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10.34 Hilbert Nullstellensatz

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

  1. For any maximal ideal $\mathfrak m \subset k[x_1, \ldots , x_ n]$ the field extension $\kappa (\mathfrak m)/k$ 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 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. $\square$

Lemma 10.34.2. Let $R$ be a ring. Let $K$ be a field. If $R \subset K$ and $K$ is of finite type over $R$, then there exists an $f \in R$ such that $R_ f$ is a field, and $K/R_ f$ is a finite field extension.

Proof. By Lemma 10.30.2 there exist a nonempty open $U \subset \mathop{\mathrm{Spec}}(R)$ contained in the image $\{ (0)\} $ of $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(R)$. Choose $f \in R$, $f \not= 0$ such that $D(f) \subset U$, i.e., $D(f) = \{ (0)\} $. Then $R_ f$ is a domain whose spectrum has exactly one point and $R_ f$ is a field. Then $K$ is a finitely generated algebra over the field $R_ f$ and hence a finite field extension of $R_ f$ by the Hilbert Nullstellensatz (Theorem 10.34.1). $\square$

Comments (4)

Comment #7360 by Zongzhu Lin on

The first part (1) of Theorem 10.34.1 is called by many as Zeriski's Lemma, which is the H_3 in Zriski's 1947 paper "A new proof of Hilbert's Nullstellensatz" in Bulletin of AMS. The proof presented is roughly the same as Zariski's original proof except the case . Zariski's specifically constructed a non-zero element has the property that any nonzero element has the property that is a factor of in for some (thus k[x_n]$ has only finitely irreducible elements etc,) which is impossible. Zariski's proof is much more self-contained with out using the Chevalley's theorem. There are also other ways to prove Zariski's Lemma using Krull's dimension, or Noether's Normalization Theorem.

By the way I am not clear (in the proof of case) why the constructible set containing a point has to contain a standard open . Constructible sets need not be open. A reference would be helpful.

Comment #7361 by Laurent Moret-Bailly on

Strangely, in "tags" mode, parts (1) and (2) are also converted to tags in the proof (but not in the statement).

Comment #7362 by on

@7361: That is a limitation of how things are being displayed. There is a work-around possible, the question is whether there are enough cases of this causing confusion to put in the effort. I'll put it on the possible features list though, thanks for noticing!

Comment #7366 by David Holmes on

Hi Zongzhu Lin, I think 005K might be the reference you are looking for? Alternatively, I think it's not so hard to see the claim from the definition of constructibility.

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