fixed some wrong linebreaks (introduced by misuse of latexindent?)
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@ -5,31 +5,29 @@
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Then the following sets coincide
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\begin{enumerate}
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\item
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$\left\{ \sum_{s \in
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S'} r_{s} \cdot s ~ |~ S
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\subseteq S' \text{finite}, r_s \in R, \right\}$
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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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$\bigcap_{\substack{S \subseteq N \subseteq M\\N \text{submodule}}} N$,
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\item
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The $\subseteq$-smallest submodule of $M$ containing $S$
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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!
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Submodule]{submodule of $M $ generated by $S$}.
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If $N= M$ we say that \vocab[Module!
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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
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generated by $S$.
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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 if the
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following equivalent conditions hold:
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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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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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@ -37,9 +35,9 @@
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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, then the polynomial rings
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$R[X_1,\ldots, X_n]$ in
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finitely many variables are Noetherian.
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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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@ -61,8 +59,8 @@
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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, there exists a surjective homomorphism $R^n
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\to M$.
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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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@ -73,49 +71,47 @@
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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 $M \xrightarrow{p}
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M''$ is surjective.
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If $M$ is finitely generated (resp.
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Noetherian), then so is $M''$.
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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 $M' \xrightarrow{f}
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M \xrightarrow{p} M'' \to 0$ be exact.
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If $M'$ and $M ''$ are finitely generated (reps.
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Noetherian), so is $M$.
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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
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sequence.
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If $M$ is generated by $S, |S| < \omega$, then $M''$ is generated by $p(S)$.
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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 $f(M')$ and assume $0 \to M'
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\xrightarrow{f}
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M \xrightarrow{p} M'' \to 0$ to be exact.
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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 $N' \coloneqq
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f^{-1}(N)$ and $N''\coloneqq p(N)$ are finitely
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generated.
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Since $0 \to N' \to N \to N'' \to 0$ is exact, $N$ is finitely generated.
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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 $A$ with a ring homomorphism $R
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\xrightarrow{\alpha} A$.
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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 $\alpha(R) \subseteq A' \subseteq
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A$.\\
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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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@ -123,9 +119,9 @@
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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)
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X^{\beta}
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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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@ -134,33 +130,31 @@
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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!
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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 $a_i
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\in I$.
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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 $S$ \\ $=$ the union of subalgebras
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generated by finite $S' \subseteq S$\\ $= $ the image of
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$R[X_s | s \in S]$
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under $P \mapsto (\alpha(P))(S)$.
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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. $A$ is a
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module over itself and the ringhomomorphism $R \to A$ allows us to derive an
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$R$-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
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finite if $A$ is finitely generated as an $R$-module.
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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 iff it is finite as a field
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extension.
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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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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 and $B$ by
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$b_1,\ldots,b_n$ as an $A$-module.
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For every $b$ there exist $\alpha_j \in A$ such that $b =
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\sum_{j=1}^{n}
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\alpha_j b_j$.
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We have $\alpha_j = \sum_{i=1}^{m} \rho_{ij} a_i$ for some
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$\rho_{ij} \in R$
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thus $b = \sum_{i=1}^{m} \sum_{j=1}^{n} \rho_{ij}
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a_i b_j$ and the $a_ib_j$
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generate $B$ as an $R$-module.
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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$, since $1$ simply can be adjoined.}
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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 $A = (a_{ij})
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\Mat(n,n,R)$.
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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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\[
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\det(A) \coloneqq \sum_{\pi
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\in S_n} \sgn(\pi)
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\prod_{i=1}^{n} a_{i, \pi(i)}
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\det(A) \coloneqq \sum_{\pi \in S_n} \sgn(\pi) \prod_{i=1}^{n} a_{i, \pi(i)}.
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\]
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Define $\text{Adj}(A)$ by $\text{Adj}(A)^{T}_{ij}
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\coloneqq (-1)^{i+j} \cdot
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M_{ij}$, where $M_{ij}$ is the
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determinant of the matrix resulting from $A$
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after deleting the $i^{\text{th}}$ row and the
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$j^{\text{th}}$ column.
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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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$\det(AB) = \det(A)\det(B)$.
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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: $A \cdot \text{Adj}(A) = \text{Adj}(A) \cdot
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A = \det(A) \cdot \mathbf{1}_n$. $A$ is invertible
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iff $\det(A)$ is a unit.
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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: If $P_A = \det(T \cdot \mathbf{1}_n - A)$ \footnote{$T \cdot \mathbf{1}_n -A \in \Mat(n,n,A[T])$}, then
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$P_A(A) = 0$.
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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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@ -234,20 +222,17 @@ commutative rings with $1$.
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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, thus Caley-Hamilton holds
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for fields.
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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
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$(X_{i,j})_{i,j = 1}^{n} \in
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\Mat(n,n,S)$ where
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$S \coloneqq \Z[X_{i,j} | 1 \le i, j \le n]$ (this is a domain) under the
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morphism $S \to A$ of evaluation defined by $X_{i,j}
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\mapsto a_{i,j}$.
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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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%TODO: lernen
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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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@ -256,88 +241,77 @@ commutative rings with $1$.
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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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$\exists n \in
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\N, (r_i)_{i=0}^{n-1}, r_i \in R: a^n =
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\sum_{i=0}^{n-1} r_i a^i$
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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
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exists a subalgebra $B \subseteq A$ finite over $R$ and containing $a$.
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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, there is a subalgebra of
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$A$ finite over $R$ and containing all $a_i$.
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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
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\ref{propinte} are called
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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}, if all $a \in A$ are 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
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closure} of $R$ in $A$.
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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}
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\item[B $\implies$ A]
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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
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$R$-module.
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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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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
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$a b_i = q(\rho_i)$.
