282 lines
19 KiB
TeX
282 lines
19 KiB
TeX
\subsection{Finitely generated and Noetherian modules}
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\begin{definition}[Generated submodule]
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Let $R$ be a ring, $M$ an $R$-module, $S \subseteq M$.
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Then the following sets coincide
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\begin{enumerate}
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\item $\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{finite}, r_s \in R, \right\}$
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\item $\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$
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\item The $\subseteq$-smallest submodule of $M$ containing $S$
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\end{enumerate}
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This subset of $N \subseteq M$ is called the \vocab[Module!Submodule]{submodule of $M $ generated by $S$}. If $N= M$ we say that \vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
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$M$ is finitely generated $:\iff \exists S \subseteq M$ finite such that $M$ is generated by $S$.
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\end{definition}
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\begin{definition}[Noetherian $R$-module]
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$M$ is a \vocab{Noetherian} $R$-module if the following equivalent conditions hold:
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\begin{enumerate}
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\item Every submodule $N \subseteq M$ is finitely generated.
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\item Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates
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\item Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a $\subseteq$-largest element.
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\end{enumerate}
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\end{definition}
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\begin{proposition}[Hilbert's Basissatz]\label{basissatz}
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If $R$ is a Noetherian ring, then the polynomial rings $R[X_1,\ldots, X_n]$ in finitely many variables are Noetherian.
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\end{proposition}
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\subsubsection{Properties of finite generation and Noetherianness}
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\begin{fact}[Properties of Noetherian modules]
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\begin{enumerate}
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\item Every Noetherian module over an arbitrary ring is finitely generated.
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\item If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is finitely generated.
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\item Every submodule of a Noetherian module is Noetherian.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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\begin{enumerate}
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\item By definition, $M$ is a submodule of itself. Thus it is finitely generated.
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\item Since $M$ is finitely generated, there exists a surjective homomorphism $R^n \to M$. As $R$ is Noetherian, $R^n$ is Noethrian as well.
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\item trivial
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\end{enumerate}
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\end{proof}
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\begin{fact}
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Let $M, M', M''$ be $R$-modules.
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\begin{enumerate}
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\item Suppose $M \xrightarrow{p} M''$ is surjective. If $M$ is finitely generated (resp. Noetherian), then so is $M''$.
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\item Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact. If $M'$ and $M ''$ are finitely generated (reps. Noetherian), so is $M$.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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\begin{enumerate}
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\item Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$. Then $p^{-1} M_i''$ yields a strictly ascending sequence.
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If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$.
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\item Because of 1. we can replace $M'$ by $f(M')$ and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ to be exact. The fact about finite generation follows from EInführung in die Algebra.
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If $M', M''$ are Noetherian, $N \subseteq M$ a submodule, then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely generated. Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated.
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\end{enumerate}
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\end{proof}
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\subsection{Ring extensions of finite type}
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\begin{definition}[$R$-algebra]
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Let $R$ be a ring. An $R$-algebra $(A, \alpha)$ is a ring $A$ with a ring homomorphism $R \xrightarrow{\alpha} A$.
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$\alpha$ will usually be omitted. In general $\alpha$ is not assumed to be injective.\\
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\\
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An $R$-subalgebra is a subring $\alpha(R) \subseteq A' \subseteq A$.\\
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A morphism of $R$-algebras $A \xrightarrow{f} \tilde{A}$ is a ring homomorphism with $\tilde{\alpha} = f \alpha$.
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\end{definition}
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\begin{definition}[Generated (sub)algebra, algebra of finite type]
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Let $(A, \alpha)$ be an $R$-algebra.
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\begin{align}
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\alpha: R[X_1,\ldots,X_m] &\longrightarrow A[X_1,\ldots,X_m] \\
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P = \sum_{\beta \in \N^m} p_\beta X^{\beta} &\longmapsto \sum_{\beta \in \N^m} \alpha(p_\beta) X^{\beta}
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\end{align}
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is a ring homomorphism. We will sometimes write $P(a_1,\ldots,a_m)$ instead of $(\alpha(P))(a_1,\ldots,a_m)$.
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Fix $a_1,\ldots,a_m \in A^m$. Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$. The image of this ring homomorphism is the $R$-subalgebra of $A$ \vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
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$A$ is \vocab[Algebra!of finite type]{of finite type} if it can be generated by finitely many $a_i \in I$.
