Lectures 21A

ib/linalg/07_eigenvectors_and_eigenvalues.tex

@@ -161,7 +161,7 @@ \subsection{Polynomials for matrices and endomorphisms}
 \]
 For an endomorphism \( \alpha \in L(V) \),
 \[
-	p(\alpha) = a_n \alpha^n + \dots + a_0 I \in L(V);\quad \alpha^k \equiv \underbrace{\alpha \circ \dots \circ \alpha}_{k \text{ times}}
+	p(\alpha) = a_n \alpha^n + \dots + a_0 I \in L(V);\quad \alpha^k \equiv \underbrace{\alpha \circ \cdots \circ \alpha}_{k \text{ times}}
 \]
 
 \subsection{Sharp criterion of diagonalisability}

iii/commalg/04_integrality_finiteness_finite_generation.tex

@@ -3,7 +3,7 @@ \subsection{Nakayama's lemma}
     Let \( M \) be a finitely generated \( R \)-module, and let \( f : M \to M \) be an \( R \)-linear endomorphism.
     Let \( \mathfrak a \) be an ideal in \( R \) such that \( f(M) \subseteq \mathfrak a M \).
     Then, we have an equality in \( \End_R M \)
-    \[ f^n + a_1 f^{n-1} + \dots + a_n f^0 = 0;\quad f^r = \underbrace{f \circ \dots \circ f}_{r \text{ times}} \]
+    \[ f^n + a_1 f^{n-1} + \dots + a_n f^0 = 0;\quad f^r = \underbrace{f \circ \cdots \circ f}_{r \text{ times}} \]
     where \( a_i \in \mathfrak a \).
 \end{proposition}
 \begin{proof}

