Lectures 06A

iii/commalg/02_tensor_products.tex

@@ -638,7 +638,7 @@ \subsection{Restriction and extension of scalars}
         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)
+        M \otimes_R \qty(\bigoplus_i N_i) &\simeq \bigoplus_i (M \otimes_R 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 \( R \).
@@ -661,5 +661,176 @@ \subsection{Restriction and extension of scalars}
     \[ S \otimes_R (N \otimes_R N') \simeq (S \otimes_R N) \otimes_R N' \simeq (S \otimes_R N) \otimes_S (S \otimes_R N') \]
 \end{proof}
 \begin{example}
-    \[ \mathbb C \otimes_{\mathbb R} \qty(\mathbb R^\ell \otimes_{\mathbb R} \mathbb R^k) \simeq (\mathbb C \otimes_{\mathbb R} \mathbb R^\ell) \otimes_C \mathbb R^k \simeq \mathbb C^\ell \otimes_{\mathbb C} \mathbb C^k \simeq \mathbb C^{\ell k} \]
+    \[ \mathbb C \otimes_{\mathbb R} \qty(\mathbb R^\ell \otimes_{\mathbb R} \mathbb R^k) \simeq (\mathbb C \otimes_{\mathbb R} \mathbb R^\ell) \otimes_{\mathbb C} (\mathbb C \otimes_{\mathbb R} \mathbb R^k) \simeq \mathbb C^\ell \otimes_{\mathbb C} \mathbb C^k \simeq \mathbb C^{\ell k} \]
+    By induction, one can see that
+    \[ S \otimes_R (N_1 \otimes_R \dots \otimes_R N_\ell) = (S \otimes_R N_1) \otimes_S \dots \otimes_S (S \otimes_R N_\ell) \]
 \end{example}
+
+\subsection{Extension of scalars on morphisms}
+Let \( f : N \to N' \) be an \( R \)-linear map, and \( M \) be an \( S \)-module.
+Then the map
+\[ \id_M \otimes f : M \otimes_R \to M \otimes_R N' \]
+is \( S \)-linear.
+Indeed,
+\[ (\id_M \otimes f)(s(m \otimes n)) = \id_M sm \otimes f(n) = s(m \otimes f(n)) = s((\id_M \otimes f)(m \otimes n)) \]
+\begin{example}
+    Let \( T : \mathbb R^n \to \mathbb R^\ell \) be \( R \)-linear, and use bases \( e_1, \dots, e_n \) and \( f_1, \dots, f_\ell \).
+    Then
+    \[ \id_{\mathbb C} \otimes T : \mathbb C \otimes_{\mathbb R} \mathbb R^n \to \mathbb C \otimes_{\mathbb R} \mathbb R^\ell \]
+    is given by
+    \[ (\id_{\mathbb C} \otimes T)(1 \otimes e_i) = 1 \otimes T(e_i) = 1 \otimes \sum_{j=1}^\ell [T]_{ji} \cdot f_j = \sum_{j=1}^\ell [T]_{ji} (1 \otimes f_j) \]
+    This shows that the matrix \( [\id_{\mathbb C} \otimes T] \) has all real elements, and is the same as the matrix \( [T] \).
+\end{example}
+
+\subsection{Extension of scalars in algebras}
+Let \( A, B \) be \( R \)-algebras.
+Then the module \( A \otimes_R B \) is also an \( R \)-algebra.
+Furthermore, can see that \( A \otimes_R B \) is an \( A \)-algebra and a \( B \)-algebra by the maps \( a \mapsto a \otimes 1 \) and \( b \mapsto 1 \otimes b \).
+\begin{example}
+    Consider \( R[X_1, \dots, X_n] \) and \( f : R \to S \).
+    Then
+    \[ \varphi : S \otimes_R R[X_1, \dots, X_n] \similarrightarrow S[X_1, \dots, X_n] \]
+    as \( S \)-algebras.
+    Indeed, \( \varphi \) already exists as an isomorphism of \( S \)-modules given by
+    \[ \varphi(s \otimes p) = sp \]
+    and one can verify that unity and multiplication are preserved.
