# Every finitely generated algebra over a field is a Jacobson ring

Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring?

EDIT: We define a ring to be Jacobson if every prime ideal is an intersection of maximal ideals.

## 2 Answers

Your question follows immediately if you know that the polynomial ring in $n$ variables over a field is a Jacobson ring from the following trivial observation: if $A$ is a Jacobson ring and $I\subset A$ is an ideal, then $A/I$ is Jacobson.

• Thank you, sir. How can I see that any finitely generated algebra is a quotient of the polynomial algebra? – superAnnoyingUser Jun 14 '14 at 13:12

We will first establish the case when your finitely generated $k$-algebra is reduced. We will make use of the following Lemma and its corollary:

Zariski's Lemma: Let $K$ be a finitely generated $k$-algebra that is a field. Then $K$ is a finite field extension of $k$.

Corollary: For any finitely generated $k$-algebra $R$ and maximal ideal $\mathfrak{m} \subseteq R$, we have that $R/\mathfrak{m}$ is a finite field extension of $k$.

Now we can state a proposition:

Proposition: Let $R$ be a finitely generated reduced $k$-algebra and $\mathfrak{p} \subseteq R$ be a prime ideal. Then $$\mathfrak{p} = \bigcap \mathfrak{q}$$ where the intersection is over all prime ideals $\mathfrak{q}$ such that $R/\mathfrak{q}$ is finitely generated as a $k$- module.

Proof: We just need to show that $\bigcap \mathfrak{q} \subseteq \mathfrak{p}$ for the other inclusion is clear. Now suppose there is $x \in R$ such that $x\notin \mathfrak{p}$. Then in the localization $R_x$ we have that $\mathfrak{p}_x$ is a prime ideal, since $\mathfrak{p}$ does not meet $\{1,x,x^2,\ldots \}$. Now choose a maximal ideal $\mathfrak{m}_x$ containing $\mathfrak{p}_x$. We have inclusions

$$k \hookrightarrow R/(\mathfrak{m}_x \cap R) \hookrightarrow R_x/\mathfrak{m}_x$$

and since $R_x/\mathfrak{m}_x$ is a finite field extension of $k$, we have that $R_x/\mathfrak{m}_x$ is finite as a module over $R/(\mathfrak{m}_x \cap R)$. But now because $\mathfrak{m}_x$ is a maximal ideal and $\mathfrak{m}_x \cap R$ is a prime ideal, by Corollary 5.8 of Atiyah - Macdonald we get that $\mathfrak{m}_x \cap R$ is a maximal ideal. Setting $\mathfrak{m} = \mathfrak{m}_x \cap R$ shows that we have a maximal (and hence prime) ideal $\mathfrak{m}$ containing $\mathfrak{p}$ such that $\mathfrak{p} \subseteq \mathfrak{m}$. The corollary to Zariski's lemma shows that $R/\mathfrak{m}$ is finitely generated as a $k$-module, which finishes the proof of the proposition.

Corollary 1: A reduced finitely generated $k$-algebra is Jacobson.

Corollary: Since any finitely generated $k$-algebra is a quotient of a polynomial algebra in a finite number of variables, any finitely generated $k$-algebra is Jacobson.

• In your Proposition how do you know that $p\subset\cap q$ ? – user89712 Nov 23 '13 at 23:20