\documentclass{lecture}

\begin{document}

\section{Plane algebraic curves}

\begin{theorem}
    If $f \in k[x,y]$ is an irreducible polynomial such that $\mathcal{V}(f)$
    is infinite, then $\mathcal{I}(\mathcal{V}(f)) = (f)$. In particular,
    $\mathcal{V}(f)$ is irreducible in this case.
    \label{thm:plane-curve-ivf=f}
\end{theorem}

\begin{bem}[]
    \begin{enumerate}[(i)]
        \item If $k$ is algebraically closed and $n \ge 2$, then for all $f \in k[x_1, \ldots, x_n]$
            non-constant, the zero set $\mathcal{V}(f)$ is necessarily infinite.
        \item The assumption $\mathcal{V}(f)$ infinite is necessary for the conclusion of
            \ref{thm:plane-curve-ivf=f} to hold:
            The polynomial
            \[
            f(x,y) = (x^2 - 1)^2 + y^2
            \]
            is irreducible because, as a polynomial in $y$, it is monic and does not have a root
            in $\R[x]$ (for otherwise there would be a polynomial $P(x) \in \R[x]$
            such that $P(x)^2 = -(x^2-1)^2$)
            and the zero set of $f$ is
            \[
            \mathcal{V}(f) = \{ (1, 0)\} \cup \{(-1, 0)\} 
            ,\] which is reducible.
        \item \ref{thm:plane-curve-ivf=f} does not hold in this form for hypersurfaces of $k^{n}$ for $n \ge 3$.
            For instance, the polynomial
            \[
            f(x,y,z) = x^2 y^2 + z^{4} \in \R[x,y,z]
            \] is irreducible and the hypersurface
            \[
            \mathcal{V}(f) = \{ (0, y, 0)\colon y \in \R\} \cup \{(x, 0, 0)\colon x \in \R\} 
            \] is infinite. However, the function
            \[
            P\colon (x,y,z) \mapsto xy
            \] belongs to $\mathcal{I}(\mathcal{V}(f))$ but not to $(f)$. Moreover,
            $P \in \mathcal{I}(\mathcal{V}(f))$ but neither $x$ nor $y$ are in $\mathcal{I}(\mathcal{V}(f))$,
            so this ideal is not prime.
        \item Take $f(x,y) = (x-a)^2 + y^2 \in \R[x,y]$ which is irreducible. Then
            $\mathcal{V}(f) = \{ (a, 0) \} $ is irreducible, and
            $\mathcal{I}(\mathcal{V}(f)) = (x-a, y) \supsetneq (f)$. In particular, $(f)$ is a non-maximal
            prime ideal.
    \end{enumerate}
\end{bem}

We need a special case of the famous Bézout theorem, for which we need a result from algebra.
For an integral domain $R$ denote by $Q(R)$ its fraction field. If $R$ is a factorial ring then
$q \in R[T]$ is called \emph{primitve} if it is non-constant and its
coefficients are coprime in $R$.

\begin{satz}[Gauß]
    Let $R$ be a factorial ring. Then $R[T]$ is also factorial. A polynomial
    $q \in R[T]$ is prime in $R[T]$ if and only if
    \begin{enumerate}[(i)]
        \item $q \in R$ and $q$ is prime in $R$, or
        \item $q$ is primitve in $R[T]$ and prime in $Q(R)[T]$
    \end{enumerate}
    \label{satz:gauss}
\end{satz}

\begin{proof}
    Any algebra textbook.
\end{proof}

\begin{satz}
    Let $R$ be a factorial ring and $f,g \in R[X]$ coprime. Then $f$ and $g$ are
    coprime in $Q(R)[X]$.
    \label{satz:coprime-in-r-is-coprime-in-qr}
\end{satz}