iii/commalg/06_direct_and_inverse_limits.tex

@@ -1,4 +1,4 @@
-\subsection{???}
+\subsection{Limits and completions}
 \begin{definition}
     Let \( \mathcal C \) be a category.
     \begin{enumerate}
@@ -17,7 +17,7 @@ \subsection{???}
         Recall that if \( a \mid b \), then there is an embedding \( \varphi : \mathbb F_{p^a} \to \mathbb F_{p^b} \).
         The collection of embeddings \( \mathbb F_{p^a} \to \mathbb F_{p^b} \) is then given by \( x \mapsto (\varphi(x))^{p^c} \) where \( 0 \leq c < a - 1 \).
         The map \( f_{i(i+1)} : \mathbb F_{p^{i!}} \to \mathbb F_{p^{(i+1)!}} \) is defined to be one such embedding.
-        A general embedding \( f_{ij} \) is given by the composite \( f_{(j-1)j} \circ \dots \circ f_{i(i+1)} \).
+        A general embedding \( f_{ij} \) is given by the composite \( f_{(j-1)j} \circ \cdots \circ f_{i(i+1)} \).
         This creates a direct system on \( I \).
         \item Let \( Y_i = \faktor{\mathbb Z}{p^i \mathbb Z} \), and let \( h_{ij} : \faktor{\mathbb Z}{p^j \mathbb Z} \to \faktor{\mathbb Z}{p^i \mathbb Z} \) be the natural projection.
         This is an inverse system on \( I \).
@@ -82,7 +82,7 @@ \subsection{???}
 \begin{definition}
     Let \( M \) be an \( R \)-module.
     \begin{enumerate}
-        \item A \emph{filtration} of \( M \) is a sequence \( (M_n)_{n \geq 1} \) of submodules of \( M \) such that \( M_n \supseteq M_{n+1} \) for each \( n \).
+        \item A \emph{filtration} of \( M \) is a sequence \( (M_n)_{n \geq 1} \) of submodules of \( M \) such that \( M_0 = M \) and \( M_n \supseteq M_{n+1} \) for each \( n \).
         \item The \emph{completion} of \( M \) with respect to a filtration \( (M_n)_{n \geq 1} \) is \( \varprojlim \faktor{M}{M_n} \).
     \end{enumerate}
 \end{definition}
@@ -96,3 +96,111 @@ \subsection{???}
     \end{enumerate}
 \end{theorem}
 Thus \( \mathfrak a \)-adic completion is an exact functor from the category of finitely generated \( R \)-modules if \( R \) is Noetherian.
+
+\subsection{Graded rings and modules}
+\begin{definition}
+    A \emph{graded ring} is a ring \( A = \bigoplus_{n=0}^\infty A_n \), where each \( A_n \) is an additive subgroup of \( A \), such that \( A_m A_n \subseteq A_{m+n} \).
+\end{definition}
+\begin{proposition}
+    \( A_0 \) is a subring of \( A \).
+\end{proposition}
+\begin{proof}
+    It is clearly a subgroup closed under multiplication, so it suffices to check that it contains the identity element of \( A \).
+    We have
+    \[ 1_A = \sum_{i=0}^m y_i;\quad y_i \in A_i \]
+    For \( z_n \in A_n \),
+    \[ z_n = \sum_{i=0}^m y_i z_n \]
+    \( z_n \) is an element of \( A_n \), and each term \( y_i z_n \) is an element of \( A_{n+i} \).
+    But since the sum is direct, we must have \( z_n = y_0 z_n \), so \( z = y_0 z \) for all \( z \in A \).
+    Hence \( y_0 \in A_0 \) is the identity element.
+\end{proof}
+\begin{remark}
+    Each \( A_n \) is an \( A_0 \)-module as \( A_0 A_n \subseteq A_n \).
+\end{remark}
+\begin{example}
+    The polynomial ring in finitely many variables has a grading: \( k[T_1, \dots, T_m] = \bigoplus_{n=0}^\infty A_n \) where \( A_n \) is the set of homogeneous polynomials of degree \( n \).
+\end{example}
+\begin{definition}
+    Let \( A = \bigoplus_{n=0}^\infty A_n \) be a graded ring.
+    A \emph{graded \( A \)-module} is an \( A \)-module \( M = \bigoplus_{n=0}^\infty M_n \) such that \( A_m M_n \subseteq M_{m+n} \).
+\end{definition}
+For a graded ring \( A \), we define \( A_+ = \bigoplus_{n = 1}^\infty A_n = \ker (A \twoheadrightarrow A_0) \).
+This is an ideal of \( A \), and \( \faktor{A}{A_+} \simeq A_0 \).
+\begin{proposition}
+    Let \( A = \bigoplus_{i=0}^\infty A_n \) be a graded ring.
+    Then the following are equivalent:
+    \begin{enumerate}
+        \item \( A \) is Noetherian;
+        \item \( A_0 \) is Noetherian and \( A \) is finitely generated as an \( A_0 \)-algebra.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Hilbert's basis theorem shows that (ii) implies (i).
+    For the converse, \( A_0 \) is Noetherian as it is isomorphic to a quotient of the Noetherian ring \( A \).
+    Note that \( A_+ \) is generated by the set of homogeneous elements of positive degree.
+    By (i), \( A_+ \) an ideal in a Noetherian ring so is generated by a finite set \( \qty{x_1, \dots, x_s} \), and we can take each \( x_i \) to be homogeneous, say, \( x_i \in A_{k_i} \) where \( k_i > 0 \).