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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 $q(P(\mathfrak{A}) \cdot v) = P(a)q(v)$
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for all $P
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\in R[T]$.
|
||||
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.
|
||||
}
|
||||
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.}
|
||||
\item[B $\implies$ A]
|
||||
if $R$ is Noetherian.\footnote{This suffices in the exam.}
|
||||
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 $a^d = \sum_{i=0}^{d-1}
|
||||
r_ia^i$.
|
||||
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 $a^d = \sum_{i=0}^{d-1} r_ia^i$.
|
||||
\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 $0 \le \alpha_i < d_i$.
|
||||
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 $0 \le \alpha_i < d_i$.
|
||||
$B$ is closed under $a_1 \cdot $ since
|
||||
\[
|
||||
a_1a^{\alpha} =
|
||||
\begin{cases}
|
||||
a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1 \\
|
||||
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1
|
||||
a^{(\alpha_1 + 1, \alpha')} & \text{if } \alpha = (\alpha_1, \alpha'), 0 \le \alpha_1 < d_1 - 1, \\
|
||||
\sum_{j=0}^{d_1 - 1} r_{i_1,j} a^{(j, \alpha')} & \text{if } \alpha_1 = d_1 - 1.
|
||||
\end{cases}
|
||||
\]
|
||||
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.
|
||||
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.
|
||||
\end{enumerate}
|
||||
|
@ -348,13 +322,15 @@ commutative rings with $1$.
|
|||
\item[Q]
|
||||
Every finite $R$-algebra $A$ is integral.
|
||||
\item[R]
|
||||
The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$
|
||||
The integral closure of $R$ in $A$ is an $R$-subalgebra of $A$.
|
||||
\item[S]
|
||||
If $A$ is an $R$-algebra, $B$ an $A$-algebra and $b \in B$ integral over $R$,
|
||||
If $A$ is an $R$-algebra,
|
||||
$B$ an $A$-algebra and $b \in B$ integral over $R$,
|
||||
then it is integral over $A$.
|
||||
\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 $R$.
|
||||
If $A$ is an integral $R$-algebra and $B$ any $A$-algebra,
|
||||
$b \in B$ integral over $A$,
|
||||
then $b$ is integral over $R$.
|
||||
\end{enumerate}
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
|
@ -362,136 +338,126 @@ commutative rings with $1$.
|
|||
\item[Q]
|
||||
Put $ B = A $ in B.
|
||||
\item[R]
|
||||
For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$, hence integral
|
||||
over $R$.
|
||||
From B it follows, that the integral closure is closed under ring operations.
|
||||
For every $r \in R$ $\alpha(r)$ is a solution to $T - r = 0$,
|
||||
hence integral over $R$.
|
||||
From B it follows,
|
||||
that the integral closure is closed under ring operations.
|
||||
\item[S]
|
||||
trivial
|
||||
\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 $a_i \in \tilde{A}$.
|
||||
$b$ 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.
|
||||
Then there is a subalgebra $\tilde{A} \subseteq A$ finite over $R$,
|
||||
such that all $a_i \in \tilde{A}$.
|
||||
$b$ is integral over $\tilde{A}$
|
||||
Hence $\exists \tilde{B} \subseteq B$ 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.
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
|
||||
\subsection{Finiteness, finite generation and integrality} %some more remarks on finiteness, finite generation and integrality
|
||||
\subsection{Finiteness, finite generation and integrality}
|
||||
% some more remarks on finiteness, finite generation and integrality
|
||||
|
||||
\begin{fact}[Finite type and integral $\implies$ finite]
|
||||
\label{ftaiimplf}
|
||||
If $A$ is an integral $R$-algebra of finite type, then it is a finite
|
||||
$R$-algebra.
|
||||
If $A$ is an integral $R$-algebra of finite type,
|
||||
then it is a finite $R$-algebra.
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
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 $a_i \in B$.
|
||||
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 $a_i \in B$.
|
||||
We have $B = A$, as $A$ is generated by the $a_i$ as an $R$-algebra.
|
||||
\end{proof}
|
||||
\begin{fact}[Finite type in tower]
|
||||
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.
|
||||
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.
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
If $A / R$ is generated by $(a_i)_{i=1}^m$ and
|
||||
$B / A$ by $(b_j)_{j=1}^{n}$,
|
||||
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 $B$.
|
||||
\end{proof}
|
||||
{\color{red}
|
||||
\begin{fact}[About integrality and fields]
|
||||
\label{fintaf}
|
||||
Let $B$ be a domain integral over its subring $A$.
|
||||
Then $B$ is a field iff $A$ is a field.
|
||||
\end{fact}
|
||||
{
|
||||
\color{red}
|
||||
\begin{fact}[About integrality and fields]
|
||||
\label{fintaf}
|
||||
Let $B$ be a domain integral over its subring $A$.
|
||||
Then $B$ is a field iff $A$ is a field.
|
||||
\end{fact}
|
||||
}
|
||||
\begin{proof}
|
||||
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 $\alpha_i \in A$.
|
||||
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
|
||||
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i
|
||||
a^{d-1-i} \in A$.
|
||||
$a^{-1} = \sum_{i=0}^{d-1} \alpha_i a^{d-1-i} \in A$.
|
||||
|
||||
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 $\exists x \in \tilde{B} : b
|
||||
\cdot x \cdot
|
||||
1$.
|
||||
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 $\exists x \in \tilde{B} : b \cdot x \cdot 1$.
|
||||
\end{proof}
|
||||
\subsection{Noether normalization theorem}
|
||||
\begin{lemma}
|
||||
\label{nntechlemma}
|
||||
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 $\alpha \neq \beta \in S$,
|
||||
where $w_{\vec
|
||||
k}(\alpha) = \sum_{i=1}^{n} k_i
|
||||
\alpha_i$.
|
||||
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 $1$
|
||||
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 $\le
|
||||
\frac{\sqrt{n-1} }{2}$.
|
||||
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\} $.
|
||||
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
|
||||
$k_i(\alpha_i - \beta_i)$.
|
||||
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
|
||||
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)
|
||||
\ev_a: K[X_1,\ldots,X_n] & \longrightarrow A\\
|
||||
P & \longmapsto P(a_1,\ldots,a_n)
|
||||
\end{align*}
|
||||
is
|
||||
injective.
|
||||
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 $a_1, \ldots, a_n$.
|
||||
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.
|
||||
|
@ -547,5 +513,4 @@ commutative rings with $1$.
|
|||
|
||||
This contradicts the minimality of $n$, as $B$ can be generated by $< n$
|
||||
elements $b_i$.
