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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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\[ |
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(\operatorname{Spm } \mathbb{C}[x]) / \operatorname{Gal}(\mathbb{C} / \R) |
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\xlongrightarrow{\simeq} \operatorname{Spm } \R[x] |
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.\] |
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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] {$\R$}; |
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\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.} |
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\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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\[ |
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\operatorname{Spm } \R[x] = \left( \operatorname{Spm } \mathbb{C}[x] \right) |
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/ \operatorname{Gal}(\mathbb{C}/\R) |
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\] |
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where $\operatorname{Spm } \mathbb{C}[x] \simeq \mathbb{C}$ is equipped with its topology |
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of algebraic variety. |
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\item define a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$ by pushing-forward |
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the structure sheaf on $\operatorname{Spm } \mathbb{C}[x]$ |
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and then taking the $\operatorname{Gal}(\mathbb{C}/ \R)$-invariant subsheaf: |
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\[ |
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\mathcal{O}_{\operatorname{Spm } \R[x]}(U) |
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\coloneqq \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]} |
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(\pi^{-1}(U))^{\operatorname{Gal}(\mathbb{C} / \R)} |
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\] where |
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$\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$, |
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$\mathfrak{m} \mapsto \mathfrak{m} \cap \R[x]$ is the quotient map, |
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and $\operatorname{Gal}(\mathbb{C}/ \R)$ acts on |
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$\mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}(\pi^{-1}(U))$ via |
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$h \mapsto \sigma(h) = \sigma \circ h \circ \sigma ^{-1}$ (note that the open set |
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$\pi^{-1}(U)$ is $\operatorname{Gal}(\mathbb{C} / \R)$-invariant). |
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\end{enumerate} |
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Observe that |
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\[ |
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\mathcal{O}_{\operatorname{Spm } \R[x]}(\operatorname{Spm } \R[x]) |
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= \mathbb{C}[x]^{\operatorname{Gal}(\mathbb{C} / \R)} = \R[x] |
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.\] |
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Also, if $h = \frac{f}{g}$ around $x \in U$, then, around |
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$\sigma(x) \in U$, one has $\sigma(h) = \frac{\sigma(f)}{\sigma(g)}$ and, |
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for all $\lambda \in \mathbb{C}$, $\sigma(\lambda h) = \overline{\lambda} \sigma(h)$. |
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Remarkably, we will see that we can reconstruct the algebraic $\mathbb{C}$-variety |
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\[ |
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(X_{\mathbb{C}}, \mathcal{O}_{X_{\mathbb{C}}}) |
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\coloneqq (\operatorname{Spm } \mathbb{C}[x], \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}) |
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\] from the ringed space |
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\[ |
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(X, \mathcal{O}_X) \coloneqq (\operatorname{Spm } \R[x], \mathcal{O}_{\operatorname{Spm } \R[x]} |
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\] that we have just constructed. |
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\end{bsp} |
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\end{document} |
