**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.
$\square$

**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

Comment #7361 by Laurent Moret-Bailly on

Comment #7362 by Pieter Belmans on

Comment #7366 by David Holmes on