some small changes
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8 changed files with 270 additions and 339 deletions
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@ -1,4 +1,5 @@
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\documentclass[10pt,ngerman,a4paper, fancyfoot, git]{mkessler-script}
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\documentclass[english, fancyfoot% , git
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]{mkessler-script}
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\course{Algebra I}
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\lecturer{Prof.~Dr.~Jens Franke}
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@ -1,7 +1,7 @@
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\ProvidesPackage{algebra}[2022/02/10 - Style file for notes of Algebra I]
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\RequirePackage{mkessler-math}
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\RequirePackage[english]{mkessler-math}
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\RequirePackage{mkessler-refproof}
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\RequirePackage[number in = section]{fancythm}
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@ -11,7 +11,7 @@
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\RequirePackage[utf8x]{inputenc}
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\RequirePackage{babel}
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\RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
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% \RequirePackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
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\RequirePackage[normalem]{ulem}
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\RequirePackage{pdflscape}
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@ -122,11 +122,11 @@
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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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\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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\end{align}
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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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@ -285,10 +285,10 @@ commutative rings with $1$.
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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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\begin{align}
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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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\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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@ -478,10 +478,10 @@ commutative rings with $1$.
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Then there are $a = (a_i)_{i=1}^{n} \in A$ which
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are algebraically independent
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over $K$, i.e. the ring homomorphism
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\begin{align}
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\begin{align*}
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\ev_a: K[X_1,\ldots,X_n] &
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\longrightarrow A \\ P & \longmapsto P(a_1,\ldots,a_n)
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\end{align}
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\end{align*}
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is
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injective.
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$n$ and the $a_i$ can be chosen such that $A$ is finite over the image of
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@ -496,53 +496,41 @@ commutative rings with $1$.
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Let $\tilde{A}$ be the $K$-subalgebra generated by the $a_i$.
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If suffices to show that the $a_i$ are algebraically independent.
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Since $A$ is of finite type over $K$ and thus over $\tilde{A}$,
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by fact
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\ref{ftaiimplf} (integral and finite type $\implies$ finite) $A$ is finite over
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$\tilde{A}$.
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Thus we only need to show that the $a_i$ are algebraically independent over
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$K$.
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Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\} $ such
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that
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$P(a_1,\ldots,a_n) = 0$.
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by fact \ref{ftaiimplf} (integral and finite type $\implies$ finite),
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$A$ is finite over $\tilde{A}$.
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Thus we only need to show that the $a_i$ are algebraically independent
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over $K$.
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Assume there is $P \in K[X_1,\ldots,X_n] \setminus \{0\}$
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such that $P(a_1,\ldots,a_n) = 0$.
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Let $P = \sum_{\alpha \in \N^n} p_\alpha X^{\alpha}$ and
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$S = \{ \alpha \in
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\N^n | p_\alpha \neq 0\}$.
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For $\vec{k} = (k_i)_{i=1}^{n}
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\in \N^n$ and $\alpha \in \N^n$ we define
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$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n}
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k_i\alpha_i$.
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$S = \{ \alpha \in \N^n | p_\alpha \neq 0\}$.
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For $\vec{k} = (k_i)_{i=1}^{n} \in \N^n$ and $\alpha \in \N^n$ we define
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$w_{\vec{k}}(\alpha) \coloneqq \sum_{i=1}^{n} k_i\alpha_i$.
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By
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\ref{nntechlemma} it is possible to choose $\vec{k}
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\in \N^n$ such that $k_1
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= 1$ and for $\alpha \neq \beta \in S$ we have
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$w_{\vec{k}}(\alpha) \neq
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w_{\vec{k}}(\beta)$.
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By \ref{nntechlemma} it is possible to choose $\vec{k} \in \N^n$
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such that $k_1 = 1$ and for $\alpha \neq \beta \in S$ we have
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$w_{\vec{k}}(\alpha) \neq w_{\vec{k}}(\beta)$.
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Define $b_i \coloneqq a_{i+1} -
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a^{k_{i+1}}_1$ for $1 \le i < n$.
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Define $b_i \coloneqq a_{i+1} - a^{k_{i+1}}_1$ for $1 \le i < n$.
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\begin{claim}
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$A$ is integral over the subalgebra $B$ generated by the $b_i$.
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\end{claim}
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\begin{subproof}
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By the transitivity of integrality, it is sufficient to show that the $a_i$ are
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integral over $B$.
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By the transitivity of integrality,
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it is sufficient to show that the $a_i$ are integral over $B$.
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For $i > 1$ we have $a_i = b_{i-1} + a_1^{k_i}$.
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Thus it suffices to show this for $a_1$.
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Define $Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
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b_{n-1} + T^{k_n}) \in
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B[T]$.
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Define
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$Q(T) \coloneqq P(T, b_1 + T^{k_2}, \ldots,
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b_{n-1} + T^{k_n}) \in B[T]$.
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We have $0 = P(a_1,\ldots, a_n) = Q(a_1)$.
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Hence it suffices to show that the leading coefficient of $Q$ is a unit.
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We have
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\[
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T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}}
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=
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T^{\alpha_1} \prod_{i=1}^{n-1} (b_i + T^{k_i + 1})^{\alpha_{i+1}} =
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T^{w_{\vec k}(\alpha)} +
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\sum_{l = 0}^{w_{\vec k}(\alpha) - 1}
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\beta_{\alpha,
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l} T^l
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\sum_{l = 0}^{w_{\vec k}(\alpha) - 1} \beta_{\alpha, l} T^l
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\]
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with suitable $\beta_{\alpha, l} \in B$.
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@ -553,8 +541,7 @@ commutative rings with $1$.
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\]
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with $q_j \in B$ and
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$\alpha$ such that $w_{\vec k}(\alpha)$ is maximal subject
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to the condition
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$p_\alpha \neq 0$.
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to the condition $p_\alpha \neq 0$.
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Thus the leading coefficient of $Q$ is a unit.
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\end{subproof}
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@ -562,4 +549,3 @@ commutative rings with $1$.
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elements $b_i$.
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\end{proof}
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@ -6,21 +6,19 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
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\begin{definition}[zero]
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$x \in \mathfrak{k}^n$ is \vocab[Ideal!zero]{a zero of $I$}
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if $\forall x \in I: P(x) = 0$.
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Let $\Va(I)$ denote the set of zeros if $I$ in
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$\mathfrak{k}^n$.
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Let $\Va(I)$ denote the set of zeros if $I$ in $\mathfrak{k}^n$.
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The \vocab[Ideal!
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zero]{zero in a field extension $\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
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The \vocab[Ideal!zero]{zero in a field extension %
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$\mathfrak{i}$ of $\mathfrak{k}$} is defined similarly.
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\end{definition}
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\begin{remark}[Set of zeros and generators]
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Let $I$ be generated by $S$.
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Then $\{x \in R | \forall s \in S: s(x) = 0\} = \Va(I)$.
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Thus zero sets of ideals correspond to solutions sets to systems of polynomial
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equations.
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Thus zero sets of ideals correspond to solutions sets to systems of
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polynomial equations.
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If $S, \tilde{S}$ generate the same ideal $I$ they have the same
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set of
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solutions.
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set of solutions.
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Therefore we only consider zero sets of ideals.
