\documentclass{lecture}

\begin{document}

\section{Prime ideals in $k[x,y]$}

\begin{satz}
    Let $A$ be a principal ideal domain. Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. Then
    $\mathfrak{p}$ satisfies exactly one of the following three mutually exclusive possibilities:
    \begin{enumerate}[(i)]
        \item $\mathfrak{p} = (0)$
        \item $\mathfrak{p} = (f)$, where $f \in A[X]$ is irreducible
        \item $\mathfrak{p} = (a, q)$, where $a \in A$ is irreducible and
            $q \in A[X]$ such that its reduction modulo $a A$ is an irreducible element
            in $A / aA [X]$. In this case, $\mathfrak{p}$ is a maximal ideal.
    \end{enumerate}
    \label{thm:class-prim-pol-pid}
\end{satz}

\begin{proof}
    Let $\mathfrak{p} \subseteq A[X]$ be a prime ideal. If $\mathfrak{p}$ is principal, then
    $\mathfrak{p} = (f)$ for some $f \in A[X]$. If $f = 0$, we are done. Otherwise,
    since $A[X]$ is factorial by Gauß and $\mathfrak{p}$ is prime, $f$ is irreducible.

    Let now $\mathfrak{p}$ not be principal. Then there exist $f, g \in \mathfrak{p}$ without
    common factors in $A[X]$. By \ref{satz:coprime-in-r-is-coprime-in-qr}, they
    also have no common factors in the principal ideal domain $Q(A)[X]$, so
    $Mf + Ng = 1$ for some $M, N \in Q(A)[X]$. By multiplying with the denominators, we obtain
    $Pf + Qg = b$ for some $b \in A$ and $P, Q \in A[X]$. So $b \in (f, g) \subseteq \mathfrak{p}$,
    thus there is an irreducible factor $a$ of $b$ in $A$ such that $a \in \mathfrak{p}$.
    Moreover, $a A[X] \subsetneq \mathfrak{p}$ since $\mathfrak{p}$ is not principal. Now consider
    the prime ideal
    \[
    \mathfrak{p} / a A[X] \subset  A[X]/aA[X] \simeq \left( A / aA \right)[X]
    .\] Since $A$ is a PID and $a$ is irreducible, $A/aA$ is a field and $(A / aA)[X]$ a PID.
    So $\mathfrak{p}/aA[X]$ is generated by an irreducible element $\overline{q} \in (A/aA)[X]$
    for some $q \in A[X]$. Thus $\mathfrak{p} = (a, q)$. Moreover
    \[
    \faktor{A[X]}{\mathfrak{p}} \simeq
    \faktor{\left(\faktor{A}{aA}\right)[X]}{\left(\faktor{\mathfrak{p}}{aA}\right)[X]}
    =
    \faktor{\left( \faktor{A}{aA} \right)[X] }
    {\overline{q} \left( \faktor{A}{aA} \right)[X] }
    \] which is a field since $\left( \faktor{A}{aA} \right)[X]$ is a PID. So $\mathfrak{p}$ is maximal in
    $A[X]$.
\end{proof}

Using \ref{thm:class-prim-pol-pid} we can give a simple proof for the classification of maximal ideals
of $k[T_1, \ldots, T_n]$ when $k$ is algebraically closed and $n=2$.

\begin{korollar}
    If $k$ is algebraically closed, a maximal ideal $\mathfrak{m}$ of $k[x,y]$ is of the form
    $\mathfrak{m} = (x-a, y-b)$ with $(a, b) \in k^2$. In particular, principal ideals are never maximal.
    \label{kor:max-ideals-alg-closed-k2}
\end{korollar}

\begin{proof}
    Since $\mathfrak{m}$ is maximal, it is prime and $\mathfrak{m} \neq (0)$. By
    \ref{thm:class-prim-pol-pid}, $\mathfrak{m} = (P, f)$
    with $P \in k[x]$ irreducible and $f \in k[x,y]$ such that
    its image $\overline{f}$ in $(k[x]/(P))[y]$ is irreducible or
    $\mathfrak{m} = (f)$ for $f \in k[x,y]$ irreducible.

