This tag has label algebra-theorem-nullstellensatz, it is called Hilbert Nullstellensatz in the Stacks project and it points to
The corresponding content:
Theorem 9.31.1 (Hilbert Nullstellensatz). Let $k$ be a field.The same is true in any finite type $k$-algebra.
- For any maximal ideal $\mathfrak m \subset k[x_1, \ldots, x_n]$ the field extension $k \subset \kappa(\mathfrak m)$ is finite.
- Any radical ideal $I \subset k[x_1, \ldots, x_n]$ is the intersection of maximal ideals containing it.
Proof. It is enough to prove part (1) of the theorem for the case of a polynomial algebra $k[x_1, \ldots, x_n]$, because any finitely generated $k$-algebra is a quotient of such a polynomial algebra. We prove this by induction on $n$. The case $n = 0$ is clear. Suppose that $\mathfrak m$ is a maximal ideal in $k[x_1, \ldots, x_n]$. Let $\mathfrak p \subset k[x_n]$ be the intersection of $\mathfrak m$ with $k[x_n]$.
If $\mathfrak p \not = (0)$, then $\mathfrak p$ is maximal and generated by an irreducible monic polynomial $P$ (because of the Euclidean algorithm in $k[x_n]$). Then $k' = k[x_n]/\mathfrak p$ is a finite field extension of $k$ and contained in $\kappa(\mathfrak m)$. In this case we get a surjection $$ k'[x_1, \ldots, x_{n-1}] \to k'[x_1, \ldots, x_n] = k' \otimes_k k[x_1, \ldots, x_n] \longrightarrow \kappa(\mathfrak m) $$ and hence we see that $\kappa(\mathfrak m)$ is a finite extension of $k'$ by induction hypothesis. Thus $\kappa(\mathfrak m)$ is finite over $k$ as well.
If $\mathfrak p = (0)$ we consider the ring extension $k[x_n] \subset k[x_1, \ldots, x_n]/\mathfrak m$. This is a finitely generated ring extension, hence of finite presentation by Lemmas 9.29.3 and 9.29.4. Thus the image of $\mathop{\rm Spec}(k[x_1, \ldots, x_n]/\mathfrak m)$ in $\mathop{\rm Spec}(k[x_n])$ is constructible by Theorem 9.27.9. Since the image contains $(0)$ we conclude that it contains a standard open $D(f)$ for some $f\in k[x_n]$ nonzero. Since clearly $D(f)$ is infinite we get a contradiction with the assumption that $k[x_1, \ldots, x_n]/\mathfrak m$ is a field (and hence has a spectrum consisting of one point).
To prove part (2) let $I \subset R$ be radical, with $R$ of finite type over $k$. Let $f \in R$, $f \not \in I$. Pick a maximal ideal $\mathfrak m'$ in the nonzero ring $R_f/IR_f = (R/I)_f$. Let $\mathfrak m \subset R$ be the inverse image of $\mathfrak m'$ in $R$. We see that $I \subset \mathfrak m$ and $f \not \in \mathfrak m$. If we show that $\mathfrak m$ is a maximal ideal of $R$, then we are done. We clearly have $$ k \subset R/\mathfrak m \subset \kappa(\mathfrak m'). $$ By part (1) the field extension $k \subset \kappa(\mathfrak m')$ is finite. By elementary field theory we conclude that $R/\mathfrak m$ is a field. $\square$
\begin{theorem}[Hilbert Nullstellensatz]
\label{theorem-nullstellensatz}
Let $k$ be a field.
\begin{enumerate}
\item
\label{item-finite-kappa}
For any maximal ideal $\mathfrak m \subset k[x_1, \ldots, x_n]$
the field extension $k \subset \kappa(\mathfrak m)$ is finite.
\item
\label{item-polynomial-ring-Jacobson}
Any radical ideal $I \subset k[x_1, \ldots, x_n]$
is the intersection of maximal ideals containing it.
\end{enumerate}
The same is true in any finite type $k$-algebra.
\end{theorem}
\begin{proof}
It is enough to prove part (\ref{item-finite-kappa}) of
the theorem for the case of a polynomial
algebra $k[x_1, \ldots, x_n]$, because any finitely generated
$k$-algebra is a quotient of such a polynomial algebra.
We prove this by induction on $n$. The case $n = 0$ is clear.
Suppose that $\mathfrak m$ is a maximal ideal in $k[x_1, \ldots, x_n]$.
Let $\mathfrak p \subset k[x_n]$ be the intersection
of $\mathfrak m$ with $k[x_n]$.
\medskip\noindent
If $\mathfrak p \not = (0)$,
then $\mathfrak p$ is maximal and generated by an irreducible
monic polynomial $P$ (because of the Euclidean algorithm
in $k[x_n]$). Then
$k' = k[x_n]/\mathfrak p$ is a finite field extension of $k$
and contained in $\kappa(\mathfrak m)$. In this case
we get a surjection
$$
k'[x_1, \ldots, x_{n-1}]
\to
k'[x_1, \ldots, x_n] =
k' \otimes_k k[x_1, \ldots, x_n]
\longrightarrow
\kappa(\mathfrak m)
$$
and hence we see that $\kappa(\mathfrak m)$ is a finite
extension of $k'$ by induction hypothesis. Thus $\kappa(\mathfrak m)$
is finite over $k$ as well.
\medskip\noindent
If $\mathfrak p = (0)$ we consider the ring
extension $k[x_n] \subset k[x_1, \ldots, x_n]/\mathfrak m$.
This is a finitely generated ring extension, hence
of finite presentation by
Lemmas \ref{lemma-obvious-Noetherian} and
\ref{lemma-Noetherian-finite-type-is-finite-presentation}.
Thus the image of $\Spec(k[x_1, \ldots, x_n]/\mathfrak m)$
in $\Spec(k[x_n])$ is constructible by
Theorem \ref{theorem-chevalley}. Since the image
contains $(0)$ we conclude that it contains a standard
open $D(f)$ for some $f\in k[x_n]$ nonzero. Since clearly
$D(f)$ is infinite we get a contradiction with the
assumption that $k[x_1, \ldots, x_n]/\mathfrak m$ is
a field (and hence has a spectrum consisting of one point).
\medskip\noindent
To prove part (\ref{item-polynomial-ring-Jacobson}) let
$I \subset R$ be radical, with $R$ of finite type over $k$.
Let $f \in R$, $f \not \in I$. Pick a maximal ideal $\mathfrak m'$
in the nonzero ring $R_f/IR_f = (R/I)_f$. Let $\mathfrak m \subset R$
be the inverse image of $\mathfrak m'$ in $R$. We see that
$I \subset \mathfrak m$
and $f \not \in \mathfrak m$. If we show that $\mathfrak m$ is a maximal
ideal of $R$, then we are done. We clearly have
$$
k \subset R/\mathfrak m \subset \kappa(\mathfrak m').
$$
By part (\ref{item-finite-kappa}) the field extension
$k \subset \kappa(\mathfrak m')$
is finite. By elementary field theory we conclude that $R/\mathfrak m$
is a field.
\end{proof}
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