+    Further,
+    \[ S \otimes \qty(\faktor{R[X_1, \dots, X_n]}{I}) \simeq \faktor{S[X_1, \dots, X_n]}{I^e} \]
+\end{example}
+\begin{proposition}
+    Let \( A \) be an \( R \)-algebra and \( B \) be an \( S \)-algebra.
+    Then
+    \[ A \otimes_R B \simeq (A \otimes_R S) \otimes_S R \]
+    as \( S \)-algebras.
+\end{proposition}
+\begin{proposition}
+    Let \( A, B \) be \( R \)-algebras.
+    Then
+    \[ S \otimes_R (A \otimes_R B) \simeq (S \otimes_R A) \otimes_S (S \otimes_R B) \]
+    as \( S \)-algebras.
+\end{proposition}
+The proofs are omitted, but trivial.
+
+\subsection{Exactness properties of the tensor product}
+Let \( M \) be an \( R \)-module.
+There is a functor
+\[ T_M : \mathbf{Mod}_R \to \mathbf{Mod}_R \]
+from the category of \( R \)-modules to itself given by
+\[ T_M(N) = M \otimes_R N;\quad T_M(N \xrightarrow f N') = \id_M \otimes f \]
+We intend to show that if
+% https://q.uiver.app/#q=WzAsNCxbMCwwLCJBIl0sWzEsMCwiQiJdLFsyLDAsIkMiXSxbMywwLCIwIl0sWzAsMSwiZiJdLFsxLDIsImciXSxbMiwzXV0=
+\[\begin{tikzcd}
+	A & B & C & 0
+	\arrow["f", from=1-1, to=1-2]
+	\arrow["g", from=1-2, to=1-3]
+	\arrow[from=1-3, to=1-4]
+\end{tikzcd}\]
+is an exact sequence of \( R \)-modules, then
+% https://q.uiver.app/#q=WzAsNCxbMCwwLCJNIFxcb3RpbWVzX1IgQSJdLFsxLDAsIk0gXFxvdGltZXNfUiBCIl0sWzIsMCwiTSBcXG90aW1lc19SIEMiXSxbMywwLCIwIl0sWzAsMSwiXFxpZF9NIFxcb3RpbWVzIGYiXSxbMSwyLCJcXGlkX00gXFxvdGltZXMgZyJdLFsyLDNdXQ==
+\[\begin{tikzcd}
+	{M \otimes_R A} & {M \otimes_R B} & {M \otimes_R C} & 0
+	\arrow["{T_M(f)}", from=1-1, to=1-2]
+	\arrow["{T_M(g)}", from=1-2, to=1-3]
+	\arrow[from=1-3, to=1-4]
+\end{tikzcd}\]
+is also an exact sequence.
+This shows that \( T_M \) is a \emph{right exact} functor.
+\begin{definition}
+    Let \( Q, P \) be \( R \)-modules.
+    Then
+    \[ \Hom_R(Q, P) = \qty{f : Q \to P \mid f \text{ is } R\text{-linear}} \]
+    This is also an \( R \)-module: if \( \varphi \in \Hom_R(Q, P) \),
+    \[ (r \cdot \varphi)(q) = r \cdot \varphi(q) \]
+\end{definition}
+\begin{definition}
+    Let \( Q, P \) be \( R \)-modules.
+    Then \( \Hom_R(Q, -) \) and \( \Hom_R(-, P) \) are functors \( \mathbf{Mod}_R \to \mathbf{Mod}_R \), with action on morphisms \( f : N' \to N \) given by
+    \[ \Hom_R(Q, f)(\varphi) = f \circ \varphi = f_\star(\varphi) : \Hom_R(Q, N') \to \Hom_R(Q, N') \]
+    and
+    \[ \Hom_R(f, P)(\varphi) = \varphi \circ f = f^\star(\varphi) : \Hom_R(N, Q) \to \Hom_R(N', Q) \]
+\end{definition}
+\begin{proposition}
+    Suppose
+    \[\begin{tikzcd}
+        0 & A & B & C
+        \arrow[from=1-1, to=1-2]
+        \arrow["f", from=1-2, to=1-3]
+        \arrow["g", from=1-3, to=1-4]
+    \end{tikzcd}\]
+    is exact.