    \begin{enumerate}[(1)]
        \item $\mathfrak{m} = (P, f)$. Since $k$ is algebraically closed and $P \in k[x]$ is irreducible,
            $P = x - a$ for some $a \in k$.
            \[
            k[x]/(P) = k[x]/(x-a) \simeq k
            .\] Since $\overline{f} \in k[y]$ is also irreducible, $\overline{f} = y - b$ for some $b \in k$.
        \item $\mathfrak{m} = (f)$. Since $k = \overline{k}$, $\mathcal{V}(f)$ is infinite, in particular
            $\mathcal{V}(f) \neq \emptyset$. Then if $(a,b) \in \mathcal{V}(f)$,
            \[
                (x-a, y-b) = \mathcal{I}(\{(a, b)\})
                \supset \mathcal{I}(\mathcal{V}(f)) \supset (f)
            .\] Since $(f)$ is maximal, it follows that $(f) = (x-a, y-b)$, which is impossible since
            $x -a $ and $y-b$ habe no common factors in $k[x,y]$.
    \end{enumerate}
\end{proof}

\begin{bem}[]
    The ideal $(x^2 + 1, y)$ is maximal in $\R[x,y]$ and is not of the form $(x-a, y-b)$ for $(a,b) \in \R^2$.
    Indeed,
    \[
    \faktor{\R[x,y]}{(x^2 + 1, y)}
    \simeq
    \faktor{\left( \R[y]/y\R[y] \right)[x]}{(x^2 + 1)}
    \simeq \R[x]/(x^2 + 1)
    \simeq \mathbb{C}
    .\] 
\end{bem}

\begin{satz}[]
    Let $k$ be an algebraically closed field. Then the maps $V \mapsto \mathcal{I}(V)$
    and $I \mapsto \mathcal{V}(I)$ induce a bijection
    \begin{salign*}
    \{ \text{irreducible algebraic subsets of } k^2\}
    &\longleftrightarrow \{ \text{prime ideals in } k[x,y]\} 
    \intertext{through wich we have correspondences}
        \text{points } (a, b) \in k^2 &\longleftrightarrow \text{maximal ideals } (x-a, y-b) \text{ in }k[x,y] \\
        \text{proper, infinite, irreducible algebraic sets}
                                      &\longleftrightarrow \text{prime ideals } (f) \subseteq k[x,y]
                                      \text{ with } f \text{ irreducible} \\
        k^2 &\longleftrightarrow (0)
    .\end{salign*} 
    \label{satz:correspondence-irred-subsets-prime-ideals}
\end{satz}

