9

Suppose that $A$ and $B$ are commutative rings containing a field $k$, and $B$ is finitely generated $k$-algebra. Let $\phi: A\rightarrow B$ be a ring homomorphism with $\phi|_k =\mathrm{Id}$. I am trying to prove that if $M\subset B$ is a maximal ideal, then $\phi^{-1}(M)$ is a maximal ideal of $A$.

The case when $A \subset B$ is an integral extension of rings is well-known. I think I can also prove the result when $\phi$ is surjective.

Inverse Image of Maximal Ideals discusses the case when $B$ is a finitely generated $\mathbb{Z}$-algebra but I am not sure how to generalize this.

Community
  • 1
John
  • 135
  • 4

1 Answers1

4

We have $k\subset A/\phi^{-1}(M)\subset B/M$. The extension $k\subset B/M$ is finite (see here, Corollary 8.3.9), so the extension $k\subset A/\phi^{-1}(M)$ is finite, too. This shows that $A/\phi^{-1}(M)$ is a field, so $\phi^{-1}(M)$ is maximal.

  • 1
    One way to justify that $A/\phi^{-1}(M)$ is a field, is that since $k\subset A/\phi^{-1}(M)$ is finite, it is an integral extension; since $k$ is a field, then $A/\phi^{-1}(M)$ is, too. (If $C\subset D$ is an integral extension of rings, then $C$ is a field iff $D$ is). – Bruno Stonek Sep 28 '13 at 15:19