rav: update, add lec 15
This commit is contained in:
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}
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% replace all relations with align characters (&) and add the needed padding
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\regex_replace_all:nnN
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{ (\c{leq}&|&\c{leq}|\c{geq}&|&\c{geq}|\c{iff}&|&\c{iff}|\c{impliedby}&|&\c{impliedby}|\c{implies}&|&\c{implies}|\c{approx}&|&\c{approx}|\c{equiv}&|&\c{equiv}|=&|&=|\c{le}&|&\c{le}|\c{ge}&|&\c{ge}|&\c{stackrel}(\[.*?\])?{.*?}{.*?}|\c{stackrel}(\[.*?\])?{.*?}{.*?}&|&\c{neq}|\c{neq}&|>&|&>|<&|&<) }
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{ (\c{simeq}&|&\c{simeq}|\c{leq}&|&\c{leq}|\c{geq}&|&\c{geq}|\c{iff}&|&\c{iff}|\c{impliedby}&|&\c{impliedby}|\c{implies}&|&\c{implies}|\c{approx}&|&\c{approx}|\c{equiv}&|&\c{equiv}|=&|&=|\c{le}&|&\c{le}|\c{ge}&|&\c{ge}|&\c{stackrel}(\[.*?\])?{.*?}{.*?}|\c{stackrel}(\[.*?\])?{.*?}{.*?}&|&\c{neq}|\c{neq}&|>&|&>|<&|&<) }
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{ \c{kern} \u{l_tmp_dim_needed} \1 \c{kern} \u{l_tmp_dim_needed} }
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\l__lec_text_tl
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\l__lec_text_tl
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@@ -30,5 +30,6 @@ Christian Merten (\href{mailto:cmerten@mathi.uni-heidelberg.de}{cmerten@mathi.un
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\input{rav8.tex}
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\input{rav9.tex}
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\input{rav10.tex}
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\input{rav15.tex}
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\end{document}
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\documentclass{lecture}
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\begin{document}
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\usetikzlibrary{shapes.misc}
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\tikzset{cross/.style={cross out, draw=black, minimum size=2*(#1-\pgflinewidth), inner sep=0pt, outer sep=0pt},
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%default radius will be 1pt.
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cross/.default={1pt}}
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\chapter{Hilbert's Nullstellensatz and applications}
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\section{Fields of definition}
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When $k$ is an algebraically closed field, Hilbert's Nullstellensatz gives us a bijection
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between algebraic subsets of $k^{n}$ and radical ideals in $k[T_1, \ldots, T_n]$.
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This correspondence induces an anti-equivalence of categories
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\begin{salign*}
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\{\text{affine } k\text{-varieties}\} &\longleftrightarrow
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\{\text{finitely-generated reduced } k \text{-algebras}\} \\
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(X, \mathcal{O}_X) &\longmapsto \mathcal{O}_X(X) \\
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\hat{A} = \operatorname{Hom}_{k\mathrm{-alg}}(A, k) &\longmapsfrom A
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.\end{salign*}
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\begin{lemma}
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Let $k$ be algebraically closed and $A$ a finitely-generated $k$-Algebra. Then
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the map
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\begin{salign*}
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\hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k) &\longrightarrow \operatorname{Spm } A \\
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\xi &\longmapsto \text{ker } \xi
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\end{salign*}
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is a bijection.
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\end{lemma}
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\begin{proof}
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The map
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admits an inverse
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\begin{salign*}
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\operatorname{Spm } A &\longrightarrow \operatorname{Hom}_{k\text{-alg}}(A, k) \\
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\mathfrak{m} &\longmapsto (A \to A / \mathfrak{m})
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.\end{salign*}
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This is well-defined, since $A / \mathfrak{m}$ is a finite extension of the algebraically closed field
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$k$, so $k \simeq A / \mathfrak{m}$.
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\end{proof}
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Since we have defined a product on the left-hand side of the anti-equivalence, this must correspond
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to coproduct on the right-hand side. Since the coproduct in the category of commutative
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$k$-algebras with unit is given by the tensor product, we have
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\[
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\mathcal{O}_{X \times Y} (X \times Y) \simeq \mathcal{O}_X(X) \otimes_k \mathcal{O}_Y(Y)
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.\]
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\begin{korollar}
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Let $k$ be algebraically closed. Then the tensor product of two
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reduced (resp. integral) finitely-generated $k$-algebras is reduced (resp. integral).