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For arbitrary $S \subseteq A$ the subalgebra generated by $S$ is the intersection of all subalgebras containing $S$ \\
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$=$ the union of subalgebras generated by finite $S' \subseteq S$\\
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$= $ the image of $R[X_s | s \in S]$ under $P \mapsto (\alpha(P))(S)$.
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\end{definition}
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\subsection{Finite ring extensions} % LECTURE 2
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\begin{definition}[Finite ring extension]
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Let $R$ be a ring and $A$ an $R$-algebra. $A$ is a module over itself and the ringhomomorphism $R \to A$ allows us to derive an $R$-module structure on $A$.
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$A$ \vocab[Algebra!finite over]{is finite over} $R$ / the $R$-algebra $A$ is finite / $A / R$ is finite if $A$ is finitely generated as an $R$-module.
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\end{definition}
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\begin{fact}[Basic properties of finiteness]
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\begin{enumerate}[A]
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\item Every ring is finite over itself.
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\item A field extension is finite as a ring extension iff it is finite as a field extension.
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\item $A$ finite $\implies$ $A$ of finite type.
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\item $A / R$ and $B / A$ finite $\implies$ $B / R$ finite.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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\begin{enumerate}[A]
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\item $1$ generates $R$ as a module
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\item trivial
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\item Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module. Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
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\item Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
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For every $b$ there exist $\alpha_j \in A$ such that $b = \sum_{j=1}^{n} \alpha_j b_j$. We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some $\rho_{ij} \in R$ thus
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$b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$ and the $a_ib_j$ generate $B$ as an $R$-module.
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\end{enumerate}
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\end{proof}
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\subsection{Determinants and Caley-Hamilton} %LECTURE 2 TODO: move to int. elements?
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This generalizes some facts about matrices to matrices with elements from commutative rings with $1$.
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\footnote{Most of this even works in commutative rings without $ 1$, since $1$ simply can be adjoined.}
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\begin{definition}[Determinant]
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Let $A = (a_{ij}) \Mat(n,n,R)$. We define the determinant by the Leibniz formula \[
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\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}
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\]
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Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$, where $M_{ij}$ is the determinant of the matrix resulting from $A$ after deleting the $i^{\text{th}}$ row and the $j^{\text{th}}$ column.
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\end{definition}
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\begin{fact}
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\begin{enumerate}
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\item $\det(AB) = \det(A)\det(B)$
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\item Development along a row or column works.
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\item Cramer's rule: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible iff $\det(A)$ is a unit.
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\item Caley-Hamilton: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then $P_A(A) = 0$.
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\end{enumerate}
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\end{fact}
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\begin{proof}
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All rules hold for the image of a matrix under a ring homomorphism if they hold for the original matrix. The converse holds in the case of injective ring homomorphisms.
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Caley-Hamilton was shown for algebraically closed fields in LA2 using the Jordan normal form.
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Fields can be embedded into their algebraic closure, thus Caley-Hamilton holds for fields. Every domain can be embedded in its field of quotients $\implies$ Caley-Hamilton holds for domains.
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In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$ where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the morphism $S \to A$ of evaluation defined by $X_{i,j} \mapsto a_{i,j}$. Thus Caley-Hamilton holds in general.
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\end{proof} %TODO: lernen
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\subsection{Integral elements and integral ring extensions} %LECTURE 2
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\begin{proposition}[on integral elements]\label{propinte}
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Let $A$ be an $R$-algebra, $a \in A$. Then the following are equivalent:
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\begin{enumerate}[A]
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\item $\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$
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\item There exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
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\end{enumerate}
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If $a_1, \ldots, a_k \in A$ satisfy these conditions, there is a subalgebra of $A$ finite over $R$ and containing all $a_i$.
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\end{proposition}
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\begin{definition}\label{intclosure}
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Elements that satisfy the conditions from \ref{propinte} are called \vocab{integral over} $R$.
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$A / R$ is \vocab[Algebra!integral]{integral}, if all $a \in A$ are integral over $R$.
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The set of elements of $A$ integral over $R$ is called the \vocab{integral closure} of $R$ in $A$.
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\end{definition}
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\begin{proof}
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\hskip 10pt
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\begin{enumerate}
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{\color{gray} \item[B $\implies$ A] Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$ and finite over $R$.
