\documentclass{lecture}

\begin{document}

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\chapter{Hilbert's Nullstellensatz and applications}

\section{Fields of definition}

When $k$ is an algebraically closed field, Hilbert's Nullstellensatz gives us a bijection
between algebraic subsets of $k^{n}$ and radical ideals in $k[T_1, \ldots, T_n]$.

This correspondence induces an anti-equivalence of categories
\begin{salign*}
    \{\text{affine } k\text{-varieties}\} &\longleftrightarrow
    \{\text{finitely-generated reduced } k \text{-algebras}\} \\
    (X, \mathcal{O}_X) &\longmapsto \mathcal{O}_X(X) \\
    \hat{A} = \operatorname{Hom}_{k\mathrm{-alg}}(A, k) &\longmapsfrom A
.\end{salign*}

\begin{lemma}
    Let $k$ be algebraically closed and $A$ a finitely-generated $k$-Algebra. Then
    the map
    \begin{salign*}
        \hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k) &\longrightarrow \operatorname{Spm } A \\
    \xi &\longmapsto \text{ker } \xi
    \end{salign*}
    is a bijection.
\end{lemma}

\begin{proof}
    The map
    admits an inverse
    \begin{salign*}
        \operatorname{Spm } A &\longrightarrow \operatorname{Hom}_{k\text{-alg}}(A, k) \\
        \mathfrak{m} &\longmapsto (A \to A / \mathfrak{m})
    .\end{salign*}
    This is well-defined, since $A / \mathfrak{m}$ is a finite extension of the algebraically closed field
    $k$, so $k \simeq A / \mathfrak{m}$.
\end{proof}

Since we have defined a product on the left-hand side of the anti-equivalence, this must correspond
to coproduct on the right-hand side. Since the coproduct in the category of commutative
$k$-algebras with unit is given by the tensor product, we have
\[
\mathcal{O}_{X \times Y} (X \times Y) \simeq \mathcal{O}_X(X) \otimes_k \mathcal{O}_Y(Y)
.\]

\begin{korollar}
    Let $k$ be algebraically closed. Then the tensor product of two
    reduced (resp. integral) finitely-generated $k$-algebras is reduced (resp. integral).
    \label{kor:k-alg-closed-tensor-of-reduced}
\end{korollar}

\begin{proof}
    This follows from the anti-equivalence of categories: Reduced since products of affine
    $k$-varieties exist and integral since the product of two irreducible affine $k$-varieties is irreducible.
\end{proof}

\begin{bem}
    \ref{kor:k-alg-closed-tensor-of-reduced} is false in general if $k = \overline{k}$. For instance
    $\mathbb{C}$ is an integral $\R$-algebra, but
    \begin{salign*}
        \mathbb{C} \otimes_{\R} \mathbb{C}
        &= \R[x]/(x^2 + 1) \otimes_{\R} \mathbb{C} \\
        &= \mathbb{C}[x]/(x^2 + 1) \\
        &= \mathbb{C}[x]/((x-i)(x+i)) \\
        &\stackrel{(*)}{\simeq} \mathbb{C}[x]/(x-i) \times \mathbb{C}[x]/(x+i) \\
        &\simeq \mathbb{C} \times \mathbb{C}
    \end{salign*}
    is not integral, where $(*)$ follows from the Chinese remainder theorem.

    For a non-reduced example, consider $k = \mathbb{F}_{p}(t)$ and choose a $p$-th root
    $\alpha = t ^{\frac{1}{p}}$ in $\overline{\mathbb{F}_p(t)}$. Then $\alpha \not\in k$
    but $\alpha ^{n} \in k$. If we put $L = k(\alpha)$, then
    $\alpha \otimes 1 - 1 \otimes \alpha \neq 0$ in $L \otimes_k L$ since
    the elements $(\alpha ^{i} \otimes \alpha ^{j})_{0 \le i, j \le p-1}$ form a basis
    of $L \otimes_k L$ as a $k$-vector space, but
    \[
        (\alpha \otimes 1 - 1 \otimes \alpha)^{p}
        = \alpha ^{p} \otimes 1 - 1 \otimes \alpha ^{p}
        = 1 \otimes \alpha ^{p} - 1 \otimes \alpha ^{p} = 0
    .\] 
\end{bem}

We now consider more generally finitely generated reduced $k$-algebras when $k$ is not
necessarily closed.