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\label{kor:k-alg-closed-tensor-of-reduced}
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\end{korollar}
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\begin{proof}
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This follows from the anti-equivalence of categories: Reduced since products of affine
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$k$-varieties exist and integral since the product of two irreducible affine $k$-varieties is irreducible.
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\end{proof}
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\begin{bem}
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\ref{kor:k-alg-closed-tensor-of-reduced} is false in general if $k = \overline{k}$. For instance
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$\mathbb{C}$ is an integral $\R$-algebra, but
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\begin{salign*}
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\mathbb{C} \otimes_{\R} \mathbb{C}
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&= \R[x]/(x^2 + 1) \otimes_{\R} \mathbb{C} \\
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&= \mathbb{C}[x]/(x^2 + 1) \\
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&= \mathbb{C}[x]/((x-i)(x+i)) \\
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&\stackrel{(*)}{\simeq} \mathbb{C}[x]/(x-i) \times \mathbb{C}[x]/(x+i) \\
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&\simeq \mathbb{C} \times \mathbb{C}
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\end{salign*}
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is not integral, where $(*)$ follows from the Chinese remainder theorem.
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For a non-reduced example, consider $k = \mathbb{F}_{p}(t)$ and choose a $p$-th root
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$\alpha = t ^{\frac{1}{p}}$ in $\overline{\mathbb{F}_p(t)}$. Then $\alpha \not\in k$
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but $\alpha ^{n} \in k$. If we put $L = k(\alpha)$, then
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$\alpha \otimes 1 - 1 \otimes \alpha \neq 0$ in $L \otimes_k L$ since
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the elements $(\alpha ^{i} \otimes \alpha ^{j})_{0 \le i, j \le p-1}$ form a basis
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of $L \otimes_k L$ as a $k$-vector space, but
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\[
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(\alpha \otimes 1 - 1 \otimes \alpha)^{p}
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= \alpha ^{p} \otimes 1 - 1 \otimes \alpha ^{p}
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= 1 \otimes \alpha ^{p} - 1 \otimes \alpha ^{p} = 0
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.\]
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\end{bem}
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We now consider more generally finitely generated reduced $k$-algebras when $k$ is not
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necessarily closed.
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\begin{bsp}
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Let $A = \R[X]/(x^2 +1)$. Since $x^2 + 1 $ is irreducible in $\R[x]$, it
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generates a maximal ideal, thus the finitely-generated $\R$-algebra $A$ is a field and in
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particular reduced. We can equip the topogical space
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$X \coloneqq \operatorname{Spm } A = \{ (0)\} $ with a sheaf of regular functions, defined
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by $\mathcal{O}_X(\{(0)\}) = A$. In other words, $\operatorname{Spm } A$ is just a point,
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but equipped with the reduced $\R$-algebra $A$. It thus differs from the
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point $\operatorname{Spm } \R$, which is equipped with the reduced $\R$-algebra $\R$,
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since $\R[x]/(x^2 + 1) \not\simeq \R$ as $\R$-algebras. Indeed, the $\R$-algebra $\R[x]/(x^2+1)$
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is $2$ dimensional as a real vector space.
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$A$ possesses a non-trivial $\R$-algebra automorphism induced by the automorphism of $\R$-algebras,
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$P \mapsto P(-x)$ in $\R[x]$. Indeed, $\R[x]/(x^2+1) \simeq \mathbb{C}$ as $\R$-algebras,
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with the previous automorphism corresponding to the complex conjugation $z \mapsto \overline{z}$.
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\end{bsp}
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\begin{bsp}
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By analogy with the Zariski topology on maximal spectra of (finitely generated, reduced)
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$\mathbb{C}$-algebras, we can equip $X = \operatorname{Spm } A$ with
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a Zariski topology for all (finitely generated reduced) $\R$-algebras $A$: the closed subsets
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of this topology are given by
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\[
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\mathcal{V}_X(I) \coloneqq \{ \mathfrak{m} \in \operatorname{Spm } A \mid \mathfrak{m} \supset I\}
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\] for any ideal $I \subseteq A$.