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Let $(b_i)_{i=1}^{n}$ generate $B$ as an $R$-module.
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\begin{align}
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q: R^n &\longrightarrow B \\
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(r_1,\ldots,r_n) &\longmapsto \sum_{i=1}^{n} r_i b_i
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\end{align}
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is surjective. Thus there are $\rho_{i} = \left( r_{i,j} \right)_{j=1}^n \in R^n$ such that $a b_i = q(\rho_i)$. Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
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Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$. By induction it follows that $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$. Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$ and using Caley-Hamilton, we obtain $P(a) \cdot q(v) = 0$. $P$ is monic. Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$. Thus $P(a) = 0$ and $a$ satisfies A.
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}
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\item[B $\implies$ A] if $R$ is Noetherian.\footnote{This suffices in the exam.}
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Let $a \in A$ satisfy B. Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$. Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$ with $0 \le i < n$. As a finitely generated module over the Noetherian ring $R$, $B$ is a Noetherian $R$-module. Thus the ascending sequence $M_n$ stabilizes at some step $d$ and $a^d \in M_d$. Thus there are $(r_i)_{i=0}^{d-1} \in R^d$ such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
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\item[A $\implies$ B] Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A, i.e. $a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$ with $r_{i,j} \in R$. Let $B \subseteq A$ be the sub-$R$-module generated by $a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$ with $0 \le \alpha_i < d_i$.
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$B$ is closed under $a_1 \cdot $ since \[a_1a^{\alpha} = \begin{cases}
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a^{(\alpha_1 + 1, \alpha')} &\text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1\\
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\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} &\text{if } \alpha_1 = d_1 - 1
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\end{cases}\]
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By symmetry, this hold for all $a_i$. By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$, $B$ is invariant under $a^{\alpha}\cdot $. Since these generate $B$ as an $R$-module, $B$ is multiplicatively closed. Thus A holds. Furthermore we have shown the final assertion of the proposition.
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\end{enumerate}
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\end{proof}
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\begin{corollary}\label{cintclosure}
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\begin{enumerate}
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\item[Q] Every finite $R$-algebra $A$ is integral.
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\item[R] The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$
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\item[S] If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$, then it is integral over $A$.
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\item[T] If $A$ is an integral $R$-algebra and $B$ any $A$-algebra, $b \in B$ integral over $A$, then $b$ is integral over $R$.
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\end{enumerate}
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\end{corollary}
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\begin{proof}
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\begin{enumerate}
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\item[Q] Put $ B = A $ in B.
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\item[R] For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral over $R$.
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From B it follows, that the integral closure is closed under ring operations.
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\item[S] trivial
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\item[T] Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$. Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$, such that all $a_i \in \tilde{A}$.
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$b$ is integral over $\tilde{A} \implies \exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$. Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite, $\tilde{B} / R$ is finite and $b$ satisfies B.
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\end{enumerate}
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\end{proof}
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\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality
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\begin{fact}[Finite type and integral $\implies$ finite]\label{ftaiimplf}
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If $A$ is an integral $R$-algebra of finite type, then it is a finite $R$-algebra.
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\end{fact}
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\begin{proof}
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Let $A $ be generated by $\left( a_i \right) _{i=1}^{n}$ as an $R$- algebra. By the proposition on integral elements (\ref{propinte}), there is a finite $R$-algebra $B \subseteq A$ such that all $a_i \in B$.
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We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
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\end{proof}
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\begin{fact}[Finite type in tower]
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If $A$ is an $R$-algebra of finite type and $B$ an $A$-algebra of finite type, then $B$ is an $R$-algebra of finite type.
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\end{fact}
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\begin{proof}
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If $A / R$ is generated by $(a_i)_{i=1}^m$ and $B / A$ by $(b_j)_{j=1}^{n}$, then $B /R$ is generated by the $b_j$ and the images of the $a_i$ in $B$.
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\end{proof}
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{\color{red}
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\begin{fact}[About integrality and fields] \label{fintaf}
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Let $B$ be a domain integral over its subring $A$. Then $B$ is a field iff $A$ is a field.