\begin{proof}
    Let $h = \frac{a}{b} \in Q(R)[X]$ be a common irreducible factor of $f$ and $g$ with
    $a \in R[X]$ and $b \in R \setminus 0$. By Gauß $R[X]$ is factorial, thus we
    may assume $a$ irreducible. Then
    \[
    \frac{f}{1} = \frac{p_1}{q_1} \frac{a}{b} \text{ and } \frac{g}{1} = \frac{p_2}{q_2} \frac{a}{b}
    \] for some $p_1, p_2 \in R[X]$ and $q_1, q_2 \in R \setminus 0$.
    So $p_1 a = f q_1 b$ and $p_2 a = g q_2 b$. $a$ neither divides $q_1$, $q_2$ nor $b$, for otherwise
    $a \in R \setminus 0$ by the degree formula for polynomials and $h$ is a unit.
    Since $a$ divides $fq_1 b$ and
    $g q_2 b$ and, since $R[X]$ is factorial, $a$ is prime in $R[X]$ and thus
    $a  \mid f$ and $a  \mid g$.
\end{proof}

\begin{lemma}[Special case of Bézout]
    Let $f, g \in k[x,y]$ be two polynomials without common factors in $k[x,y]$. Then the set
    $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite.
    \label{lemma:coprime-finite-zero-locus}
\end{lemma}

\begin{proof}
    %\begin{enumerate}[(i)]
        %\item Claim: $f$ and $g$ have no common factors in $k(x)[y]$. Indeed, if
        %    $h(x,y) = \frac{H(x,y)}{L(x)} \in k(x)[y]$ is a common irreducible factor of $f$ and $g$,
        %    then we may assume $H$ and $L$ coprime in $k[x,y]$, with $L$ irreducible in $k[x]$
        %    and $H$ irreducible in $k[x,y]$. Thus we can write
        %    \begin{salign*}
        %        f(x,y) &= \frac{A(x,y)}{M(x)} \frac{H(x,y)}{L(x)}
        %        \intertext{and}
        %        g(x,y) &= \frac{B(x,y)}{N(x)} \frac{H(x,y)}{L(x)}
        %    \end{salign*}
        %    with $A$ and $M$ coprime, as well as $B$ and $N$ coprime in $k[x,y]$.
        %    So $A(x,y) H(x,y) = M(x) L(x) f(x,y)$
        %    and $B(x,y) H(x,y) = N(x) L(x) g(x,y)$.
        %    But $H(x,y)$ cannot divide $L(x), M(x)$ nor $N(x)$ in $k[x,y]$, for otherwise
        %    $H(x,y) \in k[x]$, making $h(x,y) = \frac{H(x,y)}{L(x)}$ a unit in $k(x)[y]$. But
        %    $H(x,y)$ is irreducible in $k[x,y]$ and divides both $M(x)L(x)f(x,y)$ and
        %    $N(x)L(x)g(x,y)$, so $H(x,y)$ divides $f(x,y)$ and $g(x,y)$ in $k[x,y]$. Contradiction.
        %\item
    Since $k(x)[y]$ is a principal ideal domain, \ref{satz:coprime-in-r-is-coprime-in-qr} implies
            $(f,g) = k(x)[y]$, hence the existence of $A(x,y), B(x,y), M(x), N(x)$ such that
            \[
            f(x,y) A(x,y) + g(x,y) B(x,y) = \underbrace{M(x) N(x)}_{=: D(x)}
            \] with $D(x) \in k[x]$. Since a common zero $(x,y)$ of $f$ and $g$ gives a zero of
            $D$, and $D$ has finitely many zeros, there are only finitely many $x$ such that
            $(x,y)$ is a zero of both $f$ and $g$. But, for fixed $x \in k$, the polynomial
            \[
            y \mapsto f(x,y) - g(x,y)
            \] has only finitely many zeros in $k$. So $\mathcal{V}(f) \cap \mathcal{V}(g)$ is finite.
    %\end{enumerate}
\end{proof}

