Merge branch 'main' of https://github.com/zeramorphic/cambridge-maths…

…-notes

iii/commalg/02_tensor_products.tex

@@ -543,3 +543,123 @@ \subsection{Tensor products of algebras}
     \[ R[X_1, \dots, X_n] \otimes R[T_1, \dots, T_r] \simeq R[X_1, \dots, X_n, T_1, \dots, T_r] \]
     as was shown above.
 \end{example}
+\begin{remark}
+    \begin{enumerate}
+        \item If \( f : A \to A', g : B \to B' \) are \( R \)-algebra homomorphisms, then \( f \otimes g : A \otimes B \to A' \otimes B' \) is not only an \( R \)-module homomorphism but is also an \( R \)-algebra homomorphism.
+        \item There are \( R \)-algebra homomorphisms
+        \begin{enumerate}
+            \item \( \faktor{R}{I} \otimes \faktor{R}{J} \simeq \faktor{R}{I+J} \);
+            \item \( A \otimes B \simeq B \otimes A \);
+            \item \( A \otimes (B \times C) \simeq (A \otimes B) \times (A \otimes C) \);
+            \item \( A \otimes B^n \simeq (A \otimes B)^n \);
+            \item \( (A \otimes B) \otimes C \simeq A \otimes (B \otimes C) \).
+        \end{enumerate}
+    \end{enumerate}
+\end{remark}
+
+\subsection{Restriction and extension of scalars}
+Let \( f : R \to S \) be a ring homomorphism.
+Let \( M \) be an \( S \)-module.
+Then we can \emph{restrict scalars} to make \( M \) into an \( R \)-module by
+\[ r \cdot m = f(r) \cdot m \]
+The composition \( R \to S \to \End M \) is a ring homomorphism, so this makes \( M \) into an \( R \)-module automatically without needing to check axioms.
+\begin{example}
+    Let \( f : \mathbb R \to \mathbb C \) be the inclusion.
+    Then any \( \mathbb C \)-module is an \( \mathbb R \)-module.
+\end{example}
+
+Now suppose \( f : R \to S \) is a ring homomorphism, \( M \) is an \( S \)-module, and \( N \) is an \( R \)-module.
+We can form the \( R \)-module \( M \otimes_R N \), as \( M \) is an \( R \)-module by restriction of scalars.
+\emph{Extension of scalars} shows that \( M \otimes_R N \) is also an \( S \)-module.
+The action of \( s \in S \) on pure tensors is
+\[ s \cdot (m \otimes n) = sm \otimes n \]
+We have an \( R \)-bilinear map \( M \times N \to M \otimes_R N \) by
+\[ (m, n) \mapsto sm \otimes n \]
+so by the universal property this gives rise to a map \( h_s : M \otimes_R N \to M \otimes_R N \) with the desired action on pure tensors.
+\( h_s \) is \( R \)-linear by the universal property.
+Defining \( \varphi : S \to \End(M \otimes_R N) \) by \( \varphi(s) = h_s \), one can check that \( h_s \) is a well-defined endomorphism and that \( \varphi \) is a ring homomorphism.
+\begin{example}
+    \( S \otimes_R R \simeq R \) as \( R \)-modules, by \( s \otimes r \mapsto s \cdot f(r) \).
+    This is also \( S \)-linear, since
+    \[ s'(s \otimes r) = (s's \otimes r) \mapsto s's \cdot f(r) = s'(s \cdot f(r)) \]
+    For example, \( \mathbb C \otimes_{\mathbb R} \mathbb R \simeq \mathbb C \) as \( \mathbb C \)-modules.
+\end{example}
+\begin{example}
+    Let \( M \) be an \( S \)-module and \( (N_i)_{i \in I} \) are \( R \)-modules.
+    Then
+    \[ M \otimes \qty(\bigoplus_i N_i) \simeq \bigoplus_i (M \otimes N_i) \]
+    as \( S \)-modules.
+    So \( \mathbb C \otimes_{\mathbb R} \mathbb R^n \simeq \mathbb C^n \) as \( \mathbb C \)-modules.
+\end{example}
+\begin{example}
+    Restrict the \( \mathbb C \)-module \( \mathbb C^n \) to an \( \mathbb R \)-module to obtain \( \mathbb R^{2n} \).
+    Then, extending to \( \mathbb C \),
+    \[ \mathbb C \otimes_{\mathbb R} \mathbb R^{2n} \simeq \mathbb C^{2n} \]
+    Similarly, extending \( \mathbb R^n \) to \( \mathbb C \), we find \( \mathbb C \otimes_{\mathbb R} \mathbb R^n \simeq \mathbb C^n \) over \( \mathbb C \).