|
||||
|
||||
\end{proof}
|
||||
|
|
|
@ -5,13 +5,12 @@
|
|||
\end{warning}
|
||||
|
||||
\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the
|
||||
MIT-License and can be obtained from \url{https://github.com/kesslermaximilian/LatexPackages}.
|
||||
% TODO
|
||||
MIT-License and can be obtained from
|
||||
\url{https://gitlab.com/latexci/LatexPackages}.
|
||||
\newline
|
||||
|
||||
\noindent $\mathfrak{k}$ is {\color{red} always} an
|
||||
algebraically closed field and $\mathfrak{k}^n$ is equipped with the
|
||||
Zariski-topology.
|
||||
\noindent $\mathfrak{k}$ is {\color{red} always} an algebraically closed field
|
||||
and $\mathfrak{k}^n$ is equipped with the Zariski-topology.
|
||||
Fields which are not assumed to be algebraically closed have been renamed
|
||||
(usually to $\mathfrak{l}$).
|
||||
|
||||
|
|
|
@ -1,64 +1,50 @@
|
|||
Let $\mathfrak{l}$ be any field.
|
||||
\begin{definition}
|
||||
For a $\mathfrak{l}$-vector space $V$, let $\mathbb{P}(V)$ be
|
||||
the set of
|
||||
For a $\mathfrak{l}$-vector space $V$,
|
||||
let $\mathbb{P}(V)$ be the set of
|
||||
one-dimensional subspaces of $V$.
|
||||
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq
|
||||
\mathbb{P}(\mathfrak{l}^{n+1})$, the
|
||||
\vocab[Projective space]{$n$-dimensional projective space over $\mathfrak{l}$}.
|
||||
Let $\mathbb{P}^n(\mathfrak{l}) \coloneqq \mathbb{P}(\mathfrak{l}^{n+1})$,
|
||||
the \vocab[Projective space]%
|
||||
{$n$-dimensional projective space over $\mathfrak{l}$}.
|
||||
|
||||
If $\mathfrak{l}$ is kept fixed, we will often write
|
||||
$\mathbb{P}^n$ for
|
||||
$\mathbb{P}^n(\mathfrak{l})$.
|
||||
If $\mathfrak{l}$ is kept fixed,
|
||||
we will often write $\mathbb{P}^n$ for $\mathbb{P}^n(\mathfrak{l})$.
|
||||
|
||||
When dealing with $\mathbb{P}^n$, the usual convention is to use
|
||||
$0$ as the
|
||||
index of the first coordinate.
|
||||
When dealing with $\mathbb{P}^n$,
|
||||
the usual convention is to use $0$ as the index of the first coordinate.
|
||||
|
||||
We denote the one-dimensional subspace generated by $(x_0,\ldots,x_n) \in
|
||||
\mathfrak{k}^{n+1} \setminus \{0\}$ by
|
||||
We denote the one-dimensional subspace generated by
|
||||
$(x_0,\ldots,x_n) \in \mathfrak{k}^{n+1} \setminus \{0\}$ by
|
||||
$[x_0,\ldots,x_n] \in \mathbb{P}^n$.
|
||||
If $x = [x_0,\ldots,x_n] \in \mathbb{P}^n$, the
|
||||
$(x_{i})_{i=0}^n$ are
|
||||
called
|
||||
\vocab{homogeneous coordinates} of $x$.
|
||||
If $x = [x_0,
|
||||
ldots,x_n] \in \mathbb{P}^n$, the $(x_{i})_{i=0}^n$ are called
|
||||
\vocab{homogeneous coordinates} of $x$.
|
||||
At least one of the $x_{i}$ must be $\neq 0$.
|
||||
\end{definition}
|
||||
\begin{remark}
|
||||
There are points $[1,0],
|
||||
[0,1] \in \mathbb{P}^1$ but there
|
||||
is no point $[0,0]
|
||||
\in \mathbb{P}^1$.
|
||||
There are points $[1,0], [0,1] \in \mathbb{P}^1$ but there
|
||||
is no point $[0,0] \in \mathbb{P}^1$.
|
||||
\end{remark}
|
||||
\begin{definition}[Infinite hyperplane]
|
||||
For $0 \le i \le n$ let $U_i \subseteq
|
||||
\mathbb{P}^n$ denote the set of $[x_0,\ldots,x_{n}]$ with
|
||||
$x_{i}\neq 0$.
|
||||
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$ and
|
||||
$[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
|
||||
same point $x \in
|
||||
\mathbb{P}^n$ differ by scaling with a $\lambda \in
|
||||
\mathfrak{l}^{\times}$,
|
||||
For $0 \le i \le n$ let $U_i \subseteq \mathbb{P}^n$
|
||||
denote the set of $[x_0,\ldots,x_{n}]$ with $x_{i}\neq 0$.
|
||||
This is a correct definition since two different sets $[x_0,\ldots,x_{n}]$
|
||||
and $[\xi_0,\ldots,\xi_n]$ of homogeneous coordinates for the
|
||||
same point $x \in \mathbb{P}^n$ differ by scaling with a
|
||||
$\lambda \in \mathfrak{l}^{\times}$,
|
||||
$x_i = \lambda \xi_i$.
|
||||
Since not all $x_i$ may be $0$, $\mathbb{P}^n =
|
||||
\bigcup_{i=0}^n U_i$.
|
||||
We identify $\mathbb{A}^n =
|
||||
\mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
|
||||
with
|
||||
$U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
|
||||
Since not all $x_i$ may be $0$, $\mathbb{P}^n = \bigcup_{i=0}^n U_i$.
|
||||
We identify $\mathbb{A}^n = \mathbb{A}^n(\mathfrak{l}) = \mathfrak{l}^n$
|
||||
with $U_0$ by identifying $(x_1,\ldots,x_n) \in \mathbb{A}^n$ with
|
||||
$[1,x_1,\ldots,x_n] \in \mathbb{P}^n$.
|
||||
Then $\mathbb{P}^1 = \mathbb{A}^1 \cup \{\infty\} $
|
||||
where $\infty=[0,1]$.
