\documentclass{lecture}

\begin{document}

\section{Geometric Noether normalisation}

Consider a plane algebraic curve $\mathcal{C}$, defined by the equation $f(x,y) = 0$.
If we fix $x = a$, then the polynomial equation $f(a, y) = 0$ has only finitely many solutions
(at most $\text{deg}_y f$). This means that the map
\begin{salign*}
    \mathcal{C} \coloneqq \mathcal{V}(f) &\to k
    (x,y) \mapsto x
\end{salign*}
has finite fibres. A priori, such a map is not surjective, e.g. for $f(x,y) = xy - 1$. If
$k$ is algebraically closed, one can always find such a surjective projection.

\begin{theorem}
    Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial
    of degree $d \ge 1$. Then there is a morphism of affine varieties
    \[
    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
    \] 
    such that:
    \begin{enumerate}[(i)]
        \item $\pi$ is surjective
        \item for $t \in k^{n-1}$, the fibre $\pi^{-1}(\{t\}) \subseteq \mathcal{V}(f)$ consists
            of at most $d$ points.
    \end{enumerate}
    \label{thm:geom-noether-norm}
\end{theorem}

\begin{proof}
    Let $f \in k[x_1, \ldots, x_n]$ be of degree $d$. We construct a change of variables
    of the form $(x_i \mapsto x_i + a_i x_n)_{1 \le i \le n-1}$ and
    $x_n \mapsto x_n$, such that the term of degree $d$ of
    $f(x_1 + a_1x_n, \ldots, x_{n-1} + a_{n-1}x_n, x_n)$ becomes
    $c x_n^{d}$ with $c \in k^{\times }$. Since
    \begin{salign*}
    f(x_1 + a_1 x_n, \ldots, x_{n-1} + a_{n-1} x_n, x_n)
    =
    \sum_{(i_1, \ldots, i_n) \in \N^{n}} \alpha_{i_1, \ldots, i_n}
    (x_1 + a_1 x_n)^{i_1} \cdots (x_{n-1} + a_{n-1} x_n)^{i_{n-1}} x_n^{i_n}
    ,\end{salign*}
    the coefficient of $x_n^{d}$ in the above equation is obtained by considering all
    $(i_1, \ldots, i_n)$ such that $i_1 + \ldots + i_n = d$, and keeping only the term
    in $x_n^{i_j}$ when expanding $(x_j + a_j x_n)^{i_j}$, so we get
    \[
    \sum_{(i_1, \ldots, i_n) \in \N \\ i_1 + \ldots + i_n = d}
    \alpha_{i_1, \ldots, i_n} a_1^{i_1} \cdots a_{n-1}^{i_{n-1}}
    ,\] which is equal to $f_d(a_1, \ldots, a_{n-1}, 1)$, where
    $f_d$ is the (homogeneous) degree $d$ part of $f$.

    Claim: There exist $a_1, \ldots, a_{n-1} \in k$ such that $f_d(a_1, \ldots, a_{n-1}, 1) \neq 0$.
    Proof of claim by induction: if $n = 1$, $f_d = c x_1^{d}$ for some $c \neq 0$, so
    $f_d(1) = c \neq 0$. If $n \ge 2$, we can write
    \[
    f_d(x_1, \ldots, x_n) = \sum_{i=0}^{d} h_i(x_2, \ldots, x_n) x_1^{i}
    \] where $h_i \in k[x_2, \ldots, x_n]$ is homogeneous of degree $d-i$.
    Since $f_d \neq 0$, there is at least one $i_0$ such that $h_{i_0} \neq 0$. By induction,
    we can find $(a_2, \ldots, a_{n-1}) \in k^{n-2}$ such that
    $h_{i_0}(a_2, \ldots, a_{n-1}, 1) \neq 0$. But then
    $f(\cdot, a_2, \ldots, a_{n-1}, 1) \in k[x_1]$ is a non zero polynomial, so it has
    only finitely many roots. As $k$ is infinite, there exists $a_1 \in k$, such that
    $f(a_1, \ldots, a_{n-1}, 1) \neq 0$.

