Lecture 02A

iii/commalg/01_chain_conditions.tex

@@ -216,3 +216,31 @@ \subsection{Algebras}
     That is, \( \varphi(r \cdot 1_A) = r \cdot 1_B \).
 \end{definition}
 An \( R \)-algebra \( A \) is finitely generated if and only if there is some \( n \geq 0 \) and a surjective algebra homomorphism \( R[T_1, \dots, T_n] \to A \).
+\begin{theorem}[Hilbert's basis theorem]
+    Every finitely generated algebra \( A \) over a Noetherian ring \( R \) is Noetherian.
+\end{theorem}
+For example, the polynomial algebra over a field is Noetherian.
+\begin{proof}
+    It suffices to prove this for a polynomial ring, as every finitely generated algebra is a quotient of a polynomial ring.
+    It further suffices to prove this for a univariate polynomial ring \( A = R[T] \) by induction.
+    Let \( \mathfrak a \) be an ideal of \( R[T] \); we need to show that \( \mathfrak a \) is finitely generated.
+    For each \( i \geq 0 \), define
+    \[ \mathfrak a(i) = \qty{c_0 \mid c_0 T^i + \dots + c_i T^0 \in \mathfrak a} \]
+    Thus \( \mathfrak a(i) \) is the set of leading coefficients of polynomials of degree \( i \) that lie in \( \mathfrak a \).
+    Each \( \mathfrak a(i) \) is an ideal in \( R \), and \( \mathfrak a(i) \subseteq \mathfrak a(i+1) \) by multiplying by \( T \).
+    As \( R \) is Noetherian, each \( \mathfrak a(i) \) is a finitely generated ideal, and this ascending chain stabilises at \( \mathfrak a(m) \), say.
+    Let
+    \[ \mathfrak a(i) = (b_{i,1}, \dots, b_{i,n_i}) \]
+    We can choose \( f_{i,j} \) of degree \( i \) with leading coefficient \( b_{i,j} \).
+    Define the ideal
+    \[ \mathfrak b = (f_{i,j})_{i \leq m, j \leq n_i} \]
+    Note that \( \mathfrak b \) is finitely generated.
+    Defining \( \mathfrak b(i) \) in the same way as \( \mathfrak a(i) \), we have
+    \[ \forall i,\, \mathfrak a(i) = \mathfrak b(i) \]
+    By construction, \( \mathfrak b \subseteq \mathfrak a \); we claim that the reverse inclusion holds, then the proof will be complete.
+    Suppose that \( \mathfrak a \nsubseteq \mathfrak b \), and take \( f \in \mathfrak a \setminus \mathfrak b \) of minimal degree \( i \).
+    As \( \mathfrak a(i) = \mathfrak b(i) \), there is a polynomial \( g \) in \( \mathfrak b \) of degree \( i \) that has the same leading coefficient.
+    Then \( f - g \) has degree less than \( i \), and lies in \( \mathfrak a \).
+    But then by minimality, \( f - g \in \mathfrak b \), giving \( f \in \mathfrak b \).
+\end{proof}
+Therefore, if \( S \subseteq \faktor{R[T_1, \dots, T_n]}{I} \) where \( R \) is Noetherian, then \( (S) = (S_0) \) where \( S_0 \subseteq S \) is finite.

