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Lectures 08A
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zeramorphic committed Oct 21, 2023
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73 changes: 73 additions & 0 deletions iii/commalg/02_tensor_products.tex
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Expand Up @@ -1132,3 +1132,76 @@ \subsection{Flat modules}
The map \( \id_M \otimes g \) is injective as \( M \) is flat, so the map \( \id_{S \otimes_R M} \otimes g \) is also injective.
Thus \( S \otimes_R M \) is a flat \( S \)-module.
\end{proof}
We now explore some further examples of tensor products.
\begin{example}
Consider \( \mathbb Q \otimes_{\mathbb Z} \faktor{\mathbb Z}{n\mathbb Z} \).
In this ring,
\[ x \otimes y = n \cdot \frac{x}{n} \otimes y = \frac{x}{n} \otimes ny = \frac{x}{n} \otimes 0 = 0 \]
So this ring is trivial.
To prove this, we used the fact that for all \( x \in \mathbb Q \) and \( n \geq 1 \), there is an element \( y \in \mathbb Q \) such that \( ny = x \).
We say that \( \mathbb Q \) is a \emph{divisible group}.
We also needed the fact that \( \faktor{\mathbb Z}{n\mathbb Z} \) is a \emph{torsion group}: all elements are of finite order.
Hence the tensor product of a divisible group with a torsion group is zero.
In particular, it follows that
\[ \faktor{\mathbb Q}{\mathbb Z} \otimes_{\mathbb Z} \faktor{\mathbb Q}{\mathbb Z} = 0 \]
However, for an \( R \)-module \( M \neq 0 \), if \( M \) is finitely generated then \( M \otimes_R M \neq 0 \).
\end{example}
\begin{example}
Let \( V \) be a vector space over \( \mathbb Q \).
Then \( \mathbb Q \otimes_{\mathbb Q} V \cong V \) as \( \mathbb Q \)-modules, given by the map \( x \otimes v \mapsto xv \).
However, \( \mathbb Q \otimes_{\mathbb Z} V \) is also isomorphic to \( V \), given by the same map.
First, note that every tensor in \( \mathbb Q \otimes_{\mathbb Z} V \) is pure.
\[ \sum \frac{a_i}{b_i} \otimes v_i = \sum \frac{1}{b_i} \otimes a_i v_i = \sum \frac{1}{b_i} \otimes b_i \frac{a_i}{b_i} v_i = \sum 1 \otimes \frac{a_i}{b_i} v_i = 1 \otimes \sum \frac{a_i}{b_i} v_i \]
Surjectivity of the map is clear as \( 1 \otimes v \to v \).
We check injectivity on pure tensors.
If \( xv = 0 \), then \( x = 0 \) or \( v = 0 \), and in any case, \( x \otimes v = 0 \).
\end{example}
\begin{example}
Consider
\[ M \otimes_R \qty(\bigoplus_{i \in I} N_i) \simeq \bigoplus_{i \in I} \qty(M \otimes_R N_i) \]
given by \( m \otimes (n_i)_{i \in I} \mapsto (m \otimes n_i)_{i \in I} \).
This is not true with the direct product.
However, we do have a map
\[ M \otimes_R \qty(\prod_{i \in I} N_i) \to \prod_{i \in I} \qty(M \otimes_R N_i) \]
given by the same formula, but this is in general not an isomorphism.
Consider
\[ \mathbb Q \otimes_{\mathbb Z} \prod_{n = 1}^\infty \faktor{\mathbb Z}{2^n \mathbb Z} \to \prod_{n = 1}^\infty \qty(\mathbb Q \otimes_{\mathbb Z} \faktor{\mathbb Z}{2^n \mathbb Z}) \]
The right-hand side is zero, as each factor is a tensor product of a divisible group by a torsion group.
However, the left-hand side is nonzero.
Let
\[ g = (1, 1, 1, \dots) \in \prod_{n = 1}^\infty \faktor{\mathbb Z}{2^n \mathbb Z} \]
This is an element of infinite order, so \( \langle g \rangle \simeq \mathbb Z \) as a subgroup of \( \prod_{n = 1}^\infty \faktor{\mathbb Z}{2^n \mathbb Z} \).
Thus
\[ \mathbb Q \otimes_{\mathbb Z} \langle g \rangle \simeq \mathbb Q \]
as \( \mathbb Z \)-modules.
But we have an injective inclusion map
\[ \langle g \rangle \to \prod_{n = 1}^\infty \faktor{\mathbb Z}{2^n \mathbb Z} \]
We will later show that \( \mathbb Q \) is a flat \( \mathbb Z \)-module.
This justifies the fact that there is an inclusion
\[ \mathbb Q \otimes_{\mathbb Z} \langle g \rangle \hookrightarrow \mathbb Q \otimes_{\mathbb Z} \prod_{n = 1}^\infty \faktor{\mathbb Z}{2^n \mathbb Z} \]
showing that in particular the module in question is nonzero.
\end{example}
\begin{example}
Consider \( \mathbb C \otimes_{\mathbb R} \mathbb C \).
We will choose to extend scalars on the left, treating the right-hand copy of \( \mathbb C \) as an \( \mathbb R \)-module isomorphic to \( \mathbb R^2 \).
As a module, \( \mathbb C \otimes_{\mathbb R} \mathbb C \simeq \mathbb C \otimes_{\mathbb R} \mathbb R^2 \) is isomorphic to \( \mathbb C^2 \).
The basis for \( \mathbb C^2 \) is given by \( 1 \otimes 1, 1 \otimes i \).

