Lectures 03B

iii/alggeom/01_introduction.tex

@@ -142,3 +142,95 @@ \subsection{Limitations of classical algebraic geometry}
 If \( D_\delta = \mathbb V(Y + \delta) \) for \( \delta \in k \), \( C \cap D_\delta \) is two points unless \( \delta = 0 \).
 This breaks a continuity property.
 Therefore, the intersection of two affine varieties is not naturally an affine variety.
+
+\subsection{Spectrum of a ring}
+Let \( A \) be a commutative unital ring.
+\begin{definition}
+    The \emph{Zariski spectrum} of \( A \) is \( \Spec A = \qty{\mathfrak p \trianglelefteq A \text{ prime}} \).
+\end{definition}
+\begin{remark}
+    Given a ring homomorphism \( \varphi : A \to B \), we have an induced map of sets \( \varphi^{-1} : \Spec B \to \Spec A \) given by \( \mathfrak q \mapsto \varphi^{-1}(\mathfrak q) \), as the preimage of a prime ideal is always prime.
+    Note, however, that this property would fail if we only considered maximal ideals, because the preimage of a maximal ideal need not be maximal.
+    % TODO: Example
+
+    Given \( f \in A \) and a point \( \mathfrak p \in \Spec A \), we have an induced \( \overline f \in \faktor{A}{\mathfrak p} \) obtained by taking the quotient.
+    We can think of this operation as `evaluating' an \( f \in A \) at a point \( \mathfrak p \in \Spec A \), with the caveat that the codomain of this evaluation depends on \( \mathfrak p \).
+\end{remark}
+\begin{example}
+    \begin{enumerate}
+        \item Let \( A = \mathbb Z \).
+        Then \( \Spec A = \Spec \mathbb Z \) is the set \( \qty{(p) \mid p \text{ prime}} \cup \qty{(0)} \).
+        Consider an element of \( \mathbb Z \), say, \( 132 \).
+        Given a prime \( p \), we can `evaluate it at \( p \)', giving \( 132 \text{ mod } p \in \faktor{\mathbb Z}{p\mathbb Z} \).
+        Thus \( \Spec Z \) is a space, \( 132 \) is a function on \( \Spec Z \), and \( 132 \text{ mod } p \) is the value of this function at \( p \).
+        \item Let \( A = \mathbb R[X] \).
+        Then \( \Spec A \) is naturally \( \mathbb C \) modulo complex conjugation, together with the zero ideal.
+        \item If \( A = \mathbb C[X] \), then \( \Spec A \) is naturally \( \mathbb C \), together with the zero ideal.
+    \end{enumerate}
+\end{example}
+% TODO: Draw Spec Z[X]. Draw Spec k[X].
+\begin{definition}
+    Let \( f \in A \).
+    Then we define
+    \[ \mathbb V(f) = \qty{\mathfrak p \in \Spec A \mid f = 0 \text{ mod } \mathfrak p \text{, or equivalently, } f \in \mathfrak p} \subseteq \Spec A \]
+    Similarly, for \( J \trianglelefteq A \) an ideal,
+    \[ \mathbb V(J) = \qty{\mathfrak p \in \Spec A \mid \forall f \in J,\, f \in \mathfrak p} = \qty{\mathfrak p \in \Spec A \mid J \subseteq \mathfrak p} \]
+\end{definition}
+\begin{proposition}
+    The sets \( \mathbb V(J) \subseteq \Spec A \) ranging over all ideals \( J \trianglelefteq A \) form the closed sets of a topology.
+\end{proposition}
+This topology is called the \emph{Zariski topology} on \( A \).
+\begin{proof}
+    We have \( \varnothing = \mathbb V(1) \) and \( \Spec A = \mathbb V(0) \), so they are closed.
+    Note that
+    \[ \mathbb V\qty(\sum_\alpha I_\alpha) = \bigcap_\alpha \mathbb V(I_\alpha) \]
+    It remains to show \( \mathbb V(I_1) \cup \mathbb V(I_2) = \mathbb V(I_1 \cap I_2) \).
+    The containment \( \mathbb V(I_1) \cup \mathbb V(I_2) \subseteq \mathbb V(I_1 \cap I_2) \) is clear.
+    Conversely, note \( I_1 I_2 \subseteq I_1 \cap I_2 \).
