516 lines
21 KiB
TeX
516 lines
21 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
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\[\left\{ \sum_{s \in S'} r_{s} \cdot s ~ |~ S \subseteq S' \text{ finite}, r_s \in R\right\},\]
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\item
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\[\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N,\]
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\item
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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
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\vocab[Module!Submodule]{submodule of $M $ generated by $S$}.
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If $N= M$ we say that
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\vocab[Module!generated by subset $S$]{$ M$ is generated by $S$}.
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$M$ is finitely generated
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$:\iff \exists S \subseteq M$ 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
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if the following equivalent conditions hold:
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\begin{enumerate}
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\item
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Every submodule $N \subseteq M$ is finitely generated.
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\item
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Every sequence $N_0 \subset N_1 \subset \ldots$ of submodules terminates.
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\item
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Every set $\mathfrak{M} \neq \emptyset$ of submodules of $M$ has a
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$\subseteq$-largest element.
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\end{enumerate}
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\end{definition}
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\begin{proposition}[Hilbert's Basissatz]
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\label{basissatz}
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If $R$ is a Noetherian ring,
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then the polynomial rings $R[X_1,\ldots, X_n]$
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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
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Every Noetherian module over an arbitrary ring is finitely generated.
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\item
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If $R$ is a Noetherian ring, then an $R$-module is Noetherian iff it is
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finitely generated.
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\item
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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
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By definition, $M$ is a submodule of itself.
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Thus it is finitely generated.
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\item
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Since $M$ is finitely generated,
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there exists a surjective homomorphism $R^n \to M$.
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As $R$ is Noetherian, $R^n$ is Noethrian as well.
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\item
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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
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Suppose $M \xrightarrow{p} M''$ is surjective.
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If $M$ is finitely generated (resp. Noetherian),
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then so is $M''$.
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\item
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Let $M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$ be exact.
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If $M'$ and $M ''$ are finitely generated (reps. Noetherian),
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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
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Consider a sequence $M_0'' \subset M_1'' \subset \ldots \subset M''$.
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Then $p^{-1} M_i''$ yields a strictly ascending sequence.
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If $M$ is generated by $S,
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|S| < \omega$, then $M''$ is generated by $p(S)$.
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\item
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Because of 1.~we can replace $M'$ by $f(M')$
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and assume $0 \to M' \xrightarrow{f} M \xrightarrow{p} M'' \to 0$
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to be exact.
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The fact about finite generation follows from
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Einführung in die Algebra.
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If $M', M''$ are Noetherian, $N \subseteq M$ a submodule,
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then $N' \coloneqq f^{-1}(N)$ and $N''\coloneqq p(N)$
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are finitely generated.
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Since $0 \to N' \to N \to N'' \to 0$ is exact,
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$N$ 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.
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An $R$-algebra $(A, \alpha)$ is a ring $A$
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with a ring homomorphism $R \xrightarrow{\alpha} A$.
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$\alpha$ will usually be omitted.
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In general $\alpha$ is not assumed to be injective.
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\\
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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
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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}
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& \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.
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We will sometimes write $P(a_1,\ldots,a_m)$ instead of
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$(\alpha(P))(a_1,\ldots,a_m)$.
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Fix $a_1,\ldots,a_m \in A^m$.
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Then we get a ring homomorphism $R[X_1,\ldots,X_m] \to A$.
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The image of this ring homomorphism is the $R$-subalgebra of $A$
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\vocab[Algebra!generated subalgebra]{generated by the $a_i$}.
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$A$ is \vocab[Algebra!of finite type]{of finite type}
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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
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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.
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$A$ is a module over itself and the ringhomomorphism $R \to A$
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allows us to derive an $R$-module structure on $A$.
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$A$ \vocab[Algebra!finite over]{is finite over} $R$ /
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the $R$-algebra $A$ is finite / $A / R$ is finite
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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
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Every ring is finite over itself.
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\item
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A field extension is finite as a ring extension
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iff it is finite as a field extension.
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\item
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$A$ finite $\implies$ $A$ of finite type.
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\item
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$A / R$ 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
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$1$ generates $R$ as a module
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\item
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trivial
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\item
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Let $A $ be generated by $a_1,\ldots,a_n$ as an $R$-module.
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Then $A$ is generated by $a_1,\ldots,a_n$ as an $R$-algebra.
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\item
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Let $A$ be generated by $a_1,\ldots,a_m$ as an $R$-module
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and $B$ by $b_1,\ldots,b_n$ as an $A$-module.
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For every $b$ there exist $\alpha_j \in A$
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such that $b = \sum_{j=1}^{n} \alpha_j b_j$.
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We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$
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for some $\rho_{ij} \in R$
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thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij} a_i b_j$
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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
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commutative rings with $1$.
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\footnote{Most of this even works in commutative rings without $ 1$,
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since $1$ simply can be adjoined.}
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\begin{definition}[Determinant]
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Let $A = (a_{ij}) \in \Mat(n,n,R)$.
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We define the determinant by the Leibniz formula
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\[
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\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}.
