1

Let $$ f: A\to B $$ be a morphism of finitely generated $\mathbb{C}$-algebras, suppose $\mathfrak{m}\unlhd B$ is a maximal ideal, I want to show that $f^{-1}(\mathfrak{m})$ is a maximal ideal of $A$. Consider the morphism $$ A\to B/\mathfrak{m}, $$ $$ a\mapsto f(a)+\mathfrak{m}, $$ this has kernel $f^{-1}(\mathfrak{m})$, by the first isomorphism theorem we get $$ A/ f^{-1}(\mathfrak{m}) \cong \frac{f(A)\cap \mathfrak{m}}{\mathfrak{m}}, $$ so, RHS is a field, so LHS is a field and $f^{-1}(\mathfrak{m})$ is maximal. I think that my proof is wrong, probably I got RHS wrong and I am not using the fact that these are finitely generated or over $\mathbb{C}$ at all. What is the correct proof?

Jimmy R
  • 2,702
  • The image of $\pi\circ f\colon A\to B/\mathfrak{m}$ ($\pi\colon B\to B/\mathfrak{m}$ the canonical projection) is a finitely generated algebra and a subring of a field. Is it a field? – egreg Nov 05 '14 at 00:30

0 Answers0