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Note that $X = \operatorname{Spm } A$ contains
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$\hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k)$ as a subset: the points
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of $\hat{A}$ correspond to maximal ideals $\mathfrak{m}$ of $A$ with residue field
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$A / \mathfrak{m} \simeq k$. But when $k \not\simeq \overline{k}$, the set
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$\operatorname{Spm } A$ is strictly larger than $\hat{A}$: it contains maximal ideals $\mathfrak{m}$
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such that $A / \mathfrak{m}$ is a non-trivial finite extension of $k$. The induced topology on
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$\hat{A} \subseteq \operatorname{Spm } A$ is the Zariski topologoy of $\hat{A}$ that was
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introduced earlier.
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Let $A = \R[x]$. Maximal ideals in the principal ring $\R[x]$ are generated
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by a single irreducible polynomial $P$, which is either of degree $1$ or of degree $2$ with
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negative discriminant.
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In the first case, $P = x-a$ for some $a \in \R$ and the residue field is $\R[x]/(x - a) \simeq \R$,
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while, in the second case, $P = x^2 + bx + c$ for $b, c \in \R$ and $b^2 - 4c < 0$ and
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by choosing a root $z_0$ of $P$ in $\mathbb{C}$, the map
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\begin{salign*}
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\eta_{z_0} \colon \R[x]/(x^2 + bx + c) &\longrightarrow \mathbb{C} \\
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\overline{P} &\longmapsto P(z_0)
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\end{salign*}
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is a field-homomorphism. In particular it is injective. Since $\mathbb{C}$ and
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$\R[x]/(x^2 + bx + c)$ are both degree $2$ extensions of $\R$, we have
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$\R[x]/(x^2 + bx + c) \simeq \mathbb{C}$.
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Note that the other root of $x^2 + bx +c $ is $\overline{z_0}$ and that
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$\eta_{\overline{z_0}} = \sigma \circ \eta_{z_0}$ where $\sigma$ is complex conjugation on $\mathbb{C}$.
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So we have to ways to identify $\R[x]/(x^2 + bx +c)$ to $\mathbb{C}$ and they are
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related by the action of $\text{Gal}(\mathbb{C}/ \R)$ on $\mathbb{C}$.
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To sum up, the difference between the two possible types of maximal ideals $\mathfrak{m} \subseteq \R[x]$
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is the residue field, which is either $\R$ or $\mathbb{C}$. When it is $\R$, we
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find exactly the points of
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\begin{salign*}
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\widehat{\R[x]} &= \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R) \\
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&\simeq \{ \mathfrak{m} \in \operatorname{Spm } \R[x] \mid \R[x]/\mathfrak{m} \simeq \R\} \\
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&\simeq \{ (x-a) \colon a \in \R\} \\
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&\simeq \R
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.\end{salign*}
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And when the residue field is $\mathbb{C}$, we have $\mathfrak{m} = (x^2 + bx + c)$ with
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$b, c \in \R$ such that $b^2 - 4c < 0$. If we choose $z_0$ to be the root
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of $x^2 + bx +c$ with $\text{Im}(z_0) > 0$, we can identify the set of these maximal ideals with
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the subset
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\[
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H \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) > 0\}
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.\]
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In other words, the following pictures emerges, where we identify
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$\operatorname{Spm } \R[x]$ with
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\[
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\hat{H} \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) \ge 0\}
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\]
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via the map
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\begin{salign*}
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\operatorname{Spm } \R[x] &\longrightarrow \hat{H} \\
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\mathfrak{m} &\longmapsto \begin{cases}
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a \in \R & \mathfrak{m} = (x-a) \\
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z_0 \in H & \mathfrak{m} = ((x-z_0)(x-\overline{z_0})) \text{ and } \text{Im}(z_0) > 0
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\end{cases}
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\end{salign*}
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which is indeed bijective.
