Dimension of a semilocal Hilbert ring is zero

Question

Is the Krull dimension of any commutative semilocal Hilbert ring equal to zero?

I appreciate any help from anyone!

Answer 1

Actually the answer is trivially yes: by definition, a prime ideal is an intersection of maximal ideals; since your ring is semilocal, there are only finitely many such maximal ideals, so the given prime contains a maximal one, end of story.

Answer 2

I think the answer is yes.

Hopefully the argument below isn't nonsense:

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Let $A$ be such a ring. To determine its Krull dimension it is no loss of generality to quotient out by the nilradical, and thus suppose that $A$ is reduced.

Thus, since $A$ is semi-local, it has finitely many maximal ideals, say $\mathfrak m_1, \ldots, \mathfrak m_n$, whose intersection is trivial (because the Jacobson radical of $A$ equals the nilradical, which is trivial). But then the map $A \mapsto A/\mathfrak m_1 \times \cdots \times A/\mathfrak m_n$ is an isomorphism, and so $A$ is a product of finitely many fields, and hence has Krull dimension zero.