If $A \rightarrow B$ is a ring hom of noetherian rings that makes $B$ into a free $A$ module (not enecessarily finitely generated), i've been led to believe that $spec(B) \rightarrow spec (A)$ is surjective. Is this true? and how to prove that? thank you.
2026-04-12 20:50:30.1776027030
spec is surjective for ring hom which is a free module?
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