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%\begin{figure}
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% \centering
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% \begin{tikzpicture}
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% \draw[red] (-2, 0) -- (2,0) node[right] {$\R \simeq \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R)$};
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% \draw[->] (0, 0) -- (0,4);
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% \end{tikzpicture}
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% \caption{$\operatorname{Spm } \R[x] \simeq \hat{H}
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% = \left\{ z \in \mathbb{C} : \text{Im}(z) \ge 0 \right\}$}
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%\end{figure}
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We see that $\operatorname{Spm } \R[x]$ contains a lot more points
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that $\R$. One could go further and add the ideal $(0)$: This would give the set
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\[
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\mathbb{A}^{1}_{\R} = \operatorname{Spec } \R[x]
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= \operatorname{Spm } \R[x] \cup \{(0)\}
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.\]
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\end{bsp}
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\begin{bem}
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If $A$ is a $k$-algebra and $\overline{k}$ is an algebraic closure of $k$, the
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group $\text{Aut}_k(\overline{k})$ acts on the $\overline{k}$-algebra
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$A_{\overline{k}} \coloneqq A \otimes_k \overline{k}$ via
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$\sigma (a \otimes \lambda) \coloneqq a \otimes \sigma(\lambda)$. Moreover, the map
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$a \mapsto a \otimes 1$ induces an injective morphism of $k$-algebras
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$A \xhookrightarrow{} A \otimes_k \overline{k}$ since
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the tensor product over fields is left-exact.
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Its image is contained in the $k$-subalgebra
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$\operatorname{Fix}_{\operatorname{Aut}_k(\overline{k})} A_{\overline{k}} \subseteq A_{\overline{k}}$. When
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$k$ is a perfect field, this inclusion is an equality.
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\end{bem}
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\begin{bsp}
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If $A = \R[x]$, then $A \otimes_{\R} \mathbb{C} \simeq \mathbb{C}[x]$. The group
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$\text{Aut}_{\R}(\mathbb{C}) = \text{Gal}(\mathbb{C}/\R) = \langle \sigma \rangle$ with
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$\sigma\colon z \mapsto \overline{z}$, acts naturally on $\mathbb{C}[x]$. This
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is an action by $\R$-algebra automorphisms. Clearly,
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$\text{Fix}_{\langle\sigma\rangle} \mathbb{C}[x] = \R[x]$. There
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is an induced action on $\operatorname{Spm } \mathbb{C}[x]$,
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defined by
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\[
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\sigma(\mathfrak{m}) = \sigma((x-z)) \coloneqq (x - \sigma(z)) = (x - \overline{z})
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.\]
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When we identify $\operatorname{Spm } \mathbb{C}[x]$ with $\mathbb{C}$
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via $(x-z) \mapsto z$, this action is just $z \mapsto \overline{z}$. This
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,,geometric action'' induces an action of $\text{Gal}(\mathbb{C}/\R)$ on
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regular functions on $\mathbb{C}$: to $h \in \mathcal{O}_{\mathbb{C}}(U)$, there
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is associated a regular function $h \in \mathcal{O}_{\mathbb{C}}(\sigma(U))$, defined for
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all $x \in \sigma(U)$, by
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\[
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\sigma(h)(z) \coloneqq \sigma \circ h \circ \sigma ^{-1}(z) = \overline{h(\overline{z})}
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.\]
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In particular, if $h = P \in \mathcal{O}_{\mathbb{C}}(\mathbb{C}) = \mathbb{C}[x]$, then
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$P \mapsto \sigma(P)$ coincides with the natural
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$\text{Gal}(\mathbb{C} / \R)$ action on $\mathbb{C}[x]$. We will see momentarily that this
|
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defines a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$. To that end,
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let us first look more closely at the $\text{Gal}(\mathbb{C}/ \R)$ action
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on $\operatorname{Spm } \mathbb{C}[x]$. Its fixed-point set
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is
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\[
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\{ \mathfrak{m} \in \operatorname{Spm } \mathbb{C}[x] \mid \mathfrak{m} = (x-a), a \in \R\}
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\simeq \R = \operatorname{Fix}_{z \mapsto \overline{z}}(\mathbb{C})
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.\]
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Moreover, there is a map
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\begin{salign*}
|
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\operatorname{Spm } \mathbb{C}[x] &\longrightarrow \operatorname{Spm } \R[x] \\
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\mathfrak{m} &\longmapsto \mathfrak{m} \cap \R[x]
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\end{salign*}
|
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sending $(x-a) \mathbb{C}[x]$ to $(x-a)\R[x]$ if $a \in \R$,
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and $(x-z)\mathbb{C}[x]$ to $(x-z)(x-\overline{z})\R[x]$ if $z \in \mathbb{C} \setminus \R$.