\begin{bsp}
    Let $A = \R[X]/(x^2 +1)$. Since $x^2 + 1 $ is irreducible in $\R[x]$, it
    generates a maximal ideal, thus the finitely-generated $\R$-algebra $A$ is a field and in
    particular reduced. We can equip the topogical space
    $X \coloneqq \operatorname{Spm } A = \{ (0)\} $ with a sheaf of regular functions, defined
    by $\mathcal{O}_X(\{(0)\}) = A$. In other words, $\operatorname{Spm } A$ is just a point,
    but equipped with the reduced $\R$-algebra $A$. It thus differs from the
    point $\operatorname{Spm } \R$, which is equipped with the reduced $\R$-algebra $\R$,
    since $\R[x]/(x^2 + 1) \not\simeq \R$ as $\R$-algebras. Indeed, the $\R$-algebra $\R[x]/(x^2+1)$
    is $2$ dimensional as a real vector space.

    $A$ possesses a non-trivial $\R$-algebra automorphism induced by the automorphism of $\R$-algebras,
    $P \mapsto P(-x)$ in $\R[x]$. Indeed, $\R[x]/(x^2+1) \simeq \mathbb{C}$ as $\R$-algebras,
    with the previous automorphism corresponding to the complex conjugation $z \mapsto \overline{z}$.
\end{bsp}

\begin{bsp}
    By analogy with the Zariski topology on maximal spectra of (finitely generated, reduced)
    $\mathbb{C}$-algebras, we can equip $X = \operatorname{Spm } A$ with
    a Zariski topology for all (finitely generated reduced) $\R$-algebras $A$: the closed subsets
    of this topology are given by
    \[
    \mathcal{V}_X(I) \coloneqq \{ \mathfrak{m} \in \operatorname{Spm } A  \mid \mathfrak{m} \supset I\} 
    \] for any ideal $I \subseteq A$.
    Note that $X = \operatorname{Spm } A$ contains
    $\hat{A} = \operatorname{Hom}_{k\text{-alg}}(A, k)$ as a subset: the points
    of $\hat{A}$ correspond to maximal ideals $\mathfrak{m}$ of $A$ with residue field
    $A / \mathfrak{m} \simeq k$. But when $k \not\simeq \overline{k}$, the set
    $\operatorname{Spm } A$ is strictly larger than $\hat{A}$: it contains maximal ideals $\mathfrak{m}$
    such that $A / \mathfrak{m}$ is a non-trivial finite extension of $k$. The induced topology on
    $\hat{A} \subseteq \operatorname{Spm } A$ is the Zariski topologoy of $\hat{A}$ that was
    introduced earlier.

    Let $A = \R[x]$. Maximal ideals in the principal ring $\R[x]$ are generated
    by a single irreducible polynomial $P$, which is either of degree $1$ or of degree $2$ with
    negative discriminant.
    
    In the first case, $P = x-a$ for some $a \in \R$ and the residue field is $\R[x]/(x - a) \simeq \R$,
    while, in the second case, $P = x^2 + bx + c$ for $b, c \in \R$ and $b^2 - 4c < 0$ and
    by choosing a root $z_0$ of $P$ in $\mathbb{C}$, the map
    \begin{salign*}
        \eta_{z_0} \colon \R[x]/(x^2 + bx + c) &\longrightarrow \mathbb{C} \\
                             \overline{P} &\longmapsto P(z_0)
    \end{salign*}
    is a field-homomorphism. In particular it is injective. Since $\mathbb{C}$ and
    $\R[x]/(x^2 + bx + c)$ are both degree $2$ extensions of $\R$, we have
    $\R[x]/(x^2 + bx + c) \simeq \mathbb{C}$.
    Note that the other root of $x^2 + bx +c $ is $\overline{z_0}$ and that
    $\eta_{\overline{z_0}} = \sigma \circ \eta_{z_0}$ where $\sigma$ is complex conjugation on $\mathbb{C}$.
    So we have to ways to identify $\R[x]/(x^2 + bx +c)$ to $\mathbb{C}$ and they are
    related by the action of $\text{Gal}(\mathbb{C}/ \R)$ on $\mathbb{C}$.

