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Let $f\colon A\longrightarrow B$ be a finite algebra (B is finite as $A$-module) and let $\mathfrak{p}\in\mathsf{Spec} \, B$ and denote $f^{-1}(\mathfrak{p})=\mathfrak{q}$. Is it true that $B_{\mathfrak{p}}$ is finitely generated as $A_{\mathfrak{q}}$-algebra?

Vincenzo Zaccaro
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    I'm a bit rusty in commutative algebra, but I guess that if ${m_1,\dotso,m_n}\subset B$ is a set of generators for $B$ as an $A$-module, then ${\frac{m_1}{1},\dotso,\frac{m_n}{1}}$ is a set of generators for $B_{\mathfrak{p}}$ as an $A_{\mathfrak{q}}$-module. Am I missing something? – Renan Mezabarba Feb 01 '18 at 17:56
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    You can also consider $A=K[X^2,X^3]$, $B=K[X]$ and $P=(X)$. The very same element $\frac{1}{X+1}$ is not integral over $A$, so $A_p\subset B_q$ is not finite. – user26857 Feb 01 '18 at 20:44
  • I'm trying to check that it is finitely generated(quotient of a polynomial ring) not finite. – Vincenzo Zaccaro Feb 01 '18 at 20:53

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