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This map is surjective and induces a bijection
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\[
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(\operatorname{Spm } \mathbb{C}[x]) / \operatorname{Gal}(\mathbb{C} / \R)
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\xlongrightarrow{\simeq} \operatorname{Spm } \R[x]
|
||||
.\]
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Geometrically, the quotient map $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$
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is the ,,folding map``
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\begin{salign*}
|
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\mathbb{C} &\longrightarrow \hat{H} \\
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z = u + iv &\longmapsto u + i |v|
|
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.\end{salign*}
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\begin{figure}
|
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\centering
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\begin{tikzpicture}
|
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\draw[red] (-2, 0) -- (2,0) node[right] {$\R$};
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||||
\fill (1, -1) circle[radius=0.75pt] node[right] {$z_0$};
|
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\draw[->] (0,-1.5) -- (0,2);
|
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\draw[->] (3.2,0) -- node[above] {$\pi$} (4.2,0);
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\draw[red] (5, 0) -- (9,0) node[right] {$\R$};
|
||||
\fill (8, 1) circle[radius=0.75pt] node[right] {$\pi(z_0)$};
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\draw[->] (7, 0) -- (7,2);
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\end{tikzpicture}
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\caption{The quotient map
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||||
$\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$ is geometrically a folding.}
|
||||
\end{figure}
|
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In view of this, it is natural to
|
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\begin{enumerate}[(i)]
|
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\item put the quotient topology on
|
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\[
|
||||
\operatorname{Spm } \R[x] = \left( \operatorname{Spm } \mathbb{C}[x] \right)
|
||||
/ \operatorname{Gal}(\mathbb{C}/\R)
|
||||
\]
|
||||
where $\operatorname{Spm } \mathbb{C}[x] \simeq \mathbb{C}$ is equipped with its topology
|
||||
of algebraic variety.
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\item define a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$ by pushing-forward
|
||||
the structure sheaf on $\operatorname{Spm } \mathbb{C}[x]$
|
||||
and then taking the $\operatorname{Gal}(\mathbb{C}/ \R)$-invariant subsheaf:
|
||||
\[
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||||
\mathcal{O}_{\operatorname{Spm } \R[x]}(U)
|
||||
\coloneqq \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}
|
||||
(\pi^{-1}(U))^{\operatorname{Gal}(\mathbb{C} / \R)}
|
||||
\] where
|
||||
$\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$,
|
||||
$\mathfrak{m} \mapsto \mathfrak{m} \cap \R[x]$ is the quotient map,
|
||||
and $\operatorname{Gal}(\mathbb{C}/ \R)$ acts on
|
||||
$\mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}(\pi^{-1}(U))$ via
|
||||
$h \mapsto \sigma(h) = \sigma \circ h \circ \sigma ^{-1}$ (note that the open set
|
||||
$\pi^{-1}(U)$ is $\operatorname{Gal}(\mathbb{C} / \R)$-invariant).
|
||||
\end{enumerate}
|
||||
Observe that
|
||||
\[
|
||||
\mathcal{O}_{\operatorname{Spm } \R[x]}(\operatorname{Spm } \R[x])
|
||||
= \mathbb{C}[x]^{\operatorname{Gal}(\mathbb{C} / \R)} = \R[x]
|
||||
.\]
|
||||
Also, if $h = \frac{f}{g}$ around $x \in U$, then, around
|
||||
$\sigma(x) \in U$, one has $\sigma(h) = \frac{\sigma(f)}{\sigma(g)}$ and,
|
||||
for all $\lambda \in \mathbb{C}$, $\sigma(\lambda h) = \overline{\lambda} \sigma(h)$.
|
||||
|
||||
Remarkably, we will see that we can reconstruct the algebraic $\mathbb{C}$-variety
|
||||
\[
|
||||
(X_{\mathbb{C}}, \mathcal{O}_{X_{\mathbb{C}}})
|
||||
\coloneqq (\operatorname{Spm } \mathbb{C}[x], \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]})
|
||||
\] from the ringed space
|
||||
\[
|
||||
(X, \mathcal{O}_X) \coloneqq (\operatorname{Spm } \R[x], \mathcal{O}_{\operatorname{Spm } \R[x]}
|
||||
\] that we have just constructed.
|
||||
\end{bsp}
|
||||
|
||||
\end{document}
|
||||
@@ -2,7 +2,7 @@
|
||||
|
||||
\begin{document}
|
||||
|
||||
\chapter{Affine varieties}
|
||||
\chapter{Algebraic varieties}
|
||||
|
||||
\section{Spaces with functions}
|
||||
|
||||
|
||||
Reference in New Issue
Block a user