    To sum up, the difference between the two possible types of maximal ideals $\mathfrak{m} \subseteq \R[x]$
    is the residue field, which is either $\R$ or $\mathbb{C}$. When it is $\R$, we
    find exactly the points of
    \begin{salign*}
        \widehat{\R[x]}  &= \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R) \\
                     &\simeq \{ \mathfrak{m} \in \operatorname{Spm } \R[x]  \mid \R[x]/\mathfrak{m} \simeq \R\} \\
                     &\simeq \{ (x-a) \colon a \in \R\} \\
                     &\simeq \R
    .\end{salign*}
    And when the residue field is $\mathbb{C}$, we have $\mathfrak{m} = (x^2 + bx + c)$ with
    $b, c \in \R$ such that $b^2 - 4c < 0$. If we choose $z_0$ to be the root
    of $x^2 + bx +c$ with $\text{Im}(z_0) > 0$, we can identify the set of these maximal ideals with
    the subset
    \[
    H \coloneqq \{ z \in \mathbb{C}  \mid \text{Im}(z) > 0\} 
    .\]
    In other words, the following pictures emerges, where we identify
    $\operatorname{Spm } \R[x]$ with
    \[
    \hat{H} \coloneqq \{ z \in \mathbb{C} \mid \text{Im}(z) \ge 0\} 
    \] 
    via the map
    \begin{salign*}
        \operatorname{Spm } \R[x] &\longrightarrow \hat{H} \\
        \mathfrak{m} &\longmapsto \begin{cases}
            a \in \R & \mathfrak{m} = (x-a) \\
            z_0 \in H & \mathfrak{m} = ((x-z_0)(x-\overline{z_0})) \text{ and } \text{Im}(z_0) > 0
        \end{cases}
    \end{salign*}
    which is indeed bijective.
    %\begin{figure}
    %    \centering
    %    \begin{tikzpicture}
    %        \draw[red] (-2, 0) -- (2,0) node[right] {$\R \simeq \operatorname{Hom}_{\R\text{-alg}}(\R[x], \R)$};
    %        \draw[->] (0, 0) -- (0,4);
    %    \end{tikzpicture}
    %    \caption{$\operatorname{Spm } \R[x] \simeq \hat{H}
    %        = \left\{ z \in \mathbb{C} : \text{Im}(z) \ge 0 \right\}$}
    %\end{figure}
    We see that $\operatorname{Spm } \R[x]$ contains a lot more points
    that $\R$. One could go further and add the ideal $(0)$: This would give the set
    \[
    \mathbb{A}^{1}_{\R} = \operatorname{Spec } \R[x]
    = \operatorname{Spm } \R[x] \cup \{(0)\}
    .\] 
\end{bsp}

\begin{bem}
    If $A$ is a $k$-algebra and $\overline{k}$ is an algebraic closure of $k$, the
    group $\text{Aut}_k(\overline{k})$ acts on the $\overline{k}$-algebra
    $A_{\overline{k}} \coloneqq A \otimes_k \overline{k}$ via
    $\sigma (a \otimes \lambda) \coloneqq a \otimes \sigma(\lambda)$. Moreover, the map
    $a \mapsto a \otimes 1$ induces an injective morphism of $k$-algebras
    $A \xhookrightarrow{} A \otimes_k \overline{k}$ since
    the tensor product over fields is left-exact.
    Its image is contained in the $k$-subalgebra
    $\operatorname{Fix}_{\operatorname{Aut}_k(\overline{k})} A_{\overline{k}} \subseteq A_{\overline{k}}$. When
    $k$ is a perfect field, this inclusion is an equality.
\end{bem}