+    Let \( A' \) be the \( A_0 \)-subalgebra of \( A \) generated by \( \qty{x_1, \dots, x_s} \); we want to show \( A' = A \).
+    It suffices to show that \( A_n \subseteq A' \) for every \( n \geq 0 \), which we will show by induction.
+    The case \( n = 0 \) is clear.
+
+    Let \( n > 0 \), and let \( y \in A_n \).
+    Note that \( y \in A_+ \), so
+    \[ y = \sum_{i=1}^s r_i x_i \]
+    where \( r_i \in A \) and \( x_i \in A_{k_i} \).
+    Applying the projection to \( A_n \),
+    \[ y = \sum_{i=1}^s a_i x_i;\quad a_i \in A_{n-k_i} \]
+    where \( a_i \) is the \( (n-k_i) \) homogeneous part of \( r_i \).
+    As \( k_i \) is positive, the inductive hypothesis implies that each \( a_i \) can be written as a polynomial in \( x_1, \dots, x_s \) with coefficients in \( A_0 \), giving \( y \in A' \) as required.
+\end{proof}
+\begin{definition}
+    Let \( \mathfrak a \) be an ideal of \( R \), and let \( M \) be an \( R \)-module.
+    Then a filtration \( (M_n)_{n \geq 0} \) is an \emph{\( \mathfrak a \)-filtration} if \( \mathfrak a M_n \subseteq M_{n+1} \) for each \( n \geq 0 \).
+    An \( \mathfrak a \)-filtration \( (M_n)_{n \geq 0} \) is \emph{stable} if there exists \( n_0 \geq 0 \) such that \( \mathfrak a M_n = M_{n+1} \) for all \( n \geq n_0 \).
+\end{definition}
+\begin{example}
+    \( (\mathfrak a^n M)_{n \geq 0} \) is a stable \( \mathfrak a \)-filtration of \( M \).
+\end{example}
+\begin{definition}
+    Let \( \mathfrak a \) be an ideal in \( R \).
+    The \emph{associated graded ring} is
+    \[ G_{\mathfrak a}(R) = \bigoplus_{n \geq 0} \faktor{\mathfrak a^n}{\mathfrak a^{n+1}};\quad \mathfrak a^0 = R \]
+    This is a ring by defining
+    \[ (x + \mathfrak a^{n+1})(y + \mathfrak a^{m+1}) = xy + \mathfrak a^{n + m + 1};\quad x \in \mathfrak a^n, y \in \mathfrak a^m \]
+\end{definition}
+\begin{definition}
+    Let \( M \) be an \( R \)-module, and let \( \mathfrak a \) be an ideal of \( R \).
+    Let \( (M_n)_{n \geq 0} \) be an \( \mathfrak a \)-filtration of \( M \).
+    The \emph{associated graded module} is
+    \[ G(M) = \bigoplus_{n \geq 0} \faktor{M_n}{M_{n+1}} \]
+    This is a module over \( G_{\mathfrak a}(R) \) by defining
+    \[ (x + \mathfrak a^{n+1})(m + M_{\ell + 1}) = xm + M_{n+\ell+1} \]
+\end{definition}
+\begin{proposition}
+    Let \( R \) be a Noetherian ring, and let \( \mathfrak a \) be an ideal of \( R \).
+    Then
+    \begin{enumerate}
+        \item the associated graded ring \( G_{\mathfrak a}(R) \) is Noetherian; and
+        \item if \( M \) is a finitely generated \( R \)-module and \( (M_n)_{n \geq 0} \) is a stable \( \mathfrak a \)-filtration of \( M \), then the associated graded module \( G(M) \) is a finitely generated \( G_{\mathfrak a}(R) \)-module.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    \emph{Part (i).}
+    Let \( R \) be Noetherian.
+    Then let \( \mathfrak a = (x_1, \dots, x_s) \), and write \( \overline x_i \) for the image of \( x_i \) in \( \faktor{\mathfrak a}{\mathfrak a^2} \).
+    Note that
+    \[ G_{\mathfrak a}(R) = \faktor{R}{\mathfrak a} \oplus \faktor{\mathfrak a}{\mathfrak a^2} \oplus \faktor{\mathfrak a^2}{\mathfrak a^3} \oplus \cdots \]
+    \( G_{\mathfrak a}(R) \) is generated as an \( \faktor{R}{\mathfrak a} \)-algebra by \( \overline x_1, \dots, \overline x_s \), by taking sums and products.
+    Note that \( \faktor{R}{\mathfrak a} \) is Noetherian, so \( G_{\mathfrak a}(R) \) is Noetherian by Hilbert's basis theorem.
+
+    \emph{Part (ii).}
+    Let \( (M_n)_{n \geq 0} \) be a stable \( \mathfrak a \)-filtration of \( M \).
+    Then there exists \( n_0 \) such that for all \( n \geq n_0 \), we have \( M_{n_0 + r} = \mathfrak a^r M_{n_0} \).
+    Thus \( G(M) \) is generated as a \( G_{\mathfrak a}(R) \)-module by
+    \[ \faktor{M_0}{M_1} \oplus \faktor{M_1}{M_2} \oplus \dots \oplus \faktor{M_{n_0}}{M_{n_0 + 1}} \]
+    Each factor \( \faktor{M_i}{M_{i+1}} \) is a Noetherian \( R \)-module, as they are quotients of Noetherian modules, and are annihilated by \( \mathfrak a \).
+    In particular, \( G(M) \) is a finitely generated \( G_{\mathfrak a}(R) \)-module, say by \( x_1, \dots, x_s \).
+\end{proof}
+\begin{definition}
+    Let \( M \) be an \( R \)-module.
+    We say that filtrations \( (M_n), (M_n') \) of \( M \) are \emph{equivalent} if there exists \( n_0 \) such that for all \( n \geq 0 \), we have \( M_{n + n_0} \subseteq M_n' \) and \( M'_{n + n_0} \subseteq M_n \).
+\end{definition}