
Theorem 10.33.1 (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 10.30.3 and 10.30.4. 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 10.28.9. 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 a radical ideal, 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. Hence $R/\mathfrak m$ is a field by Fields, Lemma 9.8.10. Thus $\mathfrak m$ is maximal and the proof is complete. $\square$

Comment #47 by Rankeya Datta on

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 on

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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