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\end{fact}
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}
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\begin{proof}
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Let $B$ be a field and $a \in A \setminus \{0\} $. Then $a^{-1} \in B$ is integral over $A$, hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$ for some $\alpha_i \in A$. Multiplication by $a^{d-1}$ yields
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$a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
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On the other hand, let $B$ be integral over the field $A$. Let $b \in B \setminus \{0\}$. As $B$ is integral over $A$, there is a sub-$A$-algebra $\tilde{B} \subseteq B, b \in \tilde{B}$ finitely generated as an $A$-module, i.e. a finite-dimensional $A$-vector space. Since $B$ is a domain, $\tilde{B} \xrightarrow{b\cdot } \tilde{B}$ is injective, hence surjective, thus $\exists x \in \tilde{B} : b \cdot x \cdot 1$.
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\end{proof}
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\subsection{Noether normalization theorem}
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\begin{lemma}\label{nntechlemma}
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Let $S \subseteq \N^n$ be finite. Then there exists $\vec k \in \N^n$ such that $k_1 =1$ and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$ for $\alpha \neq \beta \in S$,
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where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
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\end{lemma}
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\begin{proof}
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Intuitive:
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For $\alpha \neq \beta$ the equation $w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$ ($\kappa \in \R^{n-1}$)
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defines a codimension $1$ affine hyperplane in $\R^{n-1}$. It is possible to choose $\kappa$ such that all $\kappa_i$ are $> \frac{1}{2}$ and with Euclidean distance $> \frac{\sqrt{n-1} }{2}$ from the union of these hyperplanes. By choosing the closest $\kappa'$ with integral coordinates, each coordinate will be disturbed by at most $\frac{1}{2}$, thus at Euclidean distance $\le \frac{\sqrt{n-1} }{2}$.
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More formally:\footnote{The intuitive version suffices in the exam.}
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Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\} $. We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
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Suppose $\alpha \neq \beta$. Let $i$ be the maximal index such that $\alpha_i \neq \beta_i$. Then the contributions of $\alpha_j$ (resp. $\beta_j$) with $1 \le j < i$ to $w_{\vec k}(\alpha)$ (resp. $w_{\vec k}(\beta)$) cannot undo the difference $k_i(\alpha_i - \beta_i)$.
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\end{proof}
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\begin{theorem}[Noether normalization] \label{noenort}
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Let $K$ be a field and $A$ a $K$-algebra of finite type. Then there are $a = (a_i)_{i=1}^{n} \in A$ which are algebraically independent over $K$, i.e. the ring homomorphism \begin{align}
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\ev_a: K[X_1,\ldots,X_n] &\longrightarrow A \\
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P &\longmapsto P(a_1,\ldots,a_n)
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\end{align}
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is injective. $n$ and the $a_i$ can be chosen such that $A$ is finite over the image of $\ev_a$.
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\end{theorem}
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\begin{proof}
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Let $(a_i)_{i=1}^n$ be a minimal number of elements such that $A$ is integral over its $K$-subalgebra generated by $a_1, \ldots, a_n$. (Such $a_i$ exist, since $A$ is of finite type).
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Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
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If suffices to show that the $a_i$ are algebraically independent.
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Since $A$ is of finite type over $K$ and thus over $\tilde{A}$, by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over $\tilde{A}$.
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Thus we only need to show that the $a_i$ are algebraically independent over $K$.
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Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such that $P(a_1,\ldots,a_n) = 0$. Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and $S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$. For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define $w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
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By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$ such that
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$k_1 = 1$ and for $\alpha \neq \beta \in S$ we have $w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
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Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
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\begin{claim}
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$A$ is integral over the subalgebra $B$ generated by the $b_i$.
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\end{claim}
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\begin{subproof}
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By the transitivity of integrality, it is sufficient to show that the $a_i$ are integral over $B$.
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For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$. Thus it suffices to show this for $a_1$.
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Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots, b_{n-1} + T^{k_n}) \in B[T]$.
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We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$. Hence it suffices to show that the leading coefficient of $Q$ is a unit.
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We have
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\[
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T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} = T^{w_{\vec k}(\alpha)} + \sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
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\]
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with suitable $\beta_{\alpha, l} \in B$.
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By the choice of $\vec k$, we have \[
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Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)} + \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
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\]
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with $q_j \in B$ and $\alpha$ such that $w_{\vec k }(\alpha)$ is maximal subject to the condition $p_\alpha \neq 0$.
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Thus the leading coefficient of $Q$ is a unit.
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\end{subproof}
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This contradicts the minimality of $n$, as $B$ can be generated by $< n$ elements $b_i$.
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\end{proof}
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