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\end{remark}
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@ -32,16 +30,15 @@ Let $\mathfrak{k}$ be a field, $R \coloneqq
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\end{theorem}
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\begin{remark}
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Will be shown later (see proof of
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\ref{hns1b}).
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Trivial if $n = 1$: $R$ is a PID, thus $I = pR$ for some $p \in R$.
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Will be shown later (see proof of \ref{hns1b}).
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It is trivial if $n = 1$:
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$R$ is a PID, thus $I = pR$ for some $p \in R$.
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Since $I \neq R$ $p = 0$ or $P$ is non-constant.
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$\mathfrak{k}$ algebraically closed $\leadsto$ there exists a zero of
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$p$.\\
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$\mathfrak{k}$ algebraically closed
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$\leadsto$ there exists a zero of $p$.\\
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If $\mathfrak{k}$ is not algebraically closed and $n > 0$, the
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theorem fails
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(consider $I = p(X_1) R$).
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If $\mathfrak{k}$ is not algebraically closed and $n > 0$,
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the theorem fails (consider $I = p(X_1) R$).
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\end{remark}
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Equivalent\footnote{used in a vague sense here} formulation:
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\begin{proof}
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\begin{itemize}
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\item[$\implies$]
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If $(l_i)_{i=1}^{m}$ is a base of $L$ as a
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$K$-vector space, then $L$ is generated by the $l_i$ as a $K$-algebra.
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If $(l_i)_{i=1}^{m}$ is a base of $L$ as a $K$-vector space,
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then $L$ is generated by the $l_i$ as a $K$-algebra.
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\item[$\impliedby$ ]
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Apply the Noether normalization theorem (
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\ref{noenort}) to $A = L$.
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Apply the Noether normalization theorem (\ref{noenort}) to $A = L$.
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This yields an injective ring homomorphism $\ev_a:
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K[X_1,\ldots,X_n] \to A$
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such that $A$ is finite over the image of $\ev_a$.
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By the fact about integrality and fields (
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\ref{fintaf}), the
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isomorphic image
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By the fact about integrality and fields (\ref{fintaf}),
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the isomorphic image
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of $\ev_a$ is a field.
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Thus $K[X_1,\ldots, X_n]$ is a field $\implies n = 0$.
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Thus $L / K$ is a finite ring extension, hence a finite field extension.
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Thus $L / K$ is a finite ring extension,
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hence a finite field extension.
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\end{itemize}
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\end{proof}
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\begin{remark}
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We will see several additional proofs of this theorem.
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See
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\ref{hns2unc} and
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\ref{rfuncnft}.
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See \ref{hns2unc} and \ref{rfuncnft}.
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All will be accepted in the exam.
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\ref{hns3} and
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\ref{hnsp} are closely related.
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\ref{hns3} and \ref{hnsp} are closely related.
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\end{remark}
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\begin{theorem}[Hilbert's Nullstellensatz (1b)]
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\label{hns1b}
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Let $\mathfrak{l}$ be a field and $I \subset R =
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\mathfrak{l}[X_1,\ldots,X_m]$
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a proper ideal.
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Then there are a finite field extension $\mathfrak{i}$ of
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$\mathfrak{l}$ and a
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zero of $I$ in $\mathfrak{i}^m$.
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Let $\mathfrak{l}$ be a field and
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$I \subset R = \mathfrak{l}[X_1,\ldots,X_m]$ a proper ideal.
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Then there are a finite field extension $\mathfrak{i}$ of $\mathfrak{l}$
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and a zero of $I$ in $\mathfrak{i}^m$.
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\end{theorem}
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\begin{proof}
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(HNS2 (
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\ref{hns2}) $\implies$ HNS1b (
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\ref{hns1b}))
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$I \subseteq \mathfrak{m}$ for some maximal ideal. $R /
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\mathfrak{m}$ is a field, since $\mathfrak{m}$ is maximal.
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$R / \mathfrak{m}$ is of finite type, since the images of the $X_i$
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generate it as a $\mathfrak{l}$-algebra.
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There are thus a field extension $\mathfrak{i} /
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\mathfrak{l}$ and an
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isomorphism $R / \mathfrak{m} \xrightarrow{\iota}
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\mathfrak{i}$ of
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(HNS2 (\ref{hns2}) $\implies$ HNS1b (\ref{hns1b}))
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$I \subseteq \mathfrak{m}$ for some maximal ideal.
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$R /\mathfrak{m}$ is a field,
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since $\mathfrak{m}$ is maximal.
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$R / \mathfrak{m}$ is of finite type,
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since the images of the $X_i$ generate it as a $\mathfrak{l}$-algebra.
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There are thus a field extension $\mathfrak{i} / \mathfrak{l}$
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and an isomorphism $R / \mathfrak{m} \xrightarrow{\iota} \mathfrak{i}$ of
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$\mathfrak{l}$-algebras.
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By HNS2 (
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\ref{hns2}), $\mathfrak{i} /
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\mathfrak{l}$ is a finite field
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extension.
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By HNS2 (\ref{hns2}),
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$\mathfrak{i} / \mathfrak{l}$ is a finite field extension.
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Let $x_i \coloneqq \iota (X_i \mod \mathfrak{m})$.
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\[
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P(x_1,\ldots,x_m) = \iota(P \mod \mathfrak{m})
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\]
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Both sides are morphisms $R \to \mathfrak{i}$ of
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$\mathfrak{l}$-algebras.
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Both sides are morphisms $R \to \mathfrak{i}$ of $\mathfrak{l}$-algebras.
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For for $P = X_i$ the equality is trivial.
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It follows in general, since the $X_i$ generate $R$ as a
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$\mathfrak{l}$-algebra.
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Thus $(x_1,\ldots,x_m)$ is a zero of $I$ (since $P \mod \mathfrak{m}
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= 0$ for
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$P \in I \subseteq \mathfrak{m}$).
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HNS1 (
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\ref{hns1}) can easily be derived from HNS1b.
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Thus $(x_1,\ldots,x_m)$ is a zero of $I$
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(since $P \mod \mathfrak{m} = 0$ for $P \in I \subseteq \mathfrak{m}$).
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HNS1 (\ref{hns1}) can easily be derived from HNS1b.
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\end{proof}
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\subsubsection{Nullstellensatz for uncountable fields} % from lecture 5 Yet another proof of the Nullstellensatz
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\subsubsection{Nullstellensatz for uncountable fields}
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% from lecture 5 Yet another proof of the Nullstellensatz
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The following proof of the Nullstellensatz only works for uncountable fields,
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but will be accepted in the exam.
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\begin{lemma}
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\label{dimrfunc}
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If $K$ is an uncountable field, then $\dim_K K(T)$ is uncountable.
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@ -140,23 +123,20 @@ but will be accepted in the exam.
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K\right\} $ is $K$-linearly independent.
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It follows that $\dim_K K(T) \ge \#S > \aleph_0$.
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Suppose
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$(x_{\kappa})_{\kappa \in K}$ is a
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selection of coefficients from $K$
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such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}
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$ is finite and
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Suppose $(x_{\kappa})_{\kappa \in K}$ is a selection
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of coefficients from $K$
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such that $I \coloneqq \{\kappa \in K | x_{\kappa} \neq 0\}$
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is finite and
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\[
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g \coloneqq \sum_{\kappa \in K} \frac{x_\kappa}{T-\kappa} = 0
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\]
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Let $d
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\coloneqq \prod_{\kappa \in I} (T - \kappa) $.