+    If \( I_1 \cap I_2 \subseteq \mathfrak p \), then by primality of \( \mathfrak p \), either \( I_1 \subseteq \mathfrak p \) or \( I_2 \subseteq \mathfrak p \).
+\end{proof}
+\begin{example}
+    Consider \( \Spec \mathbb C[X, Y] \).
+    The point \( (0) \in \Spec \mathbb C[X, Y] \) is dense in the Zariski topology, so \( \overline{\qty{(0)}} = \Spec \mathbb C[X, Y] \).
+    This is because all prime ideals in integral domains contain the zero ideal.
+    \( (0) \) is sometimes called the \emph{generic point}.
+
+    Consider the prime ideal \( (Y^2 - X^3) \), and consider a maximal ideal \( \mathfrak m_{a,b} = (X - a, Y - b) \) corresponding to the point \( (a, b) \).
+    Then one can show that
+    \[ \mathfrak m_{a,b} \in \overline{\qty{(Y^2 - X^3)}} \iff b^2 = a^3 \]
+    In general, points are not closed.
+\end{example}
+% TODO: lowercase variables?
+
+\subsection{Functions on open sets}
+\begin{definition}
+    Let \( f \in A \).
+    Define the \emph{distinguished open} corresponding to \( f \) to be
+    \[ U_f = \Spec A \setminus \mathbb V(f) \]
+\end{definition}
+\begin{example}
+    \begin{enumerate}
+        \item Let \( A = \mathbb C[x] \), and recall that \( \Spec A \) is \( \mathbb C \cup \qty{(0)} \), where the complex number \( a \) represents the maximal ideal \( (x - a) \).
+        Let \( f = x \) and consider
+        \[ \mathbb V(x) = \qty{\mathfrak p \mid x \in \mathfrak p} = \qty{(x)} \]
+        Hence \( U_x = \Spec A \setminus \qty{(x)} \), which is \( \Spec A \) without the complex number 0.
+        \item More generally, suppose we fix \( a_1, \dots, a_r \in \mathbb C \).
+        Then
+        \[ U = \Spec A \setminus \qty{(x - a_i)}_{i=1}^r = U_f;\quad f = \prod_{i=1}^r (x-a_i) \]
+    \end{enumerate}
+\end{example}
+\begin{lemma}
+    The distinguished opens \( U_f \), taken over all \( f \in A \), form a basis for the Zariski topology on \( \Spec A \); that is, every open set in \( \Spec A \) is a union of some collection of the \( U_f \).
+\end{lemma}
+% Proof: ex sheet 1
+
+\subsection{Localisations}
+\begin{definition}
+    Let \( f \in A \).
+    The \emph{localisation} of \( A \) at \( f \) is
+    \[ A_f = \faktor{A[x]}{(xf - 1)} \]
+\end{definition}
+Informally, we adjoin \( \frac{1}{f} \) to \( A \).
+\begin{lemma}
+    The distinguished open \( U_f \subseteq \Spec A \) is naturally homeomorphic to \( \Spec A_f \).
+\end{lemma}

iii/cat/01_definitions_and_examples.tex

@@ -207,3 +207,114 @@ \subsection{Natural transformations}
         This is a natural transformation \( \pi_n \to H_n U \) where \( U \) is the forgetful functor \( \mathbf{Top}_\star \to \mathbf{Top} \).
     \end{enumerate}
 \end{example}
+
+\subsection{Equivalence of categories}
+There is a notion of isomorphism of categories, namely, isomorphism in the category \( \mathbf{Cat} \).
+For example, \( \mathbf{Rel} \cong \mathbf{Rel}^\cop \) via the functor
+\[ A \mapsto A;\quad R \mapsto R^\circ = \qty{(b, a) \mid (a, b) \in R} \]
+However, there is a weaker notion that is often more useful in practice, called equivalence.
+To define this, we need a notion of `natural isomorphism'.
+There are two obvious definitions, which we show are equivalent.
+\begin{lemma}
+    Let \( \alpha : F \to G \) be a natural transformation between functors \( \mathcal C \rightrightarrows \mathcal D \).
+    Then \( \alpha \) is an isomorphism in the functor category \( [\mathcal C, \mathcal D] \) if and only if each component \( \alpha_A \) is an isomorphism in \( \mathcal D \).