\begin{proof}
    Let $V \subseteq k^2$ be an irreducible algebraic set. By
    \ref{satz:classification-irred-alg-subsets-plane} we
    can distinguish the following cases:
    \begin{enumerate}[(i)]
        \item If $V = k^2$, then $\mathcal{I}(V) = (0)$ since $k$ is infinite and
            $\mathcal{I}(\mathcal{V}(0)) = (0)$.
        \item If $V = \{(a,b)\} $, then $\mathcal{I}(V) \supset (x-a, y-b) \eqqcolon \mathfrak{m} $. Since
            $\mathfrak{m}$ is maximal, $\mathcal{I}(V) = \mathfrak{m}$. Since $V = \mathcal{V}(\mathfrak{m})$,
            this also shows $\mathcal{I}(\mathcal{V}(\mathfrak{m})) = \mathfrak{m}$.
        \item If $V = \mathcal{V}(f)$ where $f \in k[x,y]$ is irreducible, 
            then by \ref{thm:plane-curve-ivf=f} $\mathcal{I}(\mathcal{V}(f)) = (f)$.
    \end{enumerate}
    So, every irreducible algebraic set $V \subseteq k^2$ is of the form
    $\mathcal{V}(\mathfrak{p})$ for some prime ideal $\mathfrak{p} \subseteq k[x,y]$. Moreover,
    \[
    \mathcal{I}(\mathcal{V}(\mathfrak{p})) = \mathfrak{p}
    .\]
    Let now $\mathfrak{p}$ be a prime ideal in $k[x,y]$. By \ref{thm:class-prim-pol-pid} we can dinstiguish
    the following cases:
    \begin{enumerate}[(i)]
        \item $\mathfrak{p} = (0)$: Then $\mathcal{V}(\mathfrak{p}) = k^2$ and
            since $k$ is infinite, $k^2$ is irreducible.
        \item $\mathfrak{p}$ maximal: Then by \ref{kor:max-ideals-alg-closed-k2},
            $\mathfrak{p} = (x-a, y-b)$ for some $(a, b) \in k^2$. So $\mathcal{V}(m) = \{(a, b)\}$
            is irreducible.
        \item $\mathfrak{p} = (f)$ with $f \in k[x,y]$ irreducible. Since $k = \overline{k}$,
            $\mathcal{V}(f)$ is infinite and hence by \ref{thm:plane-curve-ivf=f} irreducible.
    \end{enumerate}
    Thus the maps in the proposition are well-defined, mutually inverse and induce the stated
    correspondences.
\end{proof}

\begin{korollar}
    Assume that $k$ is algebraically closed and let $\mathfrak{p} \subseteq k[x,y]$ be a prime ideal.
    Then
    \[
    \mathfrak{p} = \bigcap_{\mathfrak{m} \; \mathrm{maximal}, \mathfrak{m} \supset \mathfrak{p}} 
    \mathfrak{m}
    .\] 
\end{korollar}

\begin{proof}
    If $\mathfrak{p}$ is maximal, there is nothing to prove. If $\mathfrak{p} = (0)$, $\mathfrak{p}$
    is contained in $(x-a, y-b)$ for $(a, b) \in k^2$. Since $k$ is infinite, the intersection
    of these ideals is $(0)$. Otherwise, by \ref{satz:correspondence-irred-subsets-prime-ideals},
    $\mathfrak{p} = (f)$ for some $f \in k[x,y]$ irreducible. Then, since $k = \overline{k}$,
    $\mathcal{V}(f)$ is infinite and with \ref{thm:plane-curve-ivf=f}:
    \[
    \mathfrak{p} = (f) = \mathcal{I}(\mathcal{V}(f))
    = \mathcal{I}\left( \bigcup_{(a, b) \in \mathcal{V}(f)}  \{(a, b)\} \right)
    \supset \bigcap_{(a,b) \in \mathcal{V}(f)}
    \mathcal{I}(\{(a,b)\})
    \supset (f) = \mathfrak{p}
    .\] By \ref{satz:correspondence-irred-subsets-prime-ideals}, the ideals
    $\mathcal{I}(\{(a,b)\})$ for $(a, b) \in \mathcal{V}(f)$ are exactly the
    maximal ideals containing $(f) = \mathfrak{p}$.
\end{proof}

\begin{korollar}
    Let $\mathfrak{p} \subseteq k[x,y]$ be a non-principal prime ideal.
    Then $\mathcal{V}(\mathfrak{p}) \subseteq k^2$ is finite.
\end{korollar}

\begin{proof}
    Since $\mathfrak{p}$ is not principal, there exist $f, g \in \mathfrak{p}$ without common factors. Since
    $(f, g) \subset \mathfrak{p}$, we have
    \[
    \mathcal{V}(f) \cap \mathcal{V}(g) =
    \mathcal{V}(f, g) \supset \mathcal{V}(\mathfrak{p})
    \] and the left hand side is finite by \ref{lemma:coprime-finite-zero-locus}.
\end{proof}

\end{document}