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Let $d \coloneqq \prod_{\kappa \in I} (T - \kappa) $.
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Then for $\lambda \in I$ we have
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\[
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0 = (dg)(\lambda) = x_\lambda \prod_{\kappa
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\in I \setminus \{\lambda\} } (\lambda - \kappa)
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0 = (dg)(\lambda) =
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x_\lambda \prod_{\kappa \in I \setminus \{\lambda\}}(\lambda - \kappa).
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\]
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This is a contradiction as
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$x_\lambda \neq 0$.
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This is a contradiction as $x_\lambda \neq 0$.
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\end{proof}
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\begin{theorem}[Hilbert's Nullstellensatz for uncountable fields]
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|
@ -165,78 +145,64 @@ but will be accepted in the exam.
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type as a $K$-algebra, then this field extension is finite.
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\end{theorem}
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\begin{proof}
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If $(x_i)_{i=1}^{n}$ generate $L$ as an
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$K$-algebra, then the countably many
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monomials $x^{\alpha} = \prod_{i = 1}^{n}
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x_i^{\alpha_i} $ in the $x_i$ with
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$\alpha \in \N^n$ generate $L$ as a $K$-vector space.
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Thus $\dim_K L \le \aleph_0$ and the same holds for any intermediate field $K
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\subseteq M \subseteq L$ .
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If $l \in L$ is transcendent over $K$ and $M = K(l)$, then $M \cong K(T)$ has
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uncountable dimension by
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\ref{dimrfunc}.
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If $(x_i)_{i=1}^{n}$ generate $L$ as an $K$-algebra,
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then the countably many monomials
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$x^{\alpha} = \prod_{i = 1}^{n} x_i^{\alpha_i}$
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in the $x_i$ with $\alpha \in \N^n$
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generate $L$ as a $K$-vector space.
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Thus $\dim_K L \le \aleph_0$
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and the same holds for any intermediate field $K \subseteq M \subseteq L$.
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If $l \in L$ is transcendent over $K$ and $M = K(l)$,
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then $M \cong K(T)$ has uncountable dimension by \ref{dimrfunc}.
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Thus $L / K$ is algebraic, hence integral, hence finite
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(
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\ref{ftaiimplf}).
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(\ref{ftaiimplf}).
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\end{proof}
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\subsection{The Zariski topology}
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\subsubsection{Operations on ideals and \texorpdfstring{$\Va\left( I \right) $}{V(I)}}
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Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in
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\Lambda$.
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Let $R$ be a ring and $I,J, I_\lambda \subseteq R$ ideals, $\lambda \in \Lambda$.
|
||||
\begin{definition}[Radical, product and sum of ideals]
|
||||
\[
|
||||
\sqrt{I} \coloneqq \bigcap_{n=0} ^{\infty} \{ f \in R | f^n \in
|
||||
I\}
|
||||
\sqrt{I} \coloneqq \bigcap_{n=0}^{\infty} \{f \in R | f^n \in I\}
|
||||
\]
|
||||
\[
|
||||
I \cdot J \coloneqq \langle\{ i \cdot j | i \in I , j \in J\}\rangle_R
|
||||
\]
|
||||
|
||||
\[
|
||||
\sum_{\lambda \in \Lambda}
|
||||
I_\lambda \coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda | \Lambda'
|
||||
\subseteq \Lambda \text{ finite}\right\}
|
||||
\sum_{\lambda \in \Lambda} I_\lambda
|
||||
\coloneqq \left\{\sum_{\lambda \in \Lambda'} i_\lambda |
|
||||
\Lambda' \subseteq \Lambda \text{ finite}\right\}
|
||||
\]
|
||||
\end{definition}
|
||||
\begin{fact}
|
||||
The
|
||||
radical is an ideal in $R$ and $\sqrt{\sqrt{I} } =
|
||||
\sqrt{I}$.
|
||||
\\
|
||||
The radical is an ideal in $R$ and $\sqrt{\sqrt{I} } = \sqrt{I}$.\\
|
||||
$I \cdot J$ is an ideal.\\
|
||||
$\sum_{\lambda \in \Lambda}
|
||||
I_\lambda$ coincides with the ideal generated by $\bigcap_{\lambda \in
|
||||
\Lambda}
|
||||
I_\lambda$ in $R$.
|
||||
\\
|
||||
$\bigcap_{\lambda \in \Lambda}
|
||||
I_\lambda$ is an ideal.
|
||||
$\sum_{\lambda \in \Lambda} I_\lambda$ coincides with
|
||||
the ideal generated by $\bigcap_{\lambda \in \Lambda} I_\lambda$ in $R$.\\
|
||||
$\bigcap_{\lambda \in \Lambda} I_\lambda$ is an ideal.
|
||||
\end{fact}
|
||||
|
||||
Let $R = \mathfrak{k}[X_1,\ldots,X_n]$ where $\mathfrak{k}$ is an
|
||||
algebraically
|
||||
closed field.
|
||||
algebraically closed field.
|
||||
|
||||
\begin{fact}
|
||||
\label{fvop}
|
||||
Let $I, J,
|
||||
(I_{\lambda})_{\lambda \in \Lambda}$ be
|
||||
Let $I, J, (I_{\lambda})_{\lambda \in \Lambda}$ be
|
||||
ideals in $R$.
|
||||
$\Lambda$ may be infinite.
|
||||
\begin{enumerate}[A]
|
||||
\item
|
||||
$\Va(I) = \Va(\sqrt{I})$
|
||||
$\Va(I) = \Va(\sqrt{I})$,
|
||||
\item
|
||||
$\sqrt{J} \subseteq \sqrt{I} \implies
|
||||
\Va(I) \subseteq \Va(J)$
|
||||
$\sqrt{J} \subseteq \sqrt{I} \implies \Va(I) \subseteq \Va(J)$,
|
||||
\item
|
||||
$\Va(R) = \emptyset, \Va(\{0\} =\mathfrak{k}^n$
|
||||
$\Va(R) = \emptyset, \Va(\{0\} = \mathfrak{k}^n$,
|
||||
\item
|
||||
$\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$
|
||||
$\Va(I \cap J) = \Va(I\cdot J) = \Va(I) \cup \Va(J)$,
|
||||
\item
|
||||
$\Va(\sum_{\lambda \in \Lambda}
|
||||
I_\lambda) = \bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
|
||||
$\Va(\sum_{\lambda \in \Lambda} I_\lambda) =
|
||||
\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$.
|
||||
\end{enumerate}
|
||||
\end{fact}
|
||||
\begin{proof}
|
||||
|
@ -244,46 +210,44 @@ closed field.
|
|||
\item[A-C]
|
||||
trivial
|
||||
\item[D]
|
||||
$I \cdot
|
||||
J \subseteq I \cap J \subseteq I$.
|
||||
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq
|
||||
\Va(I \cdot J)$.
|
||||
By symmetry we have $\Va(I) \cup \Va(J)
|
||||
\subseteq \Va(I \cap J) \subseteq \Va(I
|
||||
\cdot J)$.
|
||||
$I \cdot J \subseteq I \cap J \subseteq I$.
|
||||
Thus $\Va(I) \subseteq \Va(I \cap J) \subseteq \Va(I \cdot J)$.