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\]
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Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij} \coloneqq (-1)^{i+j} \cdot M_{ij}$,
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where $M_{ij}$ is the determinant of the matrix resulting from $A$
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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
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$\det(AB) = \det(A)\det(B)$.
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\item
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Development along a row or column works.
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\item
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Cramer's rule:
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$A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot A = \det(A) \cdot \mathbf{1}_n$.
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$A$ is invertible iff $\det(A)$ is a unit.
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\item
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Caley-Hamilton:
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If $P_A = \det(T \cdot \mathbf{1}_n - A)$%
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\footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$},
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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
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for the original matrix.
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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
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Jordan normal form.
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Fields can be embedded into their algebraic closure,
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thus Caley-Hamilton holds for fields.
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Every domain can be embedded in its field of quotients $\implies$
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Caley-Hamilton holds for domains.
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In general, $A$ is the image of $(X_{i,j})_{i,j = 1}^{n} \in \Mat(n,n,S)$
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where $S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain)
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under the morphism $S \to A$ of evaluation defined by
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$X_{i,j} \mapsto a_{i,j}$.
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Thus Caley-Hamilton holds in general.
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\end{proof}
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\subsection{Integral elements and integral ring extensions} %LECTURE 2
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\begin{proposition}[on integral elements]
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\label{propinte}
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Let $A$ be an $R$-algebra, $a \in A$.
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Then the following are equivalent:
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\begin{enumerate}[A]
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\item
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$\exists n \in \N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n = \sum_{i=0}^{n-1} r_i a^i$.
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\item
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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,
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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}
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\label{intclosure}
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Elements that satisfy the conditions from \ref{propinte} are called
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\vocab{integral over} $R$.
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$A / R$ is \vocab[Algebra!integral]{integral},
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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
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\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]
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Let $a \in A$ such that there is a subalgebra $B \subseteq A$ containing $a$
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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.
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Thus there are $\rho_{i} = \left(r_{i,j}\right)_{j=1}^n \in R^n$
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such that $a b_i = q(\rho_i)$.
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Let $\mathfrak{A}$ be the matrix with the $\rho_i$ as columns.
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Then for all $v \in R^n: q(\mathfrak{A} \cdot v) = a \cdot q(v)$.
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By induction it follows that
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$q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$ for all $P \in R[T]$.
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Applying this to $P(T) = \det(T\cdot \mathbf{1}_n - \mathfrak{A})$
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and using Caley-Hamilton,
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we obtain $P(a) \cdot q(v) = 0$. $P$ is monic.
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Since $q$ is surjective, we find $v \in R^{n} : q(v) = 1$.
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Thus $P(a) = 0$ and $a$ satisfies A.}
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\item[B $\implies$ A]
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if $R$ is Noetherian.\footnote{This suffices in the exam.}
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Let $a \in A$ satisfy B.
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Let $B$ be a subalgebra of $A$ containing $b$ and finite over $R$.
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Let $M_n \subseteq B$ be the $R$-submodule generated by the $a^i$
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with $0 \le i < n$.
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As a finitely generated module over the Noetherian ring $R$,
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$B$ is a Noetherian $R$-module.
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Thus the ascending sequence $M_n$ stabilizes at some step $d$
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and $a^d \in M_d$.
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Thus there are $(r_i)_{i=0}^{d-1} \in R^d$
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such that $a^d = \sum_{i=0}^{d-1} r_ia^i$.
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\item[A $\implies$ B]
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Let $a = (a_i)_{i=1}^n$ where all $a_i$ satisfy A,
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i.e.~$a_i^{d_i} = \sum_{j=0}^{d_i - 1} r_{i,j}a_i^j$
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with $r_{i,j} \in R$.
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Let $B \subseteq A$ be the sub-$R$-module generated by
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$a^\alpha = \prod_{i=1}^n a_i^{\alpha_i}$
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with $0 \le \alpha_i < d_i$.
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$B$ is closed under $a_1 \cdot $ since
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\[
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a_1a^{\alpha} =
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\begin{cases}
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a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1, \\
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\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1.
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\end{cases}
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\]
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By symmetry, this hold for all $a_i$.
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By induction on $|\alpha| = \sum_{i=1}^{n} \alpha_i$,
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$B$ is invariant under $a^{\alpha}\cdot $.
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Since these generate $B$ as an $R$-module,
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$B$ is multiplicatively closed.
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Thus A holds.
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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}
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\label{cintclosure}
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\begin{enumerate}
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\item[Q]
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Every finite $R$-algebra $A$ is integral.
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\item[R]
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The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$.
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\item[S]
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If $A$ is an $R$-algebra,
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$B$ an $A$-algebra and $b \in B$ integral over $R$,
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then it is integral over $A$.
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\item[T]
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If $A$ is an integral $R$-algebra and $B$ any $A$-algebra,
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$b \in B$ integral over $A$,
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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]
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Put $ B = A $ in B.
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\item[R]
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For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$,
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hence integral over $R$.