\begin{proof}[Proof of \ref{thm:plane-curve-ivf=f}]
    Let $f \in k[x,y]$ be irreducible such that $\mathcal{V}(f) \subseteq k^2$ is infinite.
    Since $f \in \mathcal{I}(\mathcal{V}(f))$, it suffices to show that
    $\mathcal{I}(\mathcal{V}(f)) \subseteq (f)$.
    Let
    $g \in \mathcal{I}(\mathcal{V}(f))$. Then $\mathcal{V}(f) \subseteq \mathcal{V}(g)$. Thus
    \[
    \mathcal{V}(f) \cap \mathcal{V}(g) = \mathcal{V}(f)
    \] which is infinite by assumption. Thus by \ref{lemma:coprime-finite-zero-locus},
    $f$ and $g$ have a common factor. Since $f$ is irreducible, this implies that $f  \mid g$, i.e.
    $g \in (f)$.
\end{proof}

We can use \ref{thm:plane-curve-ivf=f} to find the irreducible components of a
hypersurface $\mathcal{V}(P) \subseteq k^2$.

\begin{korollar}
    Let $P \in k[x,y]$ be non-constant and $P = u P_1^{n_1} \cdots P_r^{n_r}$ be the decomposition
    into irreducible factors. If each $\mathcal{V}(P_i)$ is infinite, then the algebraic sets
    $\mathcal{V}(P_i)$ are the irreducible components of $\mathcal{V}(P)$.
\end{korollar}

\begin{proof}
    Note that
    \[
    \mathcal{V}(P) = \mathcal{V}(P_1^{n_1} \cdots P_r^{n_r}) = \mathcal{V}(P_1) \cup \ldots \cup \mathcal{V}(P_r)
    .\] Since $P_i$ is irreducible and $\mathcal{V}(P_i)$ is infinite for all $i$,
    by \ref{thm:plane-curve-ivf=f} $\mathcal{V}(P_i)$ is irreducible and for $i \neq j$
    $\mathcal{V}(P_i) \not\subset \mathcal{V}(P_j)$, for otherwise
    \[
        (P_i) = \mathcal{I}(\mathcal{V}(P_i)) \supset \mathcal{I} (\mathcal{V}(P_j)) = (P_j)
    \] which is impossible for distinct irreducible elements $P_i, P_j$.
\end{proof}

\begin{bsp}[Real plane cubics]
    Let $P(x,y) = y^2 - f(x)$ with $\text{deg}_xf = 3$ in $k[x]$. Since $\text{deg}_y P \ge 1 $
    and the leading coefficient of $P$ is $1$, the polynomial $P$ is primitive in $k[x][y]$.
    It is reducible in $k(x)[y]$ if and only if there exists $a(x), b(x) \in k(x)$ such that
    $(y-a)(y-b) = y^2 - f$, i.e. $b = -a$ and $f = a^2$ in $k(x)$, therefore also in $k[x]$.
    Since $\text{deg}_xf = 3 $, this cannot happen. So, $P$ is irreducible
    by \ref{satz:gauss}.

    Moreover, when $k = \R$, the
    cubic polynomial $f(x)$ takes on an infinite number of positive values,
    so $\mathcal{V}(y^2 - f(x)) = \mathcal{V}(P)$ is infinite. In conclusion,
    real cubics of the form $y^2 - f(x) = 0$ are irreducible algebraic sets in $\R^2$
    by \ref{thm:plane-curve-ivf=f}.
\end{bsp}

\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            xmin = -1
            ]
        \algebraiccurve[red][$y^2 = x^3$]{y^2 - x^3}
        \end{axis}
    \end{tikzpicture}
    \caption{the cuspidal cubic}
\end{figure}