|
||||
By symmetry we have
|
||||
$\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
|
||||
\subseteq \Va(I \cdot J)$.
|
||||
Let $x \not\in \Va(I) \cup \Va(J)$.
|
||||
Then there are $f \in I, g \in J$ such that $f(x) \neq 0, g(x) \neq 0$ thus
|
||||
$(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$.
|
||||
Then there are $f \in I, g \in J$
|
||||
such that $f(x) \neq 0, g(x) \neq 0$
|
||||
thus $(f \cdot g)(x) \neq 0 \implies x \not\in \Va(I\cdot J)$.
|
||||
Therefore
|
||||
\[
|
||||
\Va(I) \cup \Va(J) \subseteq
|
||||
\Va(I \cap J) \subseteq \Va(I \cdot
|
||||
J) \subseteq
|
||||
\Va(I) \cup \Va(J)
|
||||
\Va(I) \cup \Va(J) \subseteq \Va(I \cap J)
|
||||
\subseteq \Va(I \cdot J)
|
||||
\subseteq \Va(I) \cup \Va(J).
|
||||
\]
|
||||
\item[E]
|
||||
$I_\lambda \subseteq \sum_{\lambda
|
||||
\in \Lambda} I_\lambda \implies
|
||||
\Va(\sum_{\lambda \in \Lambda} I_\lambda)
|
||||
$I_\lambda \subseteq \sum_{\lambda \in \Lambda} I_\lambda
|
||||
\implies \Va(\sum_{\lambda \in \Lambda} I_\lambda)
|
||||
\subseteq \Va(I_\lambda)$.
|
||||
Thus $\Va(\sum_{\lambda \in \Lambda} I_\lambda) \subseteq \bigcap_{\lambda \in
|
||||
\Lambda}
|
||||
\Va(I_\lambda)$.
|
||||
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$ we have $f =
|
||||
\sum_{\lambda \in \Lambda} f_\lambda$.
|
||||
Thus $f$ vanishes on $\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$ and we
|
||||
have $\bigcap_{\lambda \in \Lambda} \Va(I_\lambda) \subseteq
|
||||
\Va(\sum_{\lambda
|
||||
\in \Lambda} I_\lambda)$.
|
||||
Thus
|
||||
$\Va(\sum_{\lambda \in \Lambda} I_\lambda)
|
||||
\subseteq \bigcap_{\lambda \in \Lambda} \Va(I_\lambda)$.
|
||||
On the other hand if $f \in \sum_{\lambda \in \Lambda} I_\lambda$
|
||||
we have $f = \sum_{\lambda \in \Lambda} f_\lambda$.
|
||||
Thus $f$ vanishes on
|
||||
$\bigcap_{\lambda \in \Lambda} \Va(I_{\lambda})$
|
||||
and we have
|
||||
$\bigcap_{\lambda \in \Lambda} \Va(I_\lambda)
|
||||
\subseteq \Va(\sum_{\lambda \in \Lambda} I_\lambda)$.
|
||||
\end{enumerate}
|
||||
\end{proof}
|
||||
\begin{remark}
|
||||
There is no similar way to describe $\Va(\bigcap_{\lambda \in \Lambda}
|
||||
I_\lambda)$ in terms of the
|
||||
$\Va(I_{\lambda})$ when $\Lambda$ is infinite.
|
||||
There is no similar way to describe
|
||||
$\Va(\bigcap_{\lambda \in \Lambda} I_\lambda)$
|
||||
in terms of the $\Va(I_{\lambda})$ when $\Lambda$ is infinite.
|
||||
For instance if $n = 1, I_k \coloneqq X_1^k R$ then
|
||||
$\bigcap_{k=0}^\infty I_k =
|
||||
\{0\} $ but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
|
||||
$\bigcap_{k=0}^\infty I_k = \{0\}$
|
||||
but $\bigcup_{k=0}^{\infty} \Va(I_k) = \{0\}$.
|
||||
\end{remark}
|
||||
\subsubsection{Definition of the Zariski topology}
|
||||
Let $\mathfrak{k}$ be algebraically closed, $R =
|
||||
|
@ -444,11 +408,11 @@ For $f \in R$ let $V(f) = V(fR)$.
|
|||
|
||||
\begin{corollary}
|
||||
\label{antimonbij}
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \{I \subseteq R | I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n | A \text{ Zariski-closed}\} \\
|
||||
I & \longmapsto V(I) \\
|
||||
\{f \in R | A \subseteq V(f)\} & \longmapsfrom A
|
||||
\end{align}
|
||||
\end{align*}
|
||||
is a $\subseteq$-antimonotonic bijection.
|
||||
\end{corollary}
|
||||
\begin{corollary}
|
||||
|
@ -618,12 +582,12 @@ Let $X$ be a topological space.
|
|||
\label{bijiredprim}
|
||||
By
|
||||
\ref{antimonbij} there exists a bijection
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \{I \subseteq R |
|
||||
I \text{ ideal}, I = \sqrt{I} \} & \longrightarrow \{A \subseteq \mathfrak{k}^n
|
||||
| A \text{ Zariski-closed}\} \\ I & \longmapsto V(I)\\ \{f \in R | A
|
||||
\subseteq V(f)\} & \longmapsfrom A
|
||||
\end{align}
|
||||
\end{align*}
|
||||
|
||||
Under this correspondence $A \subseteq \mathfrak{k}^n$ is
|
||||
irreducible iff $I
|
||||
|
@ -708,12 +672,17 @@ Let $X$ be a topological space.
|
|||
Let $Y \subseteq X$ be irreducible and $U \subseteq X$ an open subset such that
|
||||
$U \cap Y \neq \emptyset$.
|
||||
Then we have a bijection
|
||||
\begin{align}
|
||||
f: \{A \subseteq X | A \text{
|
||||
irreducible, closed and } Y \subseteq A\} & \longrightarrow \{B \subseteq U |
|
||||
B \text{ irreducible, closed and } Y \cap U \subseteq B\} \\ A & \longmapsto A
|
||||
\cap U \\ \overline{B} & \longmapsfrom B
|
||||
\end{align}
|
||||
\begin{IEEEeqnarray*}{rl}
|
||||
f: &\{A \subseteq X |
|
||||
A \text{ irreducible, closed and } Y \subseteq A\}\\
|
||||
& \longrightarrow \{B \subseteq U |
|
||||
B \text{ irreducible, closed and } Y \cap U \subseteq B\}\\
|
||||
\end{IEEEeqnarray*}
|
||||
given by
|
||||
\begin{align*}
|
||||
A & \longmapsto A \cap U\\
|
||||
\overline{B} & \longmapsfrom B
|
||||
\end{align*}
|
||||
where $\overline{B}$ denotes
|
||||
the closure in $X$.
|
||||
\end{fact}
|
||||
|
@ -931,9 +900,8 @@ inequalities may be strict.
|
|||
\[
|
||||
Q(X,Y) \coloneqq \sum_{i=0}^{\infty} X^i Q_i(Y) = \sum_{i,j=0}^{\infty}
|
||||
q_{i,j}X^i Y^j = \sum_{j=0}^{\infty} Y^j
|
||||
\hat{Q_j}(X) \in K[X,Y]
|
||||
\hat{Q_j}(X) \in K[X,Y].
|
||||
\]
|
||||
.
