# The Stacks Project

## Tag: 00FV

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.
1. For any maximal ideal $\mathfrak m \subset k[x_1, \ldots, x_n]$ the field extension $k \subset \kappa(\mathfrak m)$ is finite.
2. Any radical ideal $I \subset k[x_1, \ldots, x_n]$ is the intersection of maximal ideals containing it.
The same is true in any finite type $k$-algebra.

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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Comment #47 by Rankeya Datta on August 30, 2012 at 5:09 am UTC

I don't know if this counts as a math error, but there seems to be some inconsistency between how the theorem is stated and the way the proof begins. The first few lines of the proof justify why it is enough to prove part (1) of the theorem for polynomial algebras k[x_1,...,x_n] instead of finitely generated k-algebras. But, part (1) is stated in terms of polynomial algebras, and not finitely generated k-algebras. So, I think the statement of part (1) should be modified accordingly.

Comment #53 by Johan (site) on August 30, 2012 at 12:13 pm UTC

Well, the beginning of the proof discusses the case where the algebra is an arbitrary finite type k-algebra. See last line statement theorem. I think it is OK (but not great).

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