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Does this hold?

Let $k \subset A \subset B$ where $k$ is a field and $A,B$ are commutative rings.

If $B$ is a finitely-generated ring over $k$ and $\dim_k(B/A) < \infty$ then $B$ is a finitely-generated $A