+    Then, so is
+    \[\begin{tikzcd}
+        0 & {\Hom_R(Q, A)} & {\Hom_R(Q, B)} & {\Hom_R(Q, C)}
+        \arrow[from=1-1, to=1-2]
+        \arrow["{f_\star}", from=1-2, to=1-3]
+        \arrow["{g_\star}", from=1-3, to=1-4]
+    \end{tikzcd}\]
+    Thus, the covariant hom-functor is \emph{left exact}.
+\end{proposition}
+\begin{proposition}
+    Suppose
+    \[\begin{tikzcd}
+        A & B & C & 0
+        \arrow["f", from=1-1, to=1-2]
+        \arrow["g", from=1-2, to=1-3]
+        \arrow[from=1-3, to=1-4]
+    \end{tikzcd}\]
+    is exact.
+    Then, so is
+    \[\begin{tikzcd}
+        0 & {\Hom_R(C, P)} & {\Hom_R(B, P)} & {\Hom_R(A, P)}
+        \arrow[from=1-1, to=1-2]
+        \arrow["{g^\star}", from=1-2, to=1-3]
+        \arrow["{f^\star}", from=1-3, to=1-4]
+    \end{tikzcd}\]
+    Thus, the contravariant hom-functor is also left-exact.
+\end{proposition}
+\begin{lemma}
+    Consider a sequence of \( R \)-modules
+    \[\begin{tikzcd}
+        A & B & C
+        \arrow["f", from=1-1, to=1-2]
+        \arrow["g", from=1-2, to=1-3]
+    \end{tikzcd}\]
+    Suppose that for each \( R \)-module \( P \),
+    \[\begin{tikzcd}
+        {\Hom_R(C, P)} & {\Hom_R(B, P)} & {\Hom_R(A, P)}
+        \arrow["{g^\star}", from=1-1, to=1-2]
+        \arrow["{f^\star}", from=1-2, to=1-3]
+    \end{tikzcd}\]
+    is exact.
+    Then the original sequence
+    \[\begin{tikzcd}
+        A & B & C
+        \arrow["f", from=1-1, to=1-2]
+        \arrow["g", from=1-2, to=1-3]
+    \end{tikzcd}\]
+    is exact.
+\end{lemma}
+\begin{proof}
+    First, take \( P = C \).
+    By hypothesis, the following sequence is exact.
+    \[\begin{tikzcd}
+        {\Hom_R(C, C)} & {\Hom_R(B, C)} & {\Hom_R(A, C)}
+        \arrow["{g^\star}", from=1-1, to=1-2]
+        \arrow["{f^\star}", from=1-2, to=1-3]
+    \end{tikzcd}\]
+    Consider
+    \[ \id_C \mapsto \id_C \circ g \mapsto \id_C \circ g \circ f \]
+    By exactness, \( \id_C \) must be mapped to zero under \( f^\star \circ g^\star \), so \( g \circ f = 0 \).
+    Hence \( \Im f \subseteq \ker g \).
+
+    Now, take \( P = \faktor{B}{\Im f} = \coker f \).
+    \[\begin{tikzcd}
+        {\Hom_R\qty(C, \faktor{B}{\Im f})} & {\Hom_R\qty(B, \faktor{B}{\Im f})} & {\Hom_R\qty(A, \faktor{B}{\Im f})}
+        \arrow["{g^\star}", from=1-1, to=1-2]
+        \arrow["{f^\star}", from=1-2, to=1-3]
+    \end{tikzcd}\]
+    Let \( h : B \to \faktor{B}{\Im f} \) be the quotient map.
+    Then,
+    \[ f^\star(h) = h \circ f;\quad h(f(x)) = 0 \]
+    Thus by exactness, \( h \) has a preimage \( e : C \to \faktor{B}{\Im f} \).
+    Then \( g^\star(e) = e \circ g = h \), so \( \ker g \subseteq \ker h = \Im f \), giving the reverse inclusion.
+\end{proof}
+By the universal property of the tensor product,
+\[ \Hom_R(M \otimes_R N, L) \simeq \operatorname{Bilin}_R(M \times N, L) \simeq \Hom_R(N, \Hom_R(M, L)) \]