+    Restricting to \( \mathbb R \), \( \mathbb C^n \simeq \mathbb R^{2n} \).
+    So the operations of restriction and extension of scalars are not inverses in either direction.
+\end{example}
+\begin{example}
+    Consider \( \mathbb Z^n \) as a \( \mathbb Z \)-module.
+    Consider the quotient map \( f : \mathbb Z \to \faktor{\mathbb Z}{2\mathbb Z} \).
+    Extending scalars to \( \faktor{\mathbb Z}{2\mathbb Z} \),
+    \[ \faktor{\mathbb Z}{2\mathbb Z} \otimes_{\mathbb Z} \mathbb Z^n \simeq \qty(\faktor{\mathbb Z}{2\mathbb Z})^n \]
+\end{example}
+\begin{example}
+    Consider \( \mathbb C^n \otimes_{\mathbb R} \mathbb R^\ell \) as a \( \mathbb C \)-module.
+    As \( \mathbb R \)-modules,
+    \[ \mathbb C^n \otimes_{\mathbb R} \mathbb R^\ell \simeq \mathbb R^{2n} \otimes_{\mathbb R} \mathbb R^\ell \simeq \mathbb R^{2n\ell} \simeq \mathbb C^{n\ell} \]
+    We would like to make this into an isomorphism of \( \mathbb C \)-modules.
+    We will show that in fact
+    \[ \mathbb C^n \otimes_{\mathbb R} \mathbb R^\ell \simeq \mathbb C^n \otimes_{\mathbb C} (\mathbb C \otimes_{\mathbb R} \mathbb R^\ell) \]
+    where
+    \[ v \otimes u \mapsto v \otimes (1 \otimes u) \]
+    giving
+    \[ \mathbb C^n \otimes_{\mathbb R} \mathbb R^\ell \simeq \mathbb C^n \otimes_{\mathbb C} \mathbb C^\ell \simeq \mathbb C^{n\ell} \]
+    as \( \mathbb C \)-modules.
+    The isomorphism
+    \[ \mathbb C^n \otimes_{\mathbb R} \mathbb R^\ell \simeq \mathbb C^n \otimes_{\mathbb C} \mathbb C^\ell \]
+    maps a pure tensor \( v \otimes u \) to \( v \otimes u \).
+\end{example}
+\begin{proposition}
+    Let \( M \) be an \( S \)-module and \( N \) be an \( R \)-module.
+    Then
+    \[ M \otimes_R N \simeq M \otimes_S (S \otimes_R N) \]
+    as \( S \)-modules, where
+    \[ m \otimes n \mapsto m \otimes (1 \otimes n);\quad m \otimes (s \otimes n) \mapsto sm \otimes n \]
+\end{proposition}
+\begin{proof}
+    %TODO/ES1
+\end{proof}
+\begin{proposition}
+    Let \( M, M' \) be \( S \)-modules and \( N, N' \) be \( R \)-modules.
+    Then we have \( S \)-module isomorphisms
+    \begin{align*}
+        M \otimes_R N &\simeq N \otimes_R M \\
+        (M \otimes_R N) \otimes_R N' &\simeq M \otimes_R (N \otimes_R N') \\
+        (M \otimes_R N) \otimes_S M' &\simeq M \otimes_S (N \otimes_R M') \\
+        M \otimes_R \qty(\bigoplus_i N_i) &\simeq \bigoplus_i (M \otimes N_i)
+    \end{align*}
+\end{proposition}
+Heuristically, the tensor products in the above isomorphisms always operate over the largest possible ring: \( S \) if both operands are \( S \)-modules, else \( R \).
+We prove only the third result.
+\begin{proof}
+    By the previous proposition,
+    \begin{align*}
+        (M \otimes_R N) \otimes_S M' &\simeq (M \otimes_S (N \otimes_R S)) \otimes_S M' \\
+        &\simeq M \otimes_S ((N \otimes_R S) \otimes_S M') \\
+        &\simeq M \otimes_S (N \otimes_R M')
+    \end{align*}
+\end{proof}
+\begin{corollary}
+    Let \( N, N' \) be \( R \)-modules.
+    Then
+    \[ S \otimes_R (N \otimes_R N') \simeq (S \otimes_R N) \otimes_S (S \otimes_R N') \]
+    as \( S \)-modules.
+\end{corollary}
+\begin{proof}
+    \[ S \otimes_R (N \otimes_R N') \simeq (S \otimes_R N) \otimes_R N' \simeq (S \otimes_R N) \otimes_S (S \otimes_R N') \]
+\end{proof}
+\begin{example}
+    \[ \mathbb C \otimes_{\mathbb R} \qty(\mathbb R^\ell \otimes_{\mathbb R} \mathbb R^k) \simeq (\mathbb C \otimes_{\mathbb R} \mathbb R^\ell) \otimes_C \mathbb R^k \simeq \mathbb C^\ell \otimes_{\mathbb C} \mathbb C^k \simeq \mathbb C^{\ell k} \]
+\end{example}