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Let $k$ be any field (may not be algebraically closed), $A$ and $B$ are two finitely-generated $k$-algebras of (Krull) dimension 1. Suppose $f : A \rightarrow B $ be a $k$-algebra homomorphism. I want to show that if $M$ is any maximal ideal in $B$, then the preimage $f^{-1}(M)$ is also a maximal ideal.

If $A$ is a domain, then it is clear since $(0)$ is a prime ideal in $A$. But the question is the case of $A$ contains zero divisors. Now, if $M$ is a maximal ideal in $B$, $P := f^{-1}(M) \in Spec(A)$. I want use the contradiction to show that $P$ is actually a maximal ideal. Now, $f$ induces a $k$-algebra monomorphism $\overline{f}: A/P \rightarrow B/M$. Because we assume that $A/P$ is not a field, it is a domain of dimension 1. And then, I have no more idea to deal the question.

Actually, I know that if $B/M$ is integral over $A/P$, then for every maximal $\mathfrak{P}$ in $B/M$, $\mathfrak{P} \cap (A/P)$ is also a maximal ideal in $A/P$. Is this information useful for solving this question?

Peter Hu
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1 Answers1

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This follows from Zariski's lemma (no need on the restriction of the Krull dimensions). Namely, you have an embedding of $k$-algebras $A/f^{-1}(M)\to B/M$. Now, since $B/M$ is a finite type field extension of $k$, it must be finite $k$-dimensional. Then, this forces $A/f^{-1}(M)$ to be a finite $k$-dimensional domain, and thus a field.

Alex Youcis
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  • Let me check whether I understand your answer correctly or not. I recall the Zariski lemma states that if $k[x_1,\dots,x_n]/M$ is a field, then $x_i + M$ is algebraic over $k$. If we regard $B = k[x_1,\dots,x_n]/I$, then by the correspondence of ideals, we can regard $B/M \simeq k[x_1,\dots,x_n]/N$ with $M = N/I$. Hence, $B/M$ is a finite dimensional $k$-vector space. Is my realization for the first part correctly? – Peter Hu Jan 31 '14 at 06:41
  • @PeterHu Yes, that sounds about right. – Alex Youcis Jan 31 '14 at 06:49