|
||||
Then $Q(x,y) = 0$.
|
||||
Let $\hat{p_j} \coloneqq \hat{Q_j}(x)$.
|
||||
Then $\hat{P}(y) = 0$.
|
||||
|
@ -974,8 +942,7 @@ transcendence degree.
|
|||
If the ring extension $B / A$ is finite (i.e. $B$ finitely generated as an
|
||||
$A$-module) then $A$ is Noetherian.
|
||||
\end{theorem}
|
||||
\begin{fact}
|
||||
+
|
||||
\begin{fact}+
|
||||
\label{noethersubalg}
|
||||
Let $R$ be Noetherian and let $B$ be a finite $R$-algebra.
|
||||
Then every $R$-subalgebra $A \subseteq B$ is finite over $R$.
|
||||
|
@ -1308,11 +1275,11 @@ $S$-saturated ideal in $R$ when $J$ is an ideal in $R_S$.
|
|||
|
||||
\begin{proposition}
|
||||
\label{idealslocbij}
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \{I \subseteq R | I \text{ $S$-saturated ideal}\} & \longrightarrow \left\{J \subseteq R_S | J \text{ ideal}\right\} \\
|
||||
I & \longmapsto I_S \coloneqq \left\{\frac{i}{s} | i \in I, s \in S\right\} \\
|
||||
J \sqcap R & \longmapsfrom J \\
|
||||
\end{align}
|
||||
\end{align*}
|
||||
is a bijection.
|
||||
Under this bijection $I$ is a prime ideal iff $f(I)$ is.
|
||||
\end{proposition}
|
||||
|
@ -1637,11 +1604,11 @@ Localization at a prime ideal is a technique to reduce a problem to this case.
|
|||
S\} $.
|
||||
|
||||
We have a bijection
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \Spec A_S & \longrightarrow \{\fq \in
|
||||
\Spec A | \fq \subseteq \fp\} \\ \fr & \longmapsto \fr \sqcap A\\ \fq_S
|
||||
\coloneqq \left\{\frac{q}{s} | q \in \fq, s \in S\right\} & \longmapsfrom \fq
|
||||
\end{align}
|
||||
\end{align*}
|
||||
\end{proposition}
|
||||
\begin{proof}
|
||||
It is clear that $S$ is a
|
||||
|
@ -1911,10 +1878,10 @@ This is part of the proof of
|
|||
\mathfrak{k}$.
|
||||
|
||||
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\mathfrak{k}[X_1,\ldots,X_d] & \longrightarrow R \\
|
||||
P & \longmapsto P(f_1,\ldots,f_d)
|
||||
\end{align}
|
||||
\end{align*}
|
||||
is an isomorphism and in $\mathfrak{k}[X_1,\ldots,X_d]$ there is a strictly
|
||||
ascending chain of prime ideals corresponding to $\mathfrak{k}^d
|
||||
\supsetneq \{0\} \times \mathfrak{k}^{d-1} \supsetneq \ldots \supsetneq
|
||||
|
@ -2445,12 +2412,12 @@ following:
|
|||
\subseteq \fp$ of $A$ contained in $\fp$ and the prime ideals and the prime
|
||||
ideals $\tilde \fr \in \Spec B$ with $\fq \subseteq \tilde \fr \subseteq \tilde
|
||||
\fp$:
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \{\fr \in \Spec A | \fr \subseteq \fp \}
|
||||
& \longrightarrow \{\tilde \fr \in \Spec B | \fq \subseteq \tilde \fr \subseteq
|
||||
\tilde \fp\} \\ \fr & \longmapsto \pi_{B, \fq}^{-1}(\fr)\\ \tilde \fr / \fq
|
||||
& \longmapsfrom \tilde \fr
|
||||
\end{align}
|
||||
\end{align*}
|
||||
By
|
||||
\ref{bijiredprim}, the $\tilde \fr$
|
||||
are in canonical bijection with the irreducible subsets $Z$ of $Y$ containing
|
||||
|
@ -2467,11 +2434,11 @@ following:
|
|||
Let $A$ be an arbitrary ring.
|
||||
One can show that there is a bijection between $\Spec A$ and the set of
|
||||
irreducible subsets $Y \subseteq \Spec A$:
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \Spec A
|
||||
& \longrightarrow \{Y \subseteq \Spec A | Y\text{irreducible}\} \\ \fp
|
||||
& \longmapsto \Vs(\fp) \\ \bigcup_{\fp \in Y} \fp & \longmapsfrom Y
|
||||
\end{align}
|
||||
\end{align*}
|
||||
Thus, the chains $\fp = \fp_0 \supsetneq \ldots \supsetneq \fp_k$ are in
|
||||
canonical bijection with the chains $V(\fp) = X_0 \subsetneq X_1 \subsetneq
|
||||
\ldots \subsetneq X_k \subseteq \Spec A$ of irreducible subsets, and
|
||||
|
@ -2800,11 +2767,11 @@ We will use the following
|
|||
\begin{proposition}
|
||||
\label{bijspecideal}
|
||||
There is a bijection
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
f: \{A \subseteq \Spec R | A\text{ closed}\}
|
||||
& \longrightarrow \{I \subseteq R | I \text{ ideal and } I = \sqrt{I} \} \\ A
|
||||
& \longmapsto \bigcap_{\fp \in A} \fp \\ \Vspec(I) & \longmapsfrom I
|
||||
\end{align}
|
||||
\end{align*}
|
||||
Under this bijection, the irreducible subsets correspond to the prime ideals
|
||||
and the closed points $\{\mathfrak{m}\}, \mathfrak{m}
|
||||
\in \Spec A$ to the
|
||||
|
@ -3012,7 +2979,7 @@ this more general theorem.
|
|||
defines a homeomorphism between $(A \times B) \cap \Delta$ and $A \cap B$.
|
||||
Thus, $C$ is homeomorphic to an irreducible component $C'$ of $(A \times B)
|
||||
\cap \Delta$ and
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\codim(C, \mathfrak{k}^n) = n - \dim(C) = n -
|
||||
\dim(C') = n - \dim(A \times B) + \codim(C', A \times B) \\
|
||||
\overset{\text{
|
||||
|
@ -3020,7 +2987,7 @@ this more general theorem.
|
|||
\overset{\text{
|
||||
\ref{dimprod}}}{=} 2n - \dim(A) - \dim(B) =
|
||||
\codim(A,\mathfrak{k}^n) + \codim(B, \mathfrak{k}^n)
|
||||
\end{align}
|
||||
\end{align*}
|
||||
by the general
|
||||
properties of dimension and codimension,
|
||||
\ref{corpithm} applied to
|
||||
|
|
|
@ -1,11 +1,7 @@
|
|||
|
||||
\begin{warning}
|
||||
This is not an official script!
|
||||
This document was written in preparation for the oral exam.
|
||||
It mostly follows the way \textsc{Prof.
|
||||
Franke} presented the material in his
|
||||
lecture rather closely.
|
||||
There are probably errors.
|
||||
This is not an official script.
|
||||
There is no guarantee for completeness or correctness.
|
||||
\end{warning}
|
||||
|
||||
\noindent The \LaTeX template by \textsc{Maximilian Kessler} is published under the
|
||||
|
|
|
@ -384,12 +384,12 @@ Let $\mathfrak{l}$ be any field.
