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Let $A$ a finite $k$-algebra (with $k$ a field) and $B$ a subalgebra of $A$. Prove that if $\mathfrak{m}$ is a maximal ideal of $A$ then $\mathfrak{m}\cap B$ is a maximal ideal of $B$.

It is easy to see that $\mathfrak{m}\cap B$ is an ideal of $B$. We have to prove that $$B' = ~^{B}\big / _{\mathfrak{m}\cap B}$$ is a field. To do so, I've tried to use the following lemma :

Lemma : If $A'$ is a field and $B'$ is a subring of $A'$ such that $A'$ is a finitely generated $B'$-module then $B'$ is a field.

Let $A'=~^{A}\big / _{\mathfrak{m}}$.

  • $\mathfrak{m}$ is maximal, so $A'$ is a field.
  • $B$ is a subring of $A$, so $B'$ is a subring of $A'$.
  • $B$ is a finite subalgebra of $A$, so $A'$ is a finite $B'$-subalgebra.
  • How do I prove that for all $x\in A'$, $x$ is an algebraic integer over $B'$? I've tried to start noticing that $A$ is a finite $k$-algebra so is $A'$ and $A'$ is a field, hence $k\subset A'$ is a finite algebraic extension but I don't see how to continue. Do we have to use Noether's Normalization lemma?

Any help will be greatly appreciate.

user26857
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Zanzi
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    Since $A'$ is a finitely generated $k$-module, then every $x \in A'$ is integral over $k$ (we can take the polynomial to be monic by dividing), and $k \subseteq B' \subseteq A'$, so $x$ is also integral over $B'$. Does that solve it, or am I missing something? – Viktor Vaughn May 15 '16 at 15:39
  • Of course, thanks, I guess I am the one who missed something.. – Zanzi May 15 '16 at 15:50

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