    Then
    \[
    \varphi\colon \begin{cases}
        x_i \mapsto x_i + a_i x_n & 1 \le i \le n-1\\
        x_n \mapsto x_n
    \end{cases}
    \] is a invertible linear transformation $k^{n} \to k^{n}$, such that
    \[
        (f \circ \varphi^{-1})(y_1, \ldots, y_n)
        = c (y_n^{d} + g_1(y_1, \ldots, y_n) y_n^{d-1} + \ldots + g_d(y_1, \ldots, y_{n-1})
    \] for $c \neq 0$. This induces an isomorphism of affine varieties
    \begin{salign*}
        \mathcal{V}(f) &\to \mathcal{V}(f \circ \varphi^{-1}) \\
    x &\mapsto \varphi(x)
    \end{salign*}
    such that
    \[
    \begin{tikzcd}
        \mathcal{V}(f) \arrow[hookrightarrow]{r}{\varphi} \arrow[dashed]{dr}{\pi} & \arrow{d} k^{n} = k^{n-1} \times k \\
                       & k^{n-1}
    \end{tikzcd}
    \] defines the morphism $\pi$ with the desired properties. Indeed:
    Let $(x_1, \ldots, x_n) \in k^{n}$ and set $y_i \coloneqq \varphi(x_i)$. Then

    $(x_1, \ldots, x_n) \in \mathcal{V}(f)$ iff $x_n = y_n$
    is a root of the polynomial
    \[
    t ^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) t ^{d-j}
    .\] Therefore for all $t = (y_1, \ldots, y_{n-1}) \in k^{n-1}$,
    $\pi^{-1}(\{t\}) \neq \emptyset$ (because $\overline{k} = k$) and
    $\pi^{-1}(\{t\})$ has at most $d$ points.
\end{proof}

\begin{definition}
    Let $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d$.
    As in the proof of \ref{thm:geom-noether-norm}, ther exists a linear coordinate transformation
    $\varphi\colon k^{n} \to k^{n}$, such that
    $f \circ \varphi^{-1}(y_1, \ldots, y_n) = c y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1})y_n^{d-j}$. For a point $x \in \pi^{-1}(y_1, \ldots, y_{n-1}) \subseteq \mathcal{V}(f)$,
    the \emph{multiplicity} of $x$ is the multiplicity of $y_n$ as a root of that polynomial.

    A point with multiplicity $\ge 2$ are called \emph{ramification point} and
    its image lies in the \emph{discriminant locus} of $\pi$.
\end{definition}

With this vocabulary, we can refine the statement of \ref{thm:geom-noether-norm}.

\begin{definition}[Geometric Noether normalisation]
    Assume $k = \overline{k}$. If $f \in k[x_1, \ldots, x_n]$ is polynomial
    of degree $d$, a morphism of affine varieties
    \[
    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
    \] such that
    \begin{enumerate}[(i)]
        \item $\pi$ is surjective
        \item for $t \in k^{n-1}$, the number of elements in $\pi^{-1}(\{t\})$, counted
            with their respective multiplicities, is exactly $d$,
    \end{enumerate}
    is called a \emph{geometric Noether normalisation}.
\end{definition}

\begin{korollar}[Geometric Noether normalisation for hypersurfaces]
    Let $k$ be an algebraically closed field and $f \in k[x_1, \ldots, x_n]$ be a polynomial
    of degree $d \ge 1$. Then there exists a geometric Noether normalisation.
\end{korollar}

\begin{bsp}
    Let $f(x,y) = y^2 - x^{3} \in \mathbb{C}[x,y]$. Then the map
    \begin{salign*}
        \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C}
        (x,y) &\mapsto y
    \end{salign*}
    is a geometric Noether normalisation, but
    $(x,y) \mapsto x$ is not (the fibres of the latter have degree $2$, while $\text{deg } f = 3$).
\end{bsp}

\begin{bem}
    In the proof of \ref{thm:geom-noether-norm}, to construct $\varphi$ and
    the $g_j$, we only used that $k$ is infinte. Thus the statement, that
    for all $f \in k[x_1, \ldots, x_n]$ there exists a linear automorphism
    $\varphi\colon k^{n} \to k^{n}$ such that
    \[
    f \circ \varphi^{-1}(y_1, \ldots, y_n)
    = c \left(y_n^{d} + \sum_{j=1}^{d} g_j(y_1, \ldots, y_{n-1}) y_n^{d-j}\right)
    \] is valid over $k$ if $k$ is infinite. The resulting map
    \[
    \pi\colon \mathcal{V}_{k^{n}}(f) \to k^{n-1}
    \] still has finite fibres, but it is no longer surjective in general, as
    the example $f(x,y) = x^2 + y^2 - 1$ shows.