+\end{lemma}
+\begin{proof}
+    The forward direction is clear as composition in \( [\mathcal C, \mathcal D] \) is pointwise; if \( \beta \) is an inverse for \( \alpha \), then \( \beta_A \) is an inverse for \( \alpha_A \).
+    Suppose \( \beta_A \) is an inverse for \( \alpha_A \) for each \( A \).
+    We show the \( \beta \) collectively form a natural transformation by verifying the naturality squares.
+    Given \( f : A \to B \) in \( \mathcal C \), consider
+    \[\begin{tikzcd}
+        FA & FB \\
+        GA & GB
+        \arrow["Ff", from=1-1, to=1-2]
+        \arrow["{\alpha_A}", shift left=2, from=1-1, to=2-1]
+        \arrow["Gf"', from=2-1, to=2-2]
+        \arrow["{\alpha_B}"', shift right=2, from=1-2, to=2-2]
+        \arrow["{\beta_B}"', shift right=2, from=2-2, to=1-2]
+        \arrow["{\beta_A}", shift left=2, from=2-1, to=1-1]
+    \end{tikzcd}\]
+    Then
+    \[ (Ff)\beta_A = \beta_B \alpha_B (Ff) \beta_A = \beta_B(Gf) \alpha_A \beta_A = \beta_B (Gf) \]
+    using naturality of \( \alpha \).
+    Thus \( \beta \) is natural, and an inverse for \( \alpha \).
+\end{proof}
+\begin{definition}
+    Let \( \mathcal C, \mathcal D \) be categories.
+    An \emph{equivalence} between \( \mathcal C \) and \( \mathcal D \) is a pair of functors
+    \[ F : \mathcal C \to \mathcal D;\quad G : \mathcal D \to \mathcal C \]
+    and a pair of natural isomorphisms
+    \[ \alpha : 1_{\mathcal C} \to GF;\quad \beta : FG \to 1_{\mathcal D} \]
+    If \( \mathcal C \) and \( \mathcal D \) are equivalent, we write \( \mathcal C \simeq \mathcal D \).
+\end{definition}
+The reason the natural isomorphisms point in opposite directions will be clarified later.
+A property \( P \) of categories that is called \emph{categorical} if whenever \( \mathcal C \) satisfies \( P \) and \( \mathcal C \simeq \mathcal D \), then \( \mathcal D \) satisfies \( P \).
+For example, the properties of being a preorder or being a groupoid are categorical.
+Being a partial order or being a group are not categorical.
+Generally, properties that rely on equality of objects, not isomorphism, will not be categorical.
+\begin{example}
+    \begin{enumerate}
+        \item Let \( \mathbf{Set}_\star \) be the category of pointed sets and functions preserving the base point.
+        Then \( \mathbf{Set}_\star \simeq \mathbf{Part} \) by
+        \[ F : \mathbf{Set}_\star \to \mathbf{Part};\quad F(A, a) = A \setminus \qty{a};\quad F((A, a) \xrightarrow f (B, b))(x) = f(x) \]
+        and
+        \[ G : \mathbf{Part} \to \mathbf{Set}_\star;\quad G(A) = A \cup \qty{A};\quad G(A \xrightarrow f B \text{ partial})(x) = \begin{cases}
+            f(x) & \text{if } f \text{ is defined at } x \\
+            V & \text{otherwise}
+        \end{cases} \]
+        Note that \( FG = 1_{\mathbf{Part}} \), but \( GF \) is not equal to \( 1_{\mathbf{Set}_\star} \).
+        It is not possible for these two categories to be isomorphic, because there is an isomorphism class of \( \mathbf{Part} \) that has only one member, namely \( \qty{\varnothing} \), but this cannot occur in \( \mathbf{Set}_\star \).
+        \item Let \( \mathbf{fdVect}_k \) be the category of finite-dimensional vector spaces over \( k \).
+        This category is equivalent to its opposite category \( \mathbf{fdVect}_k^\cop \) via the dual space functors in both directions.
+        The natural isomorphisms \( \alpha \) and \( \beta \) are both as in the double dual example given above.
+        \item We show \( \mathbf{fdVect}_k \simeq \mathbf{Mat}_k \).
+        Define
+        \[ F : \mathbf{Mat}_k \to \mathbf{fdVect}_k;\quad F(n) = k^n \]
+        and sending a matrix \( A \) to the linear map it represents in the standard basis.
+        For each finite-dimensional vector space \( V \), choose a particular basis.
