Theorem 10.34.1 (00FV): Hilbert Nullstellensatz

Theorem 10.34.1 (Hilbert Nullstellensatz). Let $k$ be a field.

For any maximal ideal $\mathfrak m \subset k[x_1, \ldots , x_ n]$ the field extension $\kappa (\mathfrak m)/k$ is finite.
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 10.31.3 and 10.31.4. Thus the image of $\mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n]/\mathfrak m)$ in $\mathop{\mathrm{Spec}}(k[x_ n])$ is constructible by Theorem 10.29.10. 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).

Proof of (2). Let $I \subset R$ be a radical ideal, with $R$ of finite type over $k$. Let $f \in R$, $f \not\in I$. We have to find a maximal ideal $\mathfrak m \subset R$ with $I \subset \mathfrak m$ and $f \not\in \mathfrak m$. The ring $(R/I)_ f$ is nonzero, since $1 = 0$ in this ring would mean $f^ n \in I$ and since $I$ is radical this would mean $f \in I$ contrary to our assumption on $f$. Thus we may choose a maximal ideal $\mathfrak m'$ in $(R/I)_ f$, see Lemma 10.17.2. 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 $\kappa (\mathfrak m')/k$ is finite. Hence $R/\mathfrak m$ is a field by Fields, Lemma 9.8.10. Thus $\mathfrak m$ is maximal and the proof is complete. $\square$

Comments (4)

Comment #47 by Rankeya Datta on August 30, 2012 at 05:09

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 on August 30, 2012 at 12:13

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).

Comment #5862 by yogesh on December 23, 2020 at 13:04

The proof of part (2) is fine, I just would have been helped with an initial sentence on the strategy of the proof, e.g. "The non-trivial containment we need to show is $I \supseteq \bigcap_{I \subseteq \mathfrak{m}} \mathfrak{m}$ and we show this by showing the contrapositive of this inclusion statement, namely that for all $f \in R$ , if $f \notin I$ then $f \notin \bigcap_{I \subseteq \mathfrak{m}} \mathfrak{m}$ , which is to say there exists some maximal ideal $\mathfrak{m}$ with $I \subseteq \mathfrak{m}$ and $f \notin \mathfrak{m}$ .

Also, the place where one is using $I$ is radical is hidden in the claim that $(R/I)_f$ is non-zero - if $1=0$ in this ring then $f^n \in I$ and since $I$ is radical that would mean $f \in I$ contrary to hypothesis.

It may not make sense to include comments on geometric intution within the proof, and obviously you know the following but if I may explain my understanding (hopefully mostly correct) of geometrically what is happening when considering $R_f/IR_f$ is you are looking at the open set $D(f) \subset \mathrm{Spec} R$ and $V(I) \cap D(f)$ and then a closed point $\mathfrak{m}'$ in $V(I) \cap D(f)$ has $\kappa(\mathfrak{m}')$ finite degree of $k$ by part (1) using $R_f$ (and hence R_f/IR_f $) is finite type over$ k $and maybe it's not too surprising it's a closed point in$ \mathrm{Spec} \ R$.

Comment #6073 by Johan on April 29, 2021 at 17:16

@yogesh OK, I have added a few words along the lines of your suggestions. See this commit.

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