|
|||
\begin{proposition}
|
||||
\label{bijproj}
|
||||
There is a bijection
|
||||
\begin{align}
|
||||
\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
|
||||
\end{align}
|
||||
\end{align*}
|
||||
Under this bijection,
|
||||
the irreducible subsets correspond to the elements of
|
||||
$\Proj(A_\bullet)$.
|
||||
|
@ -528,11 +528,11 @@ Let $\mathfrak{l}$ be any field.
|
|||
\codim(Y
|
||||
\cap \mathbb{A}^n, Z \cap \mathbb{A}^n)$.
|
||||
Thus
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\codim(X,Y) + \codim(Y,Z) & = \codim(X \cap \mathbb{A}^n, Y
|
||||
\cap \mathbb{A}^n) + \codim(Y \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & =
|
||||
\codim(X \cap \mathbb{A}^n, Z \cap \mathbb{A}^n) \\ & = \codim(X, Z)
|
||||
\end{align}
|
||||
\end{align*}
|
||||
because $\mathfrak{k}^n$ is catenary and the first point follows.
|
||||
The remaining assertions can easily be derived from the first two.
|
||||
\end{proof}
|
||||
|
@ -566,8 +566,9 @@ Let $\mathfrak{l}$ be any field.
|
|||
\begin{proof}
|
||||
The first assertion follows from
|
||||
\ref{bijproj} and
|
||||
\ref{bijiredprim} (bijection
|
||||
of irreducible subsets and prime ideals in the projective and affine case).
|
||||
\ref{bijiredprim}
|
||||
(bijection of irreducible subsets and prime ideals in the projective
|
||||
and affine case).
|
||||
|
||||
Let $d = \dim(X)$ and
|
||||
\[
|
||||
|
@ -575,101 +576,92 @@ Let $\mathfrak{l}$ be any field.
|
|||
X_{d+1} \subsetneq \ldots \subsetneq X_n =
|
||||
\mathbb{P}^n
|
||||
\]
|
||||
be a chain of
|
||||
irreducible subsets of $\mathbb{P}^n$.
|
||||
be a chain of irreducible subsets of $\mathbb{P}^n$.
|
||||
Then
|
||||
\[
|
||||
\{0\} \subsetneq C(X_0) \subsetneq \ldots \subsetneq C(X_d) = C(X)
|
||||
\subsetneq \ldots \subsetneq C(X_n) = \mathfrak{k}^{n+1}
|
||||
\]
|
||||
is a chain of
|
||||
irreducible subsets of $\mathfrak{k}^{n+1}$.
|
||||
Hence $\dim(C(X)) \ge 1 + d$ and $\codim(C(X), \mathfrak{k}^{n+1}) \ge
|
||||
n-d$.
|
||||
Since $\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) =
|
||||
\dim(\mathfrak{k}^{n+1})
|
||||
= n+1$, the two inequalities must be equalities.
|
||||
is a chain of irreducible subsets of $\mathfrak{k}^{n+1}$.
|
||||
Hence $\dim(C(X)) \ge 1 + d$ and
|
||||
$\codim(C(X), \mathfrak{k}^{n+1}) \ge n-d$.
|
||||
Since
|
||||
$\dim(C(X)) + \codim(C(X), \mathfrak{k}^{n+1}) = \dim(\mathfrak{k}^{n+1})
|
||||
= n+1$,
|
||||
the two inequalities must be equalities.
|
||||
\end{proof}
|
||||
\subsubsection{Application to hypersurfaces in $\mathbb{P}^n$}
|
||||
\begin{definition}[Hypersurface]
|
||||
Let $n > 0$.
|
||||
By a \vocab{hypersurface} in $\mathbb{P}^n$ or
|
||||
$\mathbb{A}^n$ we understand an
|
||||
irreducible closed subset of codimension $1$.
|
||||
$\mathbb{A}^n$ we understand an irreducible closed subset
|
||||
of codimension $1$.
|
||||
\end{definition}
|
||||
|
||||
\begin{corollary}
|
||||
If $P \in A_d$ is a prime element, then $H = \Vp(P)$ is a
|
||||
hypersurface in
|
||||
$\mathbb{P}^n$ and every hypersurface $H$ in
|
||||
$\mathbb{P}^n$ can be obtained in
|
||||
this way.
|
||||
If $P \in A_d$ is a prime element,
|
||||
then $H = \Vp(P)$ is a hypersurface in $\mathbb{P}^n$
|
||||
and every hypersurface $H$ in $\mathbb{P}^n$ can be obtained in this way.
|
||||
\end{corollary}
|
||||
\begin{proof}
|
||||
If $H = \Vp(P)$ then $C(H) = \Va(P)$ is a
|
||||
hypersurface in $\mathfrak{k}^{n+1}$
|
||||
by
|
||||
\ref{irredcodimone}.
|
||||
By
|
||||
\ref{conedim}, $H$ is irreducible and of codimension $1$.
|
||||
If $H = \Vp(P)$ then $C(H) = \Va(P)$
|
||||
is a hypersurface in $\mathfrak{k}^{n+1}$
|
||||
by \ref{irredcodimone}.
|
||||
By \ref{conedim}, $H$ is irreducible and of codimension $1$.
|
||||
|
||||
Conversely, let $H$ be a hypersurface in $\mathbb{P}^n$.
|
||||
By
|
||||
\ref{conedim}, $C(H)$ is a hypersurface in
|
||||
$\mathfrak{k}^{n+1}$, hence $C(H)
|
||||
= \Vp(P)$ for some prime element $P \in A$ (again by
|
||||
\ref{irredcodimone}).
|
||||
We have $H = \Vp(\fp)$ for some $\fp \in
|
||||
\Proj(A)$ and $C(H) = \Va(\fp)$.
|
||||
By \ref{conedim}, $C(H)$ is a hypersurface in
|
||||
$\mathfrak{k}^{n+1}$,
|
||||
hence $C(H) = \Vp(P)$ for some prime element $P \in A$
|
||||
(again by \ref{irredcodimone}).
|
||||
We have $H = \Vp(\fp)$ for some $\fp \in \Proj(A)$ and $C(H) = \Va(\fp)$.
|
||||
By the bijection between closed subsets of $\mathfrak{k}^{n+1}$ and ideals
|
||||
$I =
|
||||
\sqrt{I} \subseteq A$ (
|
||||
\ref{antimonbij}), $\fp = P \cdot
|
||||
A$.
|
||||
$I = \sqrt{I} \subseteq A$ (\ref{antimonbij}),
|
||||
$\fp = P \cdot A$.
|
||||
Let $P = \sum_{k=0}^{d}P_k$ with $P_d \neq 0$ be the decomposition
|
||||
into
|
||||
homogeneous components.
|
||||
If $P_e $ with $e < d$ was $\neq 0$, it could not be a multiple of $P$
|
||||
contradicting the homogeneity of $\fp = P \cdot A$.
|
||||
into homogeneous components.
|
||||
If $P_e $ with $e < d$ was $\neq 0$,
|
||||
it could not be a multiple of $P$ contradicting the homogeneity of
|
||||
$\fp = P \cdot A$.
|
||||
Thus, $P$ is homogeneous of degree $d$.