+        Define
+        \[ G : \mathbf{fdVect}_k \to \mathbf{Mat}_k;\quad G(V) = \dim V \]
+        and let \( G(\theta) \) be the matrix representing \( \theta \) with respect to the particular bases chosen above.
+        Then \( GF = 1_{\mathbf{Mat}_k} \), as long as we chose the bases above in such a way that the \( k^n \) have the standard basis.
+        Further, \( FG \) is naturally isomorphic to \( 1_{\mathbf{fdVect}_k} \), since the chosen bases define isomorphisms \( k^{\dim V} \to V \), which are natural in \( V \).
+    \end{enumerate}
+\end{example}
+In line with the idea that we do not want to consider equality of objects but only equality of morphisms, we make the following definitions.
+\begin{definition}
+    Let \( F : \mathcal C \to \mathcal D \) be a functor.
+    We say that \( F \) is
+    \begin{enumerate}
+        \item \emph{faithful}, if for each \( f, g \in \mor \mathcal C \) with equal domain and codomain, \( Ff = Fg \) implies \( f = g \);
+        \item \emph{full}, if for each \( FA \xrightarrow g FB \), there exists a morphism \( A \xrightarrow f B \) such that \( Ff = g \);
+        \item \emph{essentially surjective}, if every \( B \in \ob \mathcal D \) is isomorphic to some \( FA \) for \( A \in \ob \mathcal C \).
+    \end{enumerate}
+\end{definition}
+Note that if \( F \) is full and faithful, it is \emph{essentially injective}: if \( FA \xrightarrow g FB \) is an isomorphism, the unique \( A \xrightarrow f B \) with \( Ff = g \) is an isomorphism, because its inverse is the unique \( B \to A \) mapped to \( g^{-1} \).
+\begin{lemma}
+    Let \( F : \mathcal C \to \mathcal D \) be a functor.
+    Then \( F \) is part of an equivalence \( \mathcal C \simeq \mathcal D \) if and only if \( F \) is full, faithful, and essentially surjective.
+\end{lemma}
+\begin{proof}
+    Suppose \( G, \alpha, \beta \) make \( F \) into an equivalence.
+    The existence of \( \beta \) ensures that \( B \simeq FGB \) for any \( B \in \ob \mathcal D \), giving essential surjectivity.
+    For faithfulness, for any \( A \xrightarrow f B \) in \( \mathcal C \), we have \( f = \alpha_B^{-1} (GFf) \alpha_A \), allowing us to reproduce \( f \) from its domain, codomain, and image under \( F \).
+    For fullness, consider \( FA \xrightarrow g FB \), and define \( f = \alpha_B^{-1} (Gg) \alpha_A : A \to B \).
+    Then, \( GFf = Gg \).
+    As \( G \) is faithful by symmetry, \( Ff = g \).
+
+    For the converse, for each object \( B \in \mathcal D \), we choose an isomorphism \( \beta_B : FGB \to B \), thus defining the action of \( G \) on objects.
+    Then we define \( G \) on morphisms by letting \( G(B \xrightarrow g C) \) be the unique \( GB \to GC \) whose image under \( F \) is
+    \[ FGB \xrightarrow{\beta_B} B \xrightarrow g C \xrightarrow{\beta_C^{-1}} FGC \]
+    This is functorial: given \( h : C \to D \), we can form \( G(hg) \) and \( (Gh)(Gg) \) which have the same image under \( F \), so must be equal.
+    By construction, \( \beta \) is a natural isomorphism \( FG \to 1_{\mathcal D} \).
+    It suffices to construct the natural isomorphism \( \alpha : 1_{\mathcal C} \to GF \).
+    Its component at \( A \), is the unique isomorphism whose image under \( F \) is
+    \[ FA \xrightarrow{\beta_{FA}} FGFA \]
+    The naturality squares for \( \alpha \) are mapped by \( F \) to naturality squares for \( \beta^{-1} \), so they commute.
+\end{proof}
+We call a subcategory full if its inclusion functor is full.
+\begin{definition}
+    A category of \emph{skeletal} if every isomorphism class has a single member.
+    A \emph{skeleton} of \( \mathcal C \) is a full subcategory \( \mathcal C' \) containing exactly one object for each isomorphism class.
+\end{definition}
+Note that an equivalence of skeletal categories is bijective on objects, and hence is an isomorphism of categories.