|
||||
More generally, when $n > 0$ $\mathbb{P}^n \setminus
|
||||
\mathbb{A}^n$ can be
|
||||
More generally, when $n > 0$ $\mathbb{P}^n \setminus \mathbb{A}^n$ can be
|
||||
identified with $\mathbb{P}^{n-1}$ identifying
|
||||
$[0,x_1,\ldots,x_n] \in
|
||||
\mathbb{P}^n \setminus \mathbb{A}^n$ with
|
||||
$[x_1,\ldots,x_n] \in
|
||||
\mathbb{P}^{n-1}$.
|
||||
$[0,x_1,\ldots,x_n] \in \mathbb{P}^n \setminus \mathbb{A}^n$ with
|
||||
$[x_1,\ldots,x_n] \in \mathbb{P}^{n-1}$.
|
||||
|
||||
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong
|
||||
\mathfrak{l}^n$ with a copy of
|
||||
Thus $\mathbb{P}^n$ is $\mathbb{A}^n \cong \mathfrak{l}^n$ with a copy of
|
||||
$\mathbb{P}^{n-1}$ added as an \vocab{infinite hyperplane} .
|
||||
\end{definition}
|
||||
|
||||
|
@ -67,40 +53,37 @@ Let $\mathfrak{l}$ be any field.
|
|||
Let $\mathbb{I} = \N$ or $\mathbb{I} = \Z$.
|
||||
\end{notation}
|
||||
\begin{definition}
|
||||
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$ we understand a
|
||||
ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
|
||||
subgroups of the
|
||||
additive group $(A, +)$ such that $A_a \cdot A_b \subseteq
|
||||
A_{a + b}$ for $a,b
|
||||
\in \mathbb{I}$ and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
|
||||
the
|
||||
sense that every $r \in A$ has a unique decomposition $r =
|
||||
\sum_{d \in
|
||||
\mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many $r_d
|
||||
\neq 0$.
|
||||
By an \vocab[Graded ring]{$\mathbb{I}$-graded ring} $A_\bullet$
|
||||
we understand a ring $A$ with a collection $(A_d)_{d \in \mathbb{I}}$ of
|
||||
subgroups of the additive group $(A, +)$
|
||||
such that $A_a \cdot A_b \subseteq A_{a + b}$ for $a,b \in \mathbb{I}$
|
||||
and such that $A = \bigoplus_{d \in \mathbb{I}} A_d$ in
|
||||
the sense that every $r \in A$ has a unique decomposition
|
||||
$r = \sum_{d \in \mathbb{I}} r_d$ with $r_d \in A_d$ and but finitely many
|
||||
$r_d \neq 0$.
|
||||
|
||||
We call the $r_d$ the \vocab{homogeneous components} of $r$.
|
||||
|
||||
An ideal $I \subseteq A$ is called \vocab{homogeneous} if $r \in I
|
||||
\implies
|
||||
\forall d \in \mathbb{I} ~ r_d \in I_d$ where $I_d \coloneqq I
|
||||
\cap A_d$.
|
||||
An ideal $I \subseteq A$ is called \vocab{homogeneous} if
|
||||
$r \in I \implies \forall d \in \mathbb{I} ~ r_d \in I_d$
|
||||
where $I_d \coloneqq I \cap A_d$.
|
||||
|
||||
By a \vocab{graded ring} we understand an $\N$-graded ring.
|
||||
Tin this case, $A_{+} \coloneqq \bigoplus_{d=1}^{\infty}
|
||||
A_d = \{r \in A | r_0
|
||||
= 0\} $ is called the \vocab{augmentation ideal} of $A$.
|
||||
Tin this case,
|
||||
$A_{+} \coloneqq \bigoplus_{d=1}^{\infty} A_d = \{r \in A | r_0 = 0\} $
|
||||
is called the \vocab{augmentation ideal} of $A$.
|
||||
\end{definition}
|
||||
\begin{remark}[Decomposition of $1$]
|
||||
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$ is the
|
||||
decomposition into homogeneous components, then $\varepsilon_a = 1 \cdot
|
||||
\varepsilon_a = \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$ with
|
||||
$\varepsilon_a\varepsilon_b \in A_{a+b}$.
|
||||
If $1 = \sum_{d \in \mathbb{I}} \varepsilon_d$
|
||||
is the decomposition into homogeneous components,
|
||||
then $\varepsilon_a = 1 \cdot \varepsilon_a
|
||||
= \sum_{b \in \mathbb{I}} \varepsilon_a\varepsilon_b$
|
||||
with $\varepsilon_a\varepsilon_b \in A_{a+b}$.
|
||||
By the uniqueness of the decomposition into homogeneous components,
|
||||
$\varepsilon_a \varepsilon_0 = \varepsilon_a$ and $b \neq 0 \implies
|
||||
\varepsilon_a \varepsilon_b = 0$.
|
||||
Applying the last equation with $a = 0$ gives $b\neq 0 \implies \varepsilon_b =
|
||||
\varepsilon_0 \varepsilon _b = 0$.
|
||||
$\varepsilon_a \varepsilon_0 = \varepsilon_a$
|
||||
and $b \neq 0 \implies \varepsilon_a \varepsilon_b = 0$.
|
||||
Applying the last equation with $a = 0$ gives
|
||||
$b\neq 0 \implies \varepsilon_b = \varepsilon_0 \varepsilon _b = 0$.
|
||||
Thus $1 = \varepsilon_0 \in A_0$.
|
||||
\end{remark}
|
||||
\begin{remark}
|
||||
|
@ -123,44 +106,41 @@ Let $\mathfrak{l}$ be any field.
|
|||
\end{proposition}
|
||||
\begin{proof}
|
||||
Most assertions are trivial.
|
||||
We only show that $J$ homogeneous $\implies \sqrt{J} $
|
||||
homogeneous.
|
||||
Let $A$ be $\mathbb{I}$-graded, $f \in \sqrt{J} $ and
|
||||
$f = \sum_{d \in
|
||||
\mathbb{I}} f_d$ the decomposition.
|
||||
To show that all $f_d \in \sqrt{J} $, we use induction on $N_f
|
||||
\coloneqq \# \{d
|
||||
\in \mathbb{I} | f_d \neq 0\}$.
|
||||
We only show that $J$ homogeneous $\implies \sqrt{J}$ homogeneous.
