y$. (We use here that a regular local ring is factorial,
or in other words is a UFD.)
The restriction of $O(D)$ to the proper surface
$E\subset V_Y$ is a line bundle which has positive degree on the 1-cycle
$L_0+g(L_0)$, by our information on $D$. Since
$L_0+g(L_0)\sim 0$ on $E$, this contradicts 
that the degree of a line bundle is well-defined
on 1-cycles modulo rational equivalence on a proper scheme
over a field (Chow Homology,
Lemma \ref{chow-lemma-proper-pushforward-rational-equivalence}
and Lemma \ref{chow-lemma-factors}). Therefore the descent datum
$(V/X,\varphi)$ is in fact not effective; that is, $U$ does not exist
as a scheme.
\end{proof}

\noindent
In this example, the descent datum {\it is }effective in the category
of algebraic spaces. More precisely,
$U$ exists as a smooth separated algebraic space
of dimension 3 over ${\mathbf C}$,
for example by Algebraic Spaces, Lemma \ref{spaces-lemma-quotient}.
Hironaka's 3-fold $U$ is a small resolution of the blow-up $S'$
of the smooth quasi-projective 3-fold $S$ along the irreducible nodal curve
$(C\cup D)/G$; the 3-fold $S'$ has a node singularity. The
other small resolution of $S'$ (differing from $U$ by a ``flop'')
is again an algebraic space which is not a scheme.



\section{Derived base change}
\label{section-derived-base-change}

\noindent
Let $R \to A$ be a ring map. In
More on Algebra, Section \ref{more-algebra-section-derived-base-change}
we construct a derived base change functor
$- \otimes_R^\mathbf{L} A : D(R) \to D(A)$.
Next, let $R \to B$ be a second ring map. Picture
$$
\xymatrix{
B \ar[r] & B \otimes_R A \\
R \ar[u] \ar[r] & A \ar[u]
}
$$
Given a $B$-module $M$ the tensor product $M \otimes_R A$ is a
$B \otimes_R A$-module. In this section we show there does not exist a
``derived base change functor'' $D(B) \to D(B \otimes_R A)$.

\medskip\noindent
Let $k$ be a field. Set $R = k[x, y]$. Set $A = R/(xy)$ and $B = R/(x^2)$.
The object $B \otimes_R^\mathbf{L} A$ in $D(A)$ is represented by
$$
x^2 : A \longrightarrow A
$$
and we have $H^0(B \otimes_R^\mathbf{L} A) = B \otimes_R A$. We claim that
there does not exist an object $E$ of $D(B \otimes_R A)$ mapping to
$B \otimes_R^\mathbf{L} A$ in $D(A)$. Namely, for such an $E$ the module
$H^0(E)$ would be free, hence $E$ would decompose as
$H^0(E)[0] \oplus H^{-1}(E)[1]$. But it is easy to see that
$B \otimes_R^\mathbf{L} A$ is not isomorphic to the sum of its
cohomology groups in $D(A)$.

\begin{lemma}
\label{lemma-no-derived-base-change}
Let $R \to A$ and $R \to B$ be ring maps. In general there does not
exist a functor $T : D(B) \to D(B \otimes_R A)$
of triangulated categories such that a $B$-module $M$ gives an
object $T(M)$ of $D(B \otimes_R A)$ which maps to
$M \otimes_R^\mathbf{L} A$ under the map $D(B \otimes_R A) \to D(A)$.
\end{lemma}

\begin{proof}
See discussion above.
\end{proof}



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