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From B it follows,
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that the integral closure is closed under ring operations.
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\item[S]
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trivial
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\item[T]
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Let $b \in B$ such that $b^n = \sum_{i=0}^{n-1} a_ib^{i}$.
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Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$,
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such that all $a_i \in \tilde{A}$.
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$b$ is integral over $\tilde{A}$
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Hence $\exists \tilde{B} \subseteq B$ finite over $\tilde{A}$ and $b \in \tilde{B}$.
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Since $\tilde{B} / \tilde{A} $ and $\tilde{A} / R$ are finite,
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$\tilde{B} / R$ 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}
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% some more remarks on finiteness, finite generation and integrality
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\begin{fact}[Finite type and integral $\implies$ finite]
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\label{ftaiimplf}
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If $A$ is an integral $R$-algebra of finite type,
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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.
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By the proposition on integral elements (\ref{propinte}),
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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
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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 $(b_j)_{j=1}^{n}$,
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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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{
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\color{red}
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\begin{fact}[About integrality and fields]
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\label{fintaf}
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Let $B$ be a domain integral over its subring $A$.
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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\} $.
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Then $a^{-1} \in B$ is integral over $A$,
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hence $a^{-d} = \sum_{i=0}^{d-1} \alpha_i a^{-i}$
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for some $\alpha_i \in A$.
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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 $A$.
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Let $b \in B \setminus \{0\}$.
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As $B$ is integral over $A$,
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there is a sub-$A$-algebra $\tilde{B} \subseteq B$,
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$b \in \tilde{B}$ finitely generated as an $A$-module,
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i.e.~a finite-dimensional $A$-vector space.
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Since $B$ is a domain,
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$\tilde{B} \xrightarrow{b\cdot } \tilde{B}$
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is injective,
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hence surjective,
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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}
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\label{nntechlemma}
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Let $S \subseteq \N^n$ be finite.
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Then there exists $\vec k \in \N^n$ such that $k_1 =1$
|
|
and $w_{\vec k}(\alpha) \neq w_{\vec k}(\beta)$
|
|
for $\alpha \neq \beta \in S$,
|
|
where $w_{\vec k}(\alpha) = \sum_{i=1}^{n} k_i \alpha_i$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Intuitive:
|
|
For $\alpha \neq \beta$ the equation
|
|
$w_{(1, \vec \kappa)}(\alpha) = w_{(1, \vec \kappa)}(\beta)$
|
|
($\kappa \in \R^{n-1}$) 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 $\le \frac{\sqrt{n-1} }{2}$.
|
|
|
|
More formally:\footnote{The intuitive version suffices in the exam.}
|
|
Define $M \coloneqq \max \{\alpha_i | \alpha \in S, 1 \le i \le n\}$.
|
|
We can choose $k$ such that $k_i > (i-1) M k_{i-1}$.
|
|
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 $k_i(\alpha_i - \beta_i)$.
|
|
\end{proof}
|
|
|
|
\begin{theorem}[Noether normalization]
|
|
\label{noenort}
|
|
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*}
|
|
\ev_a: K[X_1,\ldots,X_n] & \longrightarrow A\\
|
|
P & \longmapsto P(a_1,\ldots,a_n)
|
|
\end{align*}
|
|
is injective.
|
|
$n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
|
|
$\ev_a$.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
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).
|
|
Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
|
|
If suffices to show that the $a_i$ are algebraically independent.
|
|
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 $\tilde{A}$.
|
|
Thus we only need to show that the $a_i$ are algebraically independent
|
|
over $K$.
|
|
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
|
|
$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
|
|
|
|
By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
|
|
such that $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have
|
|
$w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
|
|
|
|
Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
|
|
\begin{claim}
|
|
$A$ is integral over the subalgebra $B$ generated by the $b_i$.
|
|
\end{claim}
|
|
\begin{subproof}
|
|
By the transitivity of integrality,
|
|
it is sufficient to show that the $a_i$ are integral over $B$.
|
|
For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
|
|
Thus it suffices to show this for $a_1$.
|
|
Define
|
|
$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
|
|
b_{n-1} + T^{k_n}) \in B[T]$.
|
|
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.
|
|
|
|
We have
|
|
\[
|
|
T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
|
|
T^{w_{\vec k}(\alpha)} +
|
|
\sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
|
|
\]
|
|
with suitable $\beta_{\alpha, l} \in B$.
|
|
|
|
By the choice of $\vec k$, we have
|
|
\[
|
|
Q(T) = p_{\alpha} T^{w_{\vec k}(\alpha)}
|
|
+ \sum_{j=0}^{w_{\vec k}(\alpha) - 1} q_j T^j
|
|
\]
|
|
with $q_j \in B$ and
|
|
$\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
|
|
to the condition $p_\alpha \neq 0$.
|
|
Thus the leading coefficient of $Q$ is a unit.
|
|
\end{subproof}
|
|
|
|
This contradicts the minimality of $n$, as $B$ can be generated by $< n$
|
|
elements $b_i$.
|
|
\end{proof}
|