|
||||
Let $A$ be $\mathbb{I}$-graded,
|
||||
$f \in \sqrt{J} $ and
|
||||
$f = \sum_{d \in \mathbb{I}} f_d$ the decomposition.
|
||||
To show that all $f_d \in \sqrt{J} $,
|
||||
we use induction on $N_f \coloneqq \# \{d \in \mathbb{I} | f_d \neq 0\}$.
|
||||
$N_f = 0$ is trivial.
|
||||
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq
|
||||
0$.
|
||||
Suppose $N_f > 0$ and $e \in \mathbb{I}$ is maximal with $f_e \neq 0$.
|
||||
For $l \in \N$, the $le$-th homogeneous component of $f^l$ is $f_e^l$.
|
||||
Choosing $l$ large enough such that $f^l \in J$ and using the homogeneity of
|
||||
$J$, we find $f_e \in \sqrt{J}$.
|
||||
As $\sqrt{J} $ is an ideal, $\tilde f \coloneqq f - f_e \in
|
||||
\sqrt{J} $.
|
||||
As $N_{\tilde f} = N_f -1$, the induction assumption may be
|
||||
applied to $\tilde
|
||||
f$ and shows $f_d \in \sqrt{J} $ for $d \neq e$.
|
||||
As $\sqrt{J} $ is an ideal,
|
||||
$\tilde f \coloneqq f - f_e \in \sqrt{J} $.
|
||||
As $N_{\tilde f} = N_f -1$,
|
||||
the induction assumption may be applied to $\tilde f$
|
||||
and shows $f_d \in \sqrt{J} $ for $d \neq e$.
|
||||
\end{proof}
|
||||
\begin{fact}
|
||||
A homogeneous ideal is finitely generated iff it can be generated by finitely
|
||||
many of its homogeneous elements.
|
||||
A homogeneous ideal is finitely generated
|
||||
iff it can be generated by finitely many of its homogeneous elements.
|
||||
In particular, this is always the case when $A$ is a Noetherian ring.
|
||||
\end{fact}
|
||||
|
||||
|
||||
\subsubsection{The Zariski topology on $\mathbb{P}^n$}
|
||||
\begin{notation}
|
||||
Recall that for $\alpha \in \N^{n+1}$ $|\alpha| =
|
||||
\sum_{i=0}^{n} \alpha_i$ and
|
||||
$x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
|
||||
Recall that for
|
||||
$\alpha \in \N^{n+1}$ $|\alpha| = \sum_{i=0}^{n} \alpha_i$
|
||||
and $x^\alpha = x_0^{\alpha_0} \cdot \ldots \cdot x_n^{\alpha_n}$.
|
||||
\end{notation}
|
||||
\begin{definition}[Homogeneous polynomials]
|
||||
Let $R$ be any ring and $f = \sum_{\alpha \in \N^{n+1}}
|
||||
f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
|
||||
We say that $f$ is \vocab{homogeneous of degree $d$} if $|\alpha| \neq d
|
||||
\implies f_\alpha = 0$ .
|
||||
Let $R$ be any ring and
|
||||
$f = \sum_{\alpha \in \N^{n+1}} f_\alpha X^{\alpha}\in R[X_0,\ldots,X_n]$.
|
||||
We say that $f$ is \vocab{homogeneous of degree $d$}
|
||||
if $|\alpha| \neq d \implies f_\alpha = 0$ .
|
||||
We denote the subset of homogeneous polynomials of degree $d$ by
|
||||
$R[X_0,\ldots,X_n]_d \subseteq R[X_0,\ldots,X_n]$.
|
||||
\end{definition}
|
||||
|
@ -169,57 +149,50 @@ Let $\mathfrak{l}$ be any field.
|
|||
\end{remark}
|
||||
\begin{definition}[Zariski topology on $\mathbb{P}^n(\mathfrak{k})$]
|
||||
\label{ztoppn}
|
||||
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.
|
||||
Let $A = \mathfrak{k}[X_0,\ldots,X_n]$.%
|
||||
\footnote{As always, $\mathfrak{k}$ is algebraically closed}
|
||||
For $f \in A_d = \mathfrak{k}[X_0,\ldots,X_n]_d$, the validity of the equation
|
||||
$f(x_0,\ldots,x_{n}) = 0$ does not depend on the choice of homogeneous
|
||||
coordinates, as
|
||||
For $f \in A_d = \mathfrak{k}[X_0,\ldots,
|
||||
_n]_d$, the validity of the equation $f(x_0,\ldots,x_{n}) = 0$
|
||||
does not depend on the choice of homogeneous coordinates, as
|
||||
\[
|
||||
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d
|
||||
f(x_0,\ldots,x_n)
|
||||
f(\lambda x_0,\ldots, \lambda x_n) 0 \lambda^d f(x_0,\ldots,x_n).
|
||||
\]
|
||||
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x)
|
||||
= 0\}$.
|
||||
Let $\Vp(f) \coloneqq \{x \in \mathbb{P}^n | f(x) = 0\}$.
|
||||
|
||||
We call a subset $X \subseteq \mathbb{P}^n$ Zariski-closed if it
|
||||
can be
|
||||
represented as
|
||||
can be represented as
|
||||
\[
|
||||
X = \bigcap_{i=1}^k \Vp(f_i)
|
||||
\]
|
||||
where the $f_i \in A_{d_i}$
|
||||
are homogeneous polynomials.
|
||||
where the $f_i \in A_{d_i}$ are homogeneous polynomials.
|
||||
\end{definition}
|
||||
\pagebreak
|
||||
\begin{fact}
|
||||
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq
|
||||
\mathbb{P}^n$ is closed, then $Y
|
||||
= X \cap \mathbb{A}^n$ can be identified with the closed subset
|
||||
If $X = \bigcap_{i = 1}^k \Vp(f_i) \subseteq \mathbb{P}^n$ is closed,
|
||||
then $Y = X \cap \mathbb{A}^n$ can be identified with the closed subset
|
||||
\[
|
||||
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n | f_i(1,x_1,\ldots,x_n) = 0, 1
|
||||
\le i \le
|
||||
k\} \subseteq \mathfrak{k}^n
|
||||
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
|
||||
f_i(1,x_1,\ldots,x_n) = 0, 1 \le i \le k\}
|
||||
\subseteq \mathfrak{k}^n.