\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            ]
        \algebraiccurve[red][$y^2 = x^2(x+1)$][-2:2][-2:2]{y^2 - x^2*(x+1)}
        \end{axis}
    \end{tikzpicture}
    \caption{the nodal cubic}
\end{figure}

\begin{figure}
    \centering
    \begin{tikzpicture}[scale=0.9]
        \begin{axis}[
            xmin = -1
            ]
        \algebraiccurve[red][$y^2 = x(x^2+1)$][-2:2][-2:2]{y^2 - x*(x^2+1)}
        \end{axis}
    \end{tikzpicture}
    \hspace{.05\textwidth}
    \begin{tikzpicture}[scale=0.9]
        \begin{axis}[
            ]
        \algebraiccurve[red][$y^2 = x(x^2-1)$][-2:2][-2:2]{y^2 - x*(x^2-1)}
        \end{axis}
    \end{tikzpicture}
    \caption{the smooth cubics: the second curve demonstrates that the notion of connectedness in
        the Zariski topologoy of $\R^2$ is very different from the one in the usual topology of $\R^2$.}
\end{figure}

\begin{satz}
    Let $k$ be an algebraically closed field and let $P \in k[T_1, \ldots, T_n]$ be a non-constant polynomial
    with $n \ge 2$. Then $\mathcal{V}(P)$ is infinite.
\end{satz}

\begin{proof}
    Since $P$ is non-constant, we may assume that $\text{deg}_{x_1} P \ge 1$. Write
    \[
    P(T_1, \ldots, T_n) = \sum_{i=1}^{d} g_i(T_2, \ldots, T_n) T_1^{i}
    ,\]
    with $d \ge 1$ and $g_d \neq 0$. Then $D_{k^{n-1}}(g_d)$ is infinite: Since $g_d \neq 0$ and
    $k$ infinite, it is non-empty. Thus let $(a_2, \ldots, a_n) \in k^{n-1}$ such that
    $g_d(a) \neq 0$. Then $g_d(ta) = g_d(ta_2, \ldots, ta_n) \in k[t]$ is a non-zero polynomial and thus
    has only finitely many zeros in $k$. In particular $D_{k^{n-1}}(g_d)$ is infinite.

    For $(a_2, \ldots, a_{n-1}) \in D_{k^{n-1}}(g_d)$, $P(T_1, a_2, \ldots, a_n) \in k[T_1]$ is non-constant
    and thus has a root $a_1$ in the algebraically closed field $k$. Hence
    $(a_1, \ldots, a_n) \in \mathcal{V}(P)$.
\end{proof}

We finally give a complete classification of irreducible algebraic sets in the affine plane $k^2$ for
an infinite field $k$.

\begin{satz}
    Let $k$ be an infinite field. Then the irreducible algebraic subsets of $k^2$ are:
    \begin{enumerate}[(i)]
        \item the whole affine plane $k^2$
        \item single points $\{ (a, b) \} \subseteq k^2$
        \item infinite algebraic sets defined by an irreducible polynomial $f \in k[x,y]$.
    \end{enumerate}
    \label{satz:classification-irred-alg-subsets-plane}
\end{satz}

\begin{proof}
    Let $V \subseteq k^2$ be an irreducible algebraic subset of the affine plane. If $V$ is finite,
    it reduces to a point. So we may assume $V$ infinite. If $\mathcal{I}(V) = (0)$, then $V = k^2$.
    Otherwise, there is a non-constant polynomial $P \in k[x,y]$ such that $P$ vanishes on $V$. Since
    $V$ is irreducible, $\mathcal{I}(V)$ is prime, so it contains an irreducible factor $f$ of $P$.
    Let $g \in \mathcal{I}(V)$. Then $V \subseteq \mathcal{V}(f) \cap \mathcal{V}(g)$, but since
    $V$ is infinite, $f$ and $g$ must have a common factor. By irreducibility of $f$, it follows
    $f  \mid g$, i.e. $g \in (f)$. Hence $\mathcal{I}(V) = (f)$ and $V = \mathcal{V}(f)$.
\end{proof}

\end{document}
