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It is a common fact that when F is a surjective homomorphism from a commutative ring A to another B (with 1), we have inverse images of the maximals in B maximal in A. Could anyone be so kind to me help solve the same problem without the condition of surjectivity of F but adding that A be semilocal? Thanks in advance.

karparvar
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The statement you're asking about is false; given a ring homomorphism $f:A\to B$ where $A$ is semilocal, it need not pull back a maximal ideal $M\subset B$ to a maximal ideal $f^{-1}(M)\subset A$.

For example, consider $A=\mathbb{Z}_{(2)}$ (the localization of $\mathbb{Z}$ at the ideal $(2)$, which is in fact a local ring) and $B=\mathbb{Q}$, with $f$ the inclusion map and $M=(0)$.

Zev Chonoles
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  • Is that a surjective map? Do you know under what hypothesis you can prove the preimage of maximal ideals is ideal? I'm trying to solve it. I reached a requirement to be bijective, but seems a bit much, because I needed both to be surjective and injective. – Santropedro Jul 15 '17 at 19:12