|
||||
\]
|
||||
Conversely, if $Y \subseteq \mathfrak{k}^n$ is
|
||||
closed it has the form
|
||||
Conversely, if $Y \subseteq \mathfrak{k}^n$ is closed it has the form
|
||||
\[
|
||||
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
|
||||
\{(x_1,\ldots,x_n) \in \mathfrak{k}^n |
|
||||
g_i(x_1,\ldots,x_n) = 0, 1 \le i \le k\}
|
||||
\]
|
||||
and can thus be identified with $X
|
||||
\cap \mathbb{A}^n$ where $X \coloneqq \bigcap_{i=1}^k
|
||||
\Vp(f_i)$ is given by
|
||||
and can thus be identified with $X \cap \mathbb{A}^n$
|
||||
where $X \coloneqq \bigcap_{i=1}^k \Vp(f_i)$ is given by
|
||||
\[
|
||||
f_i(X_0,\ldots,X_n) \coloneqq X_0^{d_i} g_i(X_1 / X_0,\ldots, X_n / X_0), d_i
|
||||
\ge \deg(g_i)
|
||||
\ge \deg(g_i).
|
||||
\]
|
||||
Thus, the Zariski topology on $\mathfrak{k}^n$ can be
|
||||
identified with the topology induced by the Zariski topology on
|
||||
$\mathbb{A}^n =
|
||||
U_0$, and the same holds for $U_i$ with $0 \le i \le n$.
|
||||
$\mathbb{A}^n = U_0$,
|
||||
and the same holds for $U_i$ with $0 \le i \le n$.
|
||||
|
||||
In this sense, the Zariski topology on $\mathbb{P}^n$ can be
|
||||
thought of as
|
||||
In this sense,
|
||||
the Zariski topology on $\mathbb{P}^n$ can be thought of as
|
||||
gluing the Zariski topologies on the $U_i \cong \mathfrak{k}^n$.
|
||||
\end{fact}
|
||||
|
||||
|
@ -227,18 +200,16 @@ Let $\mathfrak{l}$ be any field.
|
|||
|
||||
\begin{definition}
|
||||
Let $I \subseteq A = \mathfrak{k}[X_0,\ldots,X_n]$ be a homogeneous ideal.
|
||||
Let $\Vp(I) \coloneqq
|
||||
\{[x_0,\ldots,_n] \in \mathbb{P}^n |
|
||||
\forall f \in I ~
|
||||
f(x_0,\ldots,x_n) = 0\}$ As $I$ is homogeneous, it is sufficient to impose this
|
||||
condition for the homogeneous elements $f \in I$.
|
||||
Because $A$ is Noetherian, $I$ can finitely generated by homogeneous elements
|
||||
$(f_i)_{i=1}^k$ and
|
||||
$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as
|
||||
in
|
||||
\ref{ztoppn}.
|
||||
Conversely, if the homogeneous $f_i$ are given, then $I = \langle
|
||||
f_1,\ldots,f_k \rangle_A$ is homogeneous.
|
||||
Let $\Vp(I) \coloneqq \{[x_0,\ldots,_n] \in \mathbb{P}^n | \forall f \in I ~%
|
||||
f(x_0,\ldots,x_n) = 0\}$
|
||||
As $I$ is homogeneous,
|
||||
it is sufficient to impose this condition for the homogeneous elements
|
||||
$f \in I$.
|
||||
Because $A$ is Noetherian,
|
||||
$I$ can finitely generated by homogeneous elements $(f_i)_{i=1}^k$ and
|
||||
$\Vp(I)=\bigcap_{i=1}^k \Vp(f_i)$ as in \ref{ztoppn}.
|
||||
Conversely, if the homogeneous $f_i$ are given,
|
||||
then $I = \langle f_1,\ldots,f_k \rangle_A$ is homogeneous.
|
||||
\end{definition}
|
||||
\begin{remark}
|
||||
Note that $V(A) = V(A_+) = \emptyset$.
|
||||
|
@ -247,52 +218,47 @@ Let $\mathfrak{l}$ be any field.
|
|||
For homogeneous ideals in $A$ and $m \in \N$, we have:
|
||||
\begin{itemize}
|
||||
\item
|
||||
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda}
|
||||
\Vp(I_\lambda)$
|
||||
$\Vp(\sum_{\lambda \in \Lambda} I_\lambda) = \bigcap_{\lambda \in \Lambda} \Vp(I_\lambda)$.
|
||||
\item
|
||||
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) =
|
||||
\bigcup_{k=1}^m \Vp(I_k)$
|
||||
$\Vp(\bigcap_{k=1}^m I_k) = \Vp(\prod_{k=1}^{m} I_k) = \bigcup_{k=1}^m \Vp(I_k)$.
|
||||
\item
|
||||
$\Vp(\sqrt{I}) = \Vp(I)$
|
||||
$\Vp(\sqrt{I}) = \Vp(I)$.
|
||||
\end{itemize}
|
||||
\end{fact}
|
||||
\begin{fact}
|
||||
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an
|
||||
open covering of a topological space then $X$ is Noetherian iff there is a
|
||||
finite subcovering and all $U_\lambda$ are Noetherian.
|
||||
If $X = \bigcup_{\lambda \in \Lambda} U_\lambda$ is an open covering
|
||||
of a topological space then $X$ is Noetherian
|
||||
iff there is a finite subcovering and all $U_\lambda$ are Noetherian.