    However, it induces a surjective map with finite fibres
    \[
    \hat{\pi}\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1}
    \] which moreover commutes with the action of $\text{Gal}(\overline{k} / k)$.
\end{bem}

\begin{theorem}
    Let $k$ be an infinite field and $\overline{k}$ an algebraic closure of $k$. Let
    $f \in k[x_1, \ldots, x_n]$ be a polynomial of degree $d \ge 1$. Then there exists
    a $\text{Gal}(\overline{k} / k)$-equivariant geometric Noether normalisation map
    $\pi\colon \mathcal{V}_{\overline{k}^{n}}(f) \to \overline{k}^{n-1}$.
\end{theorem}

\begin{bsp}[]
    Let $f(x,y) = y^2 - x^{3} \in \R[x,y]$. Then the map
    \begin{salign*}
        \pi\colon \mathcal{V}_{\mathbb{C}^2}(y^2 - x^{3}) &\to \mathbb{C} \\
        (x,y) &\mapsto y
    .\end{salign*}
    is a geometric Noether normalisation map and it is Galois-invariant:
    \[
    \pi(\overline{(x,y)}) = \pi(\overline{x}, \overline{y}) = \overline{y} = \overline{\pi(x,y)}
    .\] 
\end{bsp}

\begin{aufgabe}[]
    Show that if $y \in \R$, the group $\text{Gal}(\mathbb{C} / \R)$ acts on $\pi^{-1}(\{y\})$,
    and that the fixed point set of that action is in bijection with
    $\{x \in \R \mid y^2 - x^{3} = 0\} $.
\end{aufgabe}

Next, we want to generalise the results above beyond the case of hypersurfaces.

\begin{theorem}
    Assume $k$ is algebraically closed. Let $V \subseteq k^{n}$ be an algebraic set.
    Then there exists a natural number $r \le n$ and a morphism of algebraic sets
    \[
    p\colon V \to k^{r}
    \] such that $p$ is surjective and has finite fibres.
    \label{thm:geom-noether-norm-general}
\end{theorem}

\begin{proof}[Sketch of proof]
    If $V = k^{n}$, we take $r = n$ and $p = \text{id}_{k^{n}}$. Otherwise
    $V = \mathcal{V}(I)$ with $I \subseteq k[x_1, \ldots, x_n]$ a non-zero ideal.
    Take $f \in I \setminus \{0\} $. Then there exists a geometric Noether normalisation
    \[
    p_1\colon \mathcal{V}(f) \to k^{n-1}
    .\]
    One can now show that $V_1 \coloneqq p_1(V)$ is an algebraic set in $k^{n-1}$. Thus there are
    two cases:
    \begin{enumerate}[(1)]
        \item $p_1(V) = k^{n-1}$. Thus $p_1|_V\colon V \to k^{n-1}$ is surjective with finite fibres
            and we are done.
        \item $p_1(V) \subsetneq k^{n-1}$. In this case
            $p_1(V) = \mathcal{V}(I_1)$ with $I_1 \subseteq k[x_1, \ldots, x_{n-1}]$ a
            non-zero ideal. So we can repeat the argument.
    \end{enumerate}
    After $r \le n$ steps, the above algorithm terminates, and this happens precisely when
    $V_r = k^{n-r}$. If we set
    \[
    p\coloneqq p_r \circ \ldots \circ p_1 \colon V \to k^{n-r}
    \] then $p$ is surjective with finite fibres because $p(V) = V_r = k^{n-r}$ and
    each $p_i$ has finite fibres.
\end{proof}

\begin{bem}[]
    By the fact used in the proof of \ref{thm:geom-noether-norm-general}, $p$ is in fact
    a closed map. Note that when $r = n$, $V = p^{-1}(\{0\})$ is actually finite, in which case
    $\text{dim }V$ should indeed be $0$.
\end{bem}

\end{document}
