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Let $A$ be a finitely generated commutative ring, i.e. a $\mathbb{Z}$-algebra of finite type. It's not hard to show that if $A$ is a field, then it must be finite. I am interested in the following question:

Is it possible that $A$ contains an infinite field?

Note that if $A$ is an integral domain, then any subfield of $A$ is finitely generated; hence, it must be finite. Thus, we may assume that $A$ is not an integral domain. Note also that it's not true that any subring of $A$ is finitely generated in general.

stupid boy
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1 Answers1

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This answer by Martin Brandenburg, to a more specific question, also works to answer this question. A finitely generated ring $A$ cannot contain an infinite field.

Every quotient ring of $A$ is also finitely generated, so if $A/I$ is the quotient of $A$ by a maximal ideal, then $A/I$ is a field that is finitely generated as a ring, and is therefore finite.

If $A$ contained an infinite field $K$, then the restriction to $K$ of the natural homomorphism $A\to A/I$ would be a ring homomorphism $K\to A/I$. But there are no homomorphisms from an infinite field to a finite field.

Jeremy Rickard
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