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Let $\varphi:A\rightarrow B$ be a homomorphism of finitely generated $k$-algebras, and let $\mathfrak{m}$ be a maximal ideal of $B$. We have the injective homomorphism $$ \overline{\varphi}:A/\varphi^{-1}(\mathfrak{m})\rightarrow B/\mathfrak{m}, a+\varphi^{-1}(\mathfrak{m})\mapsto \varphi(a)+\mathfrak{m}. $$ Is $A/\varphi^{-1}(\mathfrak{m})\subseteq B/\mathfrak{m}$ an integral extension?

Carlito
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Actually this is a finite extension of fields (hence algebraic, hence integral), by Zariski's Lemma.

Martin Brandenburg
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  • Thanks for the answer, but I am trying to prove that $A/\varphi^{-1}(\mathfrak{m})$ is a field with this result, so I would not like to use Zariski's lemma. – Carlito Nov 23 '15 at 09:13
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    We have $k \to A/\varphi^{-1}(\mathfrak m) \to B/\mathfrak m$. Use it on the composition and you get that $B/\mathfrak m$ is integral over $k$, in particular over $A/\varphi^{-1}(\mathfrak m)$. (Or use the following well known fact: An intermediate ring of a finite field extension is a field) – MooS Nov 23 '15 at 09:29