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
By definition, a topological space is Noetherian $\iff$ all open subsets are
|
||||
quasi-compact.
|
||||
By definition, a topological space is Noetherian
|
||||
$\iff$ all open subsets are quasi-compact.
|
||||
\end{proof}
|
||||
\begin{corollary}
|
||||
The Zariski topology on $\mathbb{P}^n$ is indeed a topology.
|
||||
The induced topology on the open set $\mathbb{A}^n =
|
||||
\mathbb{P}^n \setminus
|
||||
\Vp(X_0) \cong \mathfrak{k}^n$ is the Zariski
|
||||
topology on $\mathfrak{k}^n$.
|
||||
The same holds for all $U_i = \mathbb{P}^n \setminus
|
||||
\Vp(X_i) \cong
|
||||
\mathfrak{k}^n$.
|
||||
The induced topology on the open set
|
||||
$\mathbb{A}^n = \mathbb{P}^n \setminus \Vp(X_0) \cong \mathfrak{k}^n$
|
||||
is the Zariski topology on $\mathfrak{k}^n$.
|
||||
The same holds for all
|
||||
$U_i = \mathbb{P}^n \setminus \Vp(X_i) \cong \mathfrak{k}^n$.
|
||||
Moreover, the topological space $\mathbb{P}^n$ is Noetherian.
|
||||
\end{corollary}
|
||||
|
||||
\subsection{Noetherianness of graded rings}
|
||||
\begin{proposition}
|
||||
For a graded ring $R_{\bullet}$, the following conditions
|
||||
are equivalent:
|
||||
For a graded ring $R_{\bullet}$,
|
||||
the following conditions are equivalent:
|
||||
\begin{enumerate}[A]
|
||||
\item
|
||||
$R$ is Noetherian.
|
||||
\item
|
||||
Every homogeneous ideal of $R_{\bullet}$ is finitely
|
||||
generated.
|
||||
Every homogeneous ideal of $R_{\bullet}$ is finitely generated.
|
||||
\item
|
||||
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of homogeneous ideals
|
||||
terminates.
|
||||
Every chain $I_0\subseteq I_1 \subseteq \ldots$ of
|
||||
homogeneous ideals terminates.
|
||||
\item
|
||||
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals has a
|
||||
$\subseteq$-maximal element.
|
||||
Every set $\mathfrak{M} \neq \emptyset$ of homogeneous ideals
|
||||
has a $\subseteq$-maximal element.
|
||||
\item
|
||||
$R_0$ is Noetherian and the ideal $R_+$ is finitely generated.
|
||||
\item
|
||||
|
@ -316,27 +282,27 @@ Let $\mathfrak{l}$ be any field.
|
|||
The $R_0$-subalgebra $\tilde R$ of $R$ generated by the $f_i$ equals $R$.
|
||||
\end{claim}
|
||||
\begin{subproof}
|
||||
It is sufficient to show that every homogeneous $f \in R_d$ belongs to $\tilde
|
||||
R$.
|
||||
It is sufficient to show that every homogeneous $f \in R_d$
|
||||
belongs to $\tilde R$.
|
||||
We use induction on $d$.
|
||||
The case of $d = 0$ is trivial.
|
||||
Let $d > 0$ and $R_e \subseteq \tilde R$ for all $e < d$.
|
||||
as $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
|
||||
Let $f_a = \sum_{i=1}^{k} g_{i, a-d_i} f_i$, where
|
||||
$g_i = \sum_{b=0}^{\infty}
|
||||
g_{i,b}$ is the decomposition into homogeneous
|
||||
components.
|
||||
As $f \in R_+$, $f = \sum_{i=1}^{k} g_if_i$.
|
||||
Let $f_a = \sum_{i=1}^{k} g_{i,
|
||||
a-d_i} f_i$, where $g_i = \sum_{b=0}^{\infty} g_{i,b}$
|
||||
is the decomposition into homogeneous components.
|
||||
Then $f = \sum_{a=0}^{\infty} f_a$ is the decomposition of $f$ into
|
||||
homogeneous
|
||||
components, hence $a \neq d \implies f_a = 0 $.
|
||||
homogeneous components,
|
||||
hence $a \neq d \implies f_a = 0 $.
|
||||
Thus we may assume $g_i \in R_{d-d_i}$.
|
||||
As $d_i > 0$, the induction assumption may now be applied to $g_i$, hence $g_i
|
||||
\in \tilde R$, hence $f \in \tilde R$.
|
||||
As $d_i > 0$,
|
||||
the induction assumption may now be applied to $g_i$,
|
||||
hence $g_i \in \tilde R$,
|
||||
hence $f \in \tilde R$.
|
||||
\end{subproof}
|
||||
|
||||
\noindent\textbf{F $\implies$ A}
|
||||
Hilbert's Basissatz (
|
||||
\ref{basissatz})
|
||||
Hilbert's Basissatz (\ref{basissatz})
|
||||
|
||||
\end{proof}
|
||||
|
||||
|
@ -348,148 +314,130 @@ Let $\mathfrak{l}$ be any field.
|
|||
% Lecture 12
|
||||
\begin{proposition}[Projective form of the Nullstellensatz]
|
||||
\label{hnsp}
|
||||
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$, then
|
||||
$\Vp(I) \subseteq \Vp(f) \iff f \in
|
||||
\sqrt{I}$.
|
||||
If $I \subseteq A$ is a homogeneous ideal and $f \in A_d$ with $d>0$,
|
||||
then $\Vp(I) \subseteq \Vp(f) \iff f \in \sqrt{I}$.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
$\impliedby$ is clear.
|
||||
Let $\Vp(I) \subseteq \Vp(f)$.
|
||||
If $x = (x_0,\ldots,x_n) \in \Va(I)$, then either $x = 0$ in
|
||||
which case $f(x) =
|
||||
0$ since $d > 0$ or the point $[x_0,\ldots,x_n] \in
|
||||
\mathbb{P}^n$ is
|
||||
well-defined and belongs to $\Vp(I) \subseteq
|
||||
\Vp(f)$, hence $f(x) = 0$.
|
||||
Thus $\Va(I) \subseteq \Va(f)$ and $f \in
|
||||
\sqrt{I}$ be the Nullstellensatz
|
||||
(
|
||||
\ref{hns3}).
|
||||
If $x = (x_0,
|
||||
ldots,x_n) \in \Va(I)$, then either $x = 0$ in
|
||||
which case $f(x) = 0$ since $d > 0$
|
||||
or the point $[x_0,\ldots,x_n] \in \mathbb{P}^n$ is
|
||||
well-defined and belongs to $\Vp(I) \subseteq \Vp(f)$,
|
||||
hence $f(x) = 0$.
|
||||
Thus $\Va(I) \subseteq \Va(f)$ and $f \in \sqrt{I}$ be the Nullstellensatz
|
||||
(\ref{hns3}).