iii/commalg/02_tensor_products.tex

@@ -0,0 +1,81 @@
+\subsection{Introduction}
+Let \( M \) and \( N \) be \( R \)-modules.
+Informally, the tensor product of \( M \) and \( N \) over \( R \) is the set \( M \otimes_R N \) of all sums
+\[ \sum_{i=1}^\ell m_i \otimes n_i;\quad m_i \in M, n_i \in N \]
+subject to the relations
+\begin{align*}
+    (m_1 + m_2) \otimes n &= m_1 \otimes n + m_2 \otimes n \\
+    m \otimes (n_1 + n_2) &= m \otimes n_1 + m \otimes n_2 \\
+    (rm) \otimes n &= r(m \otimes n) \\
+    m \otimes (rn) &= r(m \otimes n)
+\end{align*}
+This is a module that abstracts the notion of bilinearity between two modules.
+\begin{example}
+    Consider \( \faktor{\mathbb Z}{2\mathbb Z} \otimes_{\mathbb Z} \faktor{\mathbb Z}{3\mathbb Z} \).
+    In this \( \mathbb Z \)-module,
+    \[ x \otimes y = (3x) \otimes y = x \otimes (3y) = x \otimes 0 = x \otimes (0 \cdot 0) = 0 (x \otimes 0) = 0 \]
+    Hence \( \faktor{\mathbb Z}{2\mathbb Z} \otimes_{\mathbb Z} \faktor{\mathbb Z}{3\mathbb Z} = 0 \).
+\end{example}
+\begin{example}
+    Now consider \( {\mathbb R}^n \otimes_{\mathbb R} {\mathbb R}^\ell \).
+    We will show later that this is isomorphic to \( \mathbb R^{n+\ell} \).
+\end{example}
+
+\subsection{Definitions}
+\begin{definition}
+    A map of \( R \)-modules \( f : M \times N \to L \) is \emph{\( R \)-bilinear} if for each \( m_0 \in M \) and \( n_0 \in N \), the maps \( n \mapsto f(m_0, n) \) and \( m \mapsto f(m, n_0) \) are \( R \)-linear (or equivalently, a homomorphism of \( R \)-modules).
+\end{definition}
+\begin{definition}
+    Let \( M, N \) be \( R \)-modules.
+    Let \( \mathcal F = R^{\oplus(M \times N)} \) be the free \( R \)-module with coordinates indexed by \( M \times N \).
+    Define \( K \subseteq \mathcal F \) to be the submodule generated by the following set of relations:
+    \begin{align*}
+        &(m_1 + m_2, n) - (m_1, n) - (m_2, n) \\
+        &(m, n_1 + n_2) - (m, n_1) - (m, n_2) \\
+        &r (m, n) - (rm, n) \\
+        &r (m, n) - (m, rn)
+    \end{align*}
+    The \emph{tensor product} \( M \otimes_R N \) is \( \faktor{\mathcal F}{K} \).
+    We further define the \( R \)-bilinear map
+    \[ i_{M \otimes N} : M \times N \to M \otimes N;\quad i_{M \otimes N}(m, n) = m \otimes n \]
+\end{definition}
+\begin{proposition}[universal property of the tensor product]
+    The pair \( (M \otimes_R N, i_{M \otimes_R N}) \) satisfies the following universal property.
+    For every \( R \)-module \( L \) and every \( R \)-bilinear map \( f : M \times N \to L \), there exists a unique homomorphism \( h : M \otimes_R N \to L \) such that the following diagram commutes.
+    \[\begin{tikzcd}
+        {M \times N} & {M \otimes_R N} \\
+        & L
+        \arrow["{i_{M \otimes_R N}}", from=1-1, to=1-2]
+        \arrow["h", dashed, from=1-2, to=2-2]
+        \arrow["f"', from=1-1, to=2-2]
+    \end{tikzcd}\]
+    Equivalently, \( h \circ i_{M \otimes_R N} = f \).
+\end{proposition}
+\begin{proof}
+    The conclusion \( h \circ i_{M \otimes N} = f \) holds if and only if for all \( m, n \), we have
+    \[ h(m \otimes n) = f(m, n) \]
+    Note that the elements \( \qty{m \otimes n} \) generate \( M \otimes N \) as an \( R \)-module, so there is at most one \( h \).
+    We now show that the definition of \( h \) on the pure tensors \( m \otimes n \) extends to an \( R \)-linear map \( M \otimes N \to L \).
+    The map \( R^{\oplus(M \otimes N)} \to L \) given by \( (m, n) \mapsto f(m, n) \) exists by the universal property of the direct sum.
+    However, this map vanishes on the generators of \( K \), so it factors through the quotient \( \faktor{\mathcal F}{K} \) as required.
+\end{proof}
+The universal property given above characterises the tensor product up to isomorphism.
+\begin{proposition}
+    Let \( M, N \) be \( R \)-modules, and \( (T, j) \) be an \( R \)-module and an \( R \)-bilinear map \( M \times N \to T \).
+    Suppose that \( (T, j) \) satisfies the same universal property as \( M \otimes N \).
+    Then there is a unique isomorphism of \( R \)-modules \( \varphi : M \otimes N \similarrightarrow T \) such that \( \varphi \circ i_{M \otimes N} = j \).
+\end{proposition}
+\begin{proof}
+    By using the universal property of \( M \otimes N \) and \( T \), we obtain \( \varphi \) and \( \psi \) as follows.
+    \[\begin{tikzcd}
+        {M \otimes N} && T \\
+        & {M \times N}
+        \arrow["{i_{M \otimes N}}", from=2-2, to=1-1]
+        \arrow["j"', from=2-2, to=1-3]
+        \arrow["\varphi", shift left, dashed, from=1-1, to=1-3]
+        \arrow["\psi", shift left, dashed, from=1-3, to=1-1]
+    \end{tikzcd}\]
+    Then, \( \psi \circ \varphi \circ i_{M \otimes N} = j \), so \( \psi \circ \varphi \) is obtained by applying the universal property of \( M \otimes N \) to \( i_{M \otimes N} \).
+    Hence \( \psi \circ \varphi \) is the identity map by uniqueness.
+    Similarly, \( \varphi \circ \psi \) is the identity, hence \( \varphi \) and \( \psi \) are inverse homomorphisms.
+\end{proof}

iii/commalg/main.tex

@@ -12,5 +12,7 @@
 
 \section{Chain conditions}
 \input{01_chain_conditions.tex}
+\section{Tensor products}
+\input{02_tensor_products.tex}
 
 \end{document}