\begin{bsp}
    If $A = \R[x]$, then $A \otimes_{\R} \mathbb{C} \simeq \mathbb{C}[x]$. The group
    $\text{Aut}_{\R}(\mathbb{C}) = \text{Gal}(\mathbb{C}/\R) = \langle \sigma \rangle$  with
    $\sigma\colon z \mapsto \overline{z}$, acts naturally on $\mathbb{C}[x]$. This
    is an action by $\R$-algebra automorphisms. Clearly,
    $\text{Fix}_{\langle\sigma\rangle} \mathbb{C}[x] = \R[x]$. There
    is an induced action on $\operatorname{Spm } \mathbb{C}[x]$,
    defined by
    \[
    \sigma(\mathfrak{m}) = \sigma((x-z)) \coloneqq (x - \sigma(z)) = (x - \overline{z})
    .\] 
    When we identify $\operatorname{Spm } \mathbb{C}[x]$ with $\mathbb{C}$
    via $(x-z) \mapsto z$, this action is just $z \mapsto \overline{z}$. This
    ,,geometric action'' induces an action of $\text{Gal}(\mathbb{C}/\R)$ on
    regular functions on $\mathbb{C}$: to $h \in \mathcal{O}_{\mathbb{C}}(U)$, there
    is associated a regular function $h \in \mathcal{O}_{\mathbb{C}}(\sigma(U))$, defined for
    all $x \in \sigma(U)$, by
    \[
    \sigma(h)(z) \coloneqq \sigma \circ h \circ \sigma ^{-1}(z) = \overline{h(\overline{z})}
    .\]
    In particular, if $h = P \in \mathcal{O}_{\mathbb{C}}(\mathbb{C}) = \mathbb{C}[x]$, then
    $P \mapsto \sigma(P)$ coincides with the natural
    $\text{Gal}(\mathbb{C} / \R)$ action on $\mathbb{C}[x]$. We will see momentarily that this
    defines a sheaf of $\R$-algebras on $\operatorname{Spm } \R[x]$. To that end,
    let us first look more closely at the $\text{Gal}(\mathbb{C}/ \R)$ action
    on $\operatorname{Spm } \mathbb{C}[x]$. Its fixed-point set
    is
    \[
    \{ \mathfrak{m} \in \operatorname{Spm } \mathbb{C}[x]  \mid \mathfrak{m} = (x-a), a \in \R\}
    \simeq \R = \operatorname{Fix}_{z \mapsto \overline{z}}(\mathbb{C})
    .\]
    Moreover, there is a map
    \begin{salign*}
        \operatorname{Spm } \mathbb{C}[x] &\longrightarrow \operatorname{Spm } \R[x] \\
        \mathfrak{m} &\longmapsto \mathfrak{m} \cap \R[x]
    \end{salign*}
    sending $(x-a) \mathbb{C}[x]$ to $(x-a)\R[x]$ if $a \in \R$,
    and $(x-z)\mathbb{C}[x]$ to $(x-z)(x-\overline{z})\R[x]$ if $z \in \mathbb{C} \setminus \R$.
    This map is surjective and induces a bijection
    \[
        (\operatorname{Spm } \mathbb{C}[x]) / \operatorname{Gal}(\mathbb{C} / \R)
        \xlongrightarrow{\simeq} \operatorname{Spm } \R[x]
    .\]
    Geometrically, the quotient map $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$
    is the ,,folding map``
    \begin{salign*}
        \mathbb{C} &\longrightarrow \hat{H} \\
        z = u + iv &\longmapsto u + i |v|
    .\end{salign*}
    \begin{figure}
        \centering
        \begin{tikzpicture}
            \draw[red] (-2, 0) -- (2,0) node[right] {$\R$};
            \fill (1, -1) circle[radius=0.75pt] node[right] {$z_0$};
            \draw[->] (0,-1.5) -- (0,2);
            \draw[->] (3.2,0) -- node[above] {$\pi$} (4.2,0);
            \draw[red] (5, 0) -- (9,0) node[right] {$\R$};
            \fill (8, 1) circle[radius=0.75pt] node[right] {$\pi(z_0)$};
            \draw[->] (7, 0) -- (7,2);
        \end{tikzpicture}
        \caption{The quotient map
        $\pi\colon \operatorname{Spm } \mathbb{C}[x] \to \operatorname{Spm } \R[x]$ is geometrically a folding.}
    \end{figure}
    In view of this, it is natural to
    \begin{enumerate}[(i)]
        \item put the quotient topology on
            \[
            \operatorname{Spm } \R[x] = \left( \operatorname{Spm } \mathbb{C}[x] \right)
            / \operatorname{Gal}(\mathbb{C}/\R)
            \]
            where $\operatorname{Spm } \mathbb{C}[x] \simeq \mathbb{C}$ is equipped with its topology
            of algebraic variety.
        \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:
            \[
            \mathcal{O}_{\operatorname{Spm } \R[x]}(U)
            \coloneqq \mathcal{O}_{\operatorname{Spm } \mathbb{C}[x]}
            (\pi^{-1}(U))^{\operatorname{Gal}(\mathbb{C} / \R)}
            \] 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}