|
||||
\end{proof}
|
||||
|
||||
\begin{definition}
|
||||
\footnote{This definition is not too important, the characterization in the following remark suffices.}.
|
||||
For a graded ring $R_\bullet$, let $\Proj(R_\bullet)$ be the set of
|
||||
$\fp \in
|
||||
\Spec R$ such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
|
||||
For a graded ring $R_\bullet$,
|
||||
let $\Proj(R_\bullet)$ be the set of $\fp \in \Spec R$
|
||||
such that $\fp$ is a homogeneous ideal and $\fp \not\supseteq R_+$.
|
||||
\end{definition}
|
||||
\begin{remark}
|
||||
\label{proja}
|
||||
As the elements of $A_0 \setminus \{0\}$ are units in $A$ it follows that for
|
||||
every homogeneous ideal $I$ we have $I \subseteq A_+$ or $I = A$.
|
||||
In particular, $\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ |
|
||||
\fp
|
||||
\text{ is homogeneous}\} $.
|
||||
As the elements of $A_0 \setminus \{0\}$ are units in $A$
|
||||
it follows that for every homogeneous ideal $I$
|
||||
we have $I \subseteq A_+$ or $I = A$.
|
||||
In particular,
|
||||
$\Proj(A_\bullet) = \{\fp \in \Spec A \setminus A_+ | \fp \text{ is homogeneous}\} $.
|
||||
\end{remark}
|
||||
\begin{proposition}
|
||||
\label{bijproj}
|
||||
There is a bijection
|
||||
\begin{align*}
|
||||
f: \{I \subseteq A_+ | I \text{ homogeneous
|
||||
ideal}, I = \sqrt{I}\} & \longrightarrow \{X \subseteq \mathbb{P}^n | X \text{
|
||||
closed}\} \\ I & \longmapsto \Vp(I)\\ \langle \{f \in A_d | d > 0, X
|
||||
\subseteq \Vp(f)\} \rangle & \longmapsfrom X
|
||||
f: \{I \subseteq A_+ | I \text{ homogeneous ideal}, I = \sqrt{I}\} &
|
||||
\longrightarrow \{X \subseteq \mathbb{P}^n | X \text{ closed}\}\\
|
||||
I &
|
||||
\longmapsto \Vp(I)\\
|
||||
\langle \{f \in A_d | d > 0, X \subseteq \Vp(f)\} \rangle &
|
||||
\longmapsfrom X
|
||||
\end{align*}
|
||||
Under this bijection,
|
||||
the irreducible subsets correspond to the elements of
|
||||
$\Proj(A_\bullet)$.
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
From the projective form of the Nullstellensatz it follows that $f$ is
|
||||
injective and that $f^{-1}(\Vp\left( I \right))
|
||||
= \sqrt{I} = I$.
|
||||
If $X \subseteq \mathbb{P}^n$ is closed, then $X =
|
||||
\Vp(J)$ for some homogeneous
|
||||
ideal $J \subseteq A$.
|
||||
From the projective form of the Nullstellensatz it follows
|
||||
that $f$ is injective
|
||||
and that $f^{-1}(\Vp\left( I \right)) = \sqrt{I} = I$.
|
||||
If $X \subseteq \mathbb{P}^n$ is closed,
|
||||
then $X = \Vp(J)$ for some homogeneous ideal $J \subseteq A$.
|
||||
Without loss of generality loss of generality $J = \sqrt{J}$.
|
||||
If $J \not\subseteq A_+$, then $J = A$ (
|
||||
\ref{proja}), hence $X =
|
||||
\Vp(J) =
|
||||
\emptyset = \Vp(A_+)$.
|
||||
If $J \not\subseteq A_+$, then $J = A$ (\ref{proja}),
|
||||
hence $X = \Vp(J) = \emptyset = \Vp(A_+)$.
|
||||
Thus we may assume $J \subseteq A_+$, and $f$ is surjective.
|
||||
|
||||
|
||||
Suppose $\fp \in \Proj(A_\bullet)$.
|
||||
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
|
||||
proven part of
|
||||
the proposition.
|
||||
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets, where
|
||||
$X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k =
|
||||
\sqrt{I_k}$.
|
||||
Then $\fp \neq A_+$ hence $X = \Vp(\fp) \neq \emptyset$ by the
|
||||
proven part of the proposition.
|
||||
Assume $X = X_1 \cup X_2$ is a decomposition into proper closed subsets,
|
||||
where $X_k = \Vp(I_k)$ for some $I_k \subseteq A_+, I_k = \sqrt{I_k}$.
|
||||
Since $X_k$ is a proper subset of $X$, there is $f_k \in I_k \setminus \fp$.
|
||||
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq
|
||||
\Vp(I_k)$ hence $\Vp(f_1f_2)
|
||||
\supseteq \Vp(I_1) \cup \Vp(I_2) = X =
|
||||
\Vp(\fp)$ and it follows that $f_1f_2\in
|
||||
\sqrt{\fp} = \fp \lightning$.
|
||||
We have $\Vp(f_1f_2) \supseteq \Vp(f_k) \supseteq \Vp(I_k)$
|
||||
hence $\Vp(f_1f_2) \supseteq \Vp(I_1) \cup \Vp(I_2) = X = \Vp(\fp)$
|
||||
and it follows that $f_1f_2\in \sqrt{\fp} = \fp \lightning$.
|
||||
|
||||
Assume $X = \Vp(\fp)$ is irreducible, where $\fp =
|
||||
\sqrt{\fp} \in A_+$ is
|
||||
homogeneous.
|
||||
Assume $X = \Vp(\fp)$ is irreducible,
|
||||
where $\fp = \sqrt{\fp} \in A_+$ is homogeneous.
|
||||
The $\fp \neq A_+$ as $X = \emptyset$ otherwise.
|
||||
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i}
|
||||
\setminus \fp$.
|
||||
Assume that $f_1f_2 \in \fp$ but $f_i \not\in A_{d_i} \setminus \fp$.
|
||||
Then $X \not \subseteq \Vp(f_i)$ by the projective
|
||||
Nullstellensatz when $d_i >
|
||||
0$ and because $\Vp(1) = \emptyset$ when $d_i = 0$.
|
||||