As a \( \mathbb C \)-algebra, we again choose to extend scalars on the left, considering the right-hand copy of \( \mathbb C \) as an \( \mathbb R \)-algebra.
\begin{align*}
\mathbb C \otimes_{\mathbb R} \mathbb C &\simeq \mathbb C \otimes_{\mathbb R} \faktor{\mathbb R[T]}{(T^2 + 1)} \\
&\simeq \faktor{\mathbb C[T]}{(T^2 + 1)} \\
&\simeq \faktor{\mathbb C[T]}{(T-i)(T+i)} \\
&\simeq \faktor{\mathbb C[T]}{(T-i)} \times \faktor{\mathbb C[T]}{(T+i)} \\
&\simeq \mathbb C \times \mathbb C
\end{align*}
using the Chinese remainder theorem, which will be explored later.
The action of this isomorphism on a pure tensor is
\begin{align*}
x \otimes y = (a + bi) \otimes (c + di) &\mapsto (a + bi) \otimes (c + dT + (T^2 + 1)\mathbb R[T]) \\
&\mapsto (a + bi)(c + dT) + (T^2 + 1)\mathbb C[T] \\
&= \underbrace{(ac + bdiT) + (ibc + adT)}_{P} + (T^2 + 1) \mathbb C[T] \\
&\mapsto (P + (T - i) \mathbb C[T], P + (T + i) \mathbb C[T]) \\
&\mapsto ((ac - bd) + i(bc + ad), (ac + bd) + i(bc - ad)) = (xy, x \overline{y})
\end{align*}
\end{example}
29 changes: 29 additions & 0 deletions iii/commalg/03_localisation.tex
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\subsection{Definitions}
\begin{definition}
A \emph{multiplicative set} or \emph{multiplicatively closed set} \( S \subseteq R \) is a subset such that \( 1 \in S \) and if \( a, b \in S \), then \( ab \in S \).
If \( U \subseteq R \) is any set, its \emph{multiplicative closure} \( S \) of \( U \) is the set
\[ \qty{\prod_{i = 1}^n u_i \mid n \geq 0, u_i \in U} \]
which is the smallest multiplicatively closed set containing \( U \).
\end{definition}
\begin{example}
\begin{enumerate}
\item If \( R \) is an integral domain, then \( S = R \setminus \qty{0} \) is multiplicative.
\item More generally, if \( \mathfrak p \) is a prime ideal in \( R \), then \( S = R \setminus \mathfrak p \) is multiplicative.
\item If \( x \in R \), then the set \( \qty{x^n \mid n \geq 0} \) is multiplicative.
\end{enumerate}
\end{example}
\begin{remark}
\( \mathbb Q \) is obtained from \( \mathbb Z \) by adding inverses for the elements of the multiplicative subset \( \mathbb Z \setminus \qty{0} \).
We have a ring homomorphism \( \mathbb Z \hookrightarrow \mathbb Q \).
We generalise this construction to arbitrary rings and multiplicative sets.
In general, injectivity of the ring homomorphism in question may fail.
\end{remark}
\begin{definition}
Let \( S \subseteq R \) be a multiplicative set, and let \( M \) be an \( R \)-module.
Then the \emph{localisation} of \( M \) by \( S \) is the set \( S^{-1} M = \faktor{M \times S}{\sim} \) where \( (m_1, s_1) \sim (m_2, s_2) \) if and only if there exists \( u \in S \) such that \( u(s_2 m_1 - s_1 m_2) = 0 \).
We write \( \frac{m}{s} \) for the equivalence class corresponding to \( (m, s) \).
We make \( S^{-1} M \) into an \( R \)-module by defining
\[ \frac{m_1}{s_1} + \frac{m_2}{s_2} = \frac{m_1 s_2 + m_2 s_1}{s_1 s_2};\quad r \cdot \frac{m}{s} = \frac{rm}{s} \]
We can make \( S^{-1} R \) into a ring by defining
\[ \frac{r_1}{s_1} \cdot \frac{r_2}{s_2} = \frac{r_1 r_2}{s_1 s_2} \]
\end{definition}
2 changes: 2 additions & 0 deletions iii/commalg/main.tex
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Expand Up @@ -14,5 +14,7 @@ \section{Chain conditions}
\input{01_chain_conditions.tex}
\section{Tensor products}
\input{02_tensor_products.tex}
\section{Localisation}
\input{03_localisation.tex}

\end{document}

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