|
||||
\end{proof}
|
||||
\begin{definition}
|
||||
A hypersurface $H \subseteq \mathbb{P}^n$ has
|
||||
\vocab{degree $d$} if $H =
|
||||
\Vp(P)$ where $P \in A_d$ is an irreducible polynomial.
|
||||
A hypersurface $H \subseteq \mathbb{P}^n$ has \vocab{degree $d$}
|
||||
if $H = \Vp(P)$,
|
||||
where $P \in A_d$ is an irreducible polynomial.
|
||||
\end{definition}
|
||||
|
||||
\subsubsection{Application to intersections in $\mathbb{P}^n$ and Bezout's
|
||||
theorem}
|
||||
\subsubsection{Application to intersections in $\mathbb{P}^n$
|
||||
and Bezout's theorem}
|
||||
\begin{corollary}
|
||||
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq
|
||||
\mathbb{P}^n$ be irreducible
|
||||
subsets of dimensions $a$ and $b$.
|
||||
If $a+ b \ge n$, then $A \cap B \neq \emptyset$ and every irreducible component
|
||||
of $A \cap B$ as dimension $\ge a + b - n$.
|
||||
Let $A \subseteq \mathbb{P}^n$ and $B \subseteq \mathbb{P}^n$ be
|
||||
irreducible subsets of dimensions $a$ and $b$.
|
||||
If $a+ b \ge n$,
|
||||
then $A \cap B \neq \emptyset$
|
||||
and every irreducible component of $A \cap B$
|
||||
has dimension $\ge a + b - n$.
|
||||
\end{corollary}
|
||||
|
||||
\begin{remark}
|
||||
This shows that $\mathbb{P}^n$ indeed fulfilled the goal of
|
||||
allowing for nicer
|
||||
results of algebraic geometry because ``solutions at infinity'' to systems of
|
||||
algebraic equations are present in $\mathbb{P}^n$ (see
|
||||
\ref{affineproblem}).
|
||||
This shows that $\mathbb{P}^n$ indeed fulfilled the goal of allowing for
|
||||
nicer results of algebraic geometry because ``solutions at infinity''
|
||||
to systems of algebraic equations are present in $\mathbb{P}^n$
|
||||
(see \ref{affineproblem}).
|
||||
\end{remark}
|
||||
|
||||
\begin{proof}
|
||||
The lower bound on the dimension of irreducible components of $A \cap B$ is
|
||||
easily derived from the similar affine result (corollary of the principal ideal
|
||||
theorem,
|
||||
\ref{codimintersection}).
|
||||
From the definition of the affine cone it follows that $C(A \cap B) = C(A) \cap
|
||||
C(B)$.
|
||||
easily derived from the similar affine result
|
||||
(corollary of the principal ideal theorem, \ref{codimintersection}).
|
||||
|
||||
From the definition of the affine cone it follows that
|
||||
$C(A \cap B) = C(A) \cap C(B)$.
|
||||
We have $\dim(C(A)) = a+1$ and $\dim(C(B)) = b + 1$ by
|
||||
\ref{conedim}.
|
||||
If $A \cap B = \emptyset$, then $C(A) \cap C(B) = \{0\}$ with $\{0\} $ as an
|
||||
irreducible component, contradicting the lower bound $a + b + 1 - n > 0$ for
|
||||
the dimension of irreducible components of $C(A) \cap C(B)$ (again
|
||||
\ref{codimintersection}).
|
||||
the dimension of irreducible components of $C(A) \cap C(B)$
|
||||
(again \ref{codimintersection}).
|
||||
\end{proof}
|
||||
\begin{remark}[Bezout's theorem]
|
||||
If $A \neq B$ are hypersurfaces of degree $a$ and $b$
|
||||
|
@ -682,4 +674,3 @@ Let $\mathfrak{l}$ be any field.
|
|||
% SLIDE APPLICATION TO HYPERSURFACES IN $\P^n$
|
||||
%ERROR: C(H) = V_A(P)
|
||||
%If n = 0, P = 0, V_P(P) = \emptyset is a problem!
|
||||
|
||||
|
|
|
@ -29,12 +29,12 @@
|
|||
r_{U_j, U_i \cap U_j}(f_j)$ for all $i,j \in I$.
|
||||
|
||||
Consider the map
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\phi_{U, (U_i)_{i \in I}}: \mathcal{G}(U)
|
||||
& \longrightarrow \{(f_i)_{i \in I} \in \prod_{i \in I} \mathcal{G}(U_i) |
|
||||
r_{U_i, U_i \cap U_j}(f_i) = r_{U_j, U_i \cap U_j}(f_j) \text{ for } i,j \in I
|
||||
\} \\ f & \longmapsto (r_{U, U_i}( f))_{i \in I}
|
||||
\end{align}
|
||||
\end{align*}
|
||||
|
||||
A presheaf is called \vocab[Presheaf!
|
||||
separated]{separated} if $\phi_{U, (U_i)_{i \in I}}$ is
|
||||
|
@ -163,10 +163,10 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|||
Let $X = V(I)$ where $I = \sqrt{I} \subseteq R$ is an ideal.
|
||||
Let $A = R / I$.
|
||||
Then
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\phi: A & \longrightarrow \mathcal{O}_X(X) \\ f \mod I
|
||||
& \longmapsto f\defon{X}
|
||||
\end{align}
|
||||
\end{align*}
|
||||
is an isomorphism.
|
||||
\end{proposition}
|
||||
|
||||
|
@ -176,8 +176,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|||
ring homomorphism.
|
||||
Its injectivity follows from the Nullstellensatz and $I =
|
||||
\sqrt{I}$
|
||||
(
|
||||
\ref{hns3}).
|
||||
(\ref{hns3}).
|
||||
|
||||
|
||||
Let $\phi \in \mathcal{O}_X(X)$.
|
||||
|
@ -211,8 +210,7 @@ Let $R = \mathfrak{k}[X_1,\ldots,X_n]$.
|
|||
As the $U_i = X \setminus V(g_i)$ cover $X$, $V(I) \cap
|
||||
\bigcap_{i=1}^m V(g_i)
|
||||
= X \cap \bigcap_{i=1}^m V(g_i) = \emptyset$.
|
||||
By the Nullstellensatz (
|
||||
\ref{hns1}) the ideal of $R$ generated by
|
||||
By the Nullstellensatz (\ref{hns1}) the ideal of $R$ generated by
|
||||
$I$ and the
|
||||
$a_i$ equals $R$.
|
||||
There are thus $n \ge m \in \N$ and elements
|
||||
|
@ -366,8 +364,7 @@ The following is somewhat harder than in the affine case:
|
|||
The category of topological spaces
|
||||
\item
|
||||
The category $\Var_\mathfrak{k}$ of varieties over
|
||||
$\mathfrak{k}$ (see
|
||||
\ref{defvariety})
|
||||
$\mathfrak{k}$ (see \ref{defvariety})
|
||||
\item
|
||||
If $\mathcal{A}$ is a category, then the \vocab{opposite category}
|
||||
or \vocab{dual category} is defined by $\Ob(\mathcal{A}\op) =
|
||||
|
@ -469,24 +466,21 @@ The following is somewhat harder than in the affine case:
|
|||
\begin{definition}[Algebraic variety]
|
||||
\label{defvariety}
|
||||
An \vocab{algebraic variety} or \vocab{prevariety} over
|
||||
$\mathfrak{k}$ is a
|
||||
pair $(X, \mathcal{O}_X)$, where $X$ is a topological space and
|
||||
$\mathcal{O}_X$
|
||||
$\mathfrak{k}$ is a pair $(X, \mathcal{O}_X)$,
|
||||
where $X$ is a topological space and $\mathcal{O}_X$
|
||||
a subsheaf of the sheaf of $\mathfrak{k}$-valued functions on $X$
|
||||
such that
|
||||
for every $x \in X$, there are a neighbourhood $U_x$ of $x$ in $X$, an open
|
||||
subset $V_x$ of a closed subset $Y_x$ of
|
||||
$\mathfrak{k}^{n_x}$\footnote{By the
|
||||
result of
|
||||
\ref{affopensubtopbase} it can be assumed that $V_x = Y_x$ without
|
||||
altering the definition.
|
||||
} and a homeomorphism $V_x
|
||||
such that for every $x \in X$,
|
||||
there are a neighbourhood $U_x$ of $x$ in $X$,
|
||||
an open subset $V_x$ of a closed subset $Y_x$ of $\mathfrak{k}^{n_x}$%
|
||||
\footnote{By the result of \ref{affopensubtopbase},
|
||||
it can be assumed that $V_x = Y_x$ without altering the definition.}
|
||||
and a homeomorphism $V_x
|
||||
\xrightarrow{\iota_x}
|
||||
U_x$ such that for every open subset $V \subseteq U_x$ and every function
|
||||
$V\xrightarrow{f} \mathfrak{k}$, we have $f \in
|
||||
\mathcal{O}_X(V) \iff
|
||||
\iota^{\ast}_x(f) \in
|
||||
\mathcal{O}_{Y_x}(\iota_x^{-1}(V))$,
|
||||
\mathcal{O}_{Y_x}(\iota_x^{-1}(V))$.
|
||||
|
||||
In this, the \vocab{pull-back} $\iota_x^{\ast}(f)$ of $f$ is
|
||||
defined by
|
||||
|
@ -514,29 +508,26 @@ The following is somewhat harder than in the affine case:
|
|||
If $X$ is a closed subset of $\mathfrak{k}^n$ or
|
||||
$\mathbb{P}^n$, then $(X, \mathcal{O}_X)$ is a
|
||||
variety, where $\mathcal{O}_X$ is the structure sheaf on $X$
|
||||
(
|
||||
\ref{structuresheafkn}, reps.
|
||||
\ref{structuresheafpn}).
|
||||
A variety is called \vocab[Variety!
|
||||
affine]{affine} (resp. \vocab[Variety!projective]{projective}) if it is isomorphic to a variety of
|
||||
this form, with $X $ closed in $\mathfrak{k}^n$ (resp.
|
||||
$\mathbb{P}^n$).
|
||||
A variety which is isomorphic to and open subvariety of $X$ is called
|
||||
\vocab[Variety!
|
||||
quasi-affine]{quasi-affine} (resp. \vocab[Variety!quasi-projective]{quasi-projective}).
|
||||
(\ref{structuresheafkn}, reps. \ref{structuresheafpn}).
|
||||
A variety is called \vocab[Variety!affine]{affine}
|
||||
(resp. \vocab[Variety!projective]{projective})
|
||||
if it is isomorphic to a variety of this form,
|
||||
with $X $ closed in $\mathfrak{k}^n$ (resp. $\mathbb{P}^n$).
|
||||
A variety which is isomorphic to and open subvariety of $X$
|
||||
is called \vocab[Variety!quasi-affine]{quasi-affine}
|
||||
(resp. \vocab[Variety!quasi-projective]{quasi-projective}).
|
||||
\item
|
||||
If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$ then
|
||||
$\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)}
|
||||
X$ is a morphism which is a homeomorphism of topological spaces but not an
|
||||
isomorphism of varieties.
|
||||
If $X = V(X^2 - Y^3) \subseteq \mathfrak{k}^2$
|
||||
then $\mathfrak{k} \xrightarrow{t \mapsto (t^3,t^2)}
|
||||
X$ is a morphism which is a homeomorphism of topological spaces
|
||||
but not an isomorphism of varieties.
|
||||
% TODO
|
||||
|
||||
\item
|
||||
The composition of two morphisms $X \to Y \to Z$ of varieties is a morphism of
|
||||
varieties.
|
||||
The composition of two morphisms $X \to Y \to Z$ of varieties
|
||||
is a morphism of varieties.
|
||||
\item
|
||||
$X\xrightarrow{\Id_X}
|
||||
X$ is a morphism of varieties.
|
||||
$X\xrightarrow{\Id_X} X$ is a morphism of varieties.
|
||||
\end{itemize}
|
||||
\end{example}
|
||||
|
||||
|
@ -563,8 +554,7 @@ The following is somewhat harder than in the affine case:
|
|||
\item
|
||||
The property is local on $U$, hence it is sufficient to show it in the
|
||||
quasi-affine case.
|
||||
This was done in
|
||||
\ref{structuresheafcontinuous}.
|
||||
This was done in \ref{structuresheafcontinuous}.
|
||||
\item
|
||||
For the second part, let $\lambda_x \coloneqq \lambda \defon{V_x}
|
||||
$.
|
||||
|
@ -603,21 +593,21 @@ The following is somewhat harder than in the affine case:
|
|||
\item
|
||||
Let $X,Y$ be varieties over $\mathfrak{k}$.
|
||||
Then the map
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\phi: \Hom_{\Var_\mathfrak{k}}(X,Y) & \longrightarrow
|
||||
\Hom_{\Alg_\mathfrak{k}}(\mathcal{O}_Y(Y), \mathcal{O}_X(X)) \\ (X
|
||||
\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_Y(Y) \xrightarrow{f^{\ast}}
|
||||
\mathcal{O}_X(X))
|
||||
\end{align}
|
||||
\end{align*}
|
||||
is injective when $Y$ is quasi-affine and
|
||||
bijective when $Y$ is affine.
|
||||
\item
|
||||
The contravariant functor
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
F: \Var_\mathfrak{k} & \longrightarrow \Alg_\mathfrak{k} \\ X & \longmapsto
|
||||
\mathcal{O}_X(X) \\ (X\xrightarrow{f} Y) & \longmapsto (\mathcal{O}_X(X)
|
||||
\xrightarrow{f^{\ast}} \mathcal{O}_Y(Y))
|
||||
\end{align}
|
||||
\end{align*}
|
||||
restricts to an
|
||||
equivalence of categories between the category of affine varieties over
|
||||
$\mathfrak{k}$ and the full subcategory $\mathcal{A}$ of
|
||||
|
@ -899,10 +889,10 @@ The following is somewhat harder than in the affine case:
|
|||
|
||||
If $U$ is an open neighbourhood of $x \in X$, then we have a map (resp.
|
||||
homomorphism)
|
||||
\begin{align}
|
||||
\begin{align*}
|
||||
\cdot_x : \mathcal{G}(U) & \longrightarrow \mathcal{G}_x\\
|
||||
\gamma & \longmapsto \gamma_x \coloneqq (U, \gamma) / \sim
|
||||
\end{align}
|
||||
\end{align*}
|
||||
|
||||
\end{definition}
|
||||
\begin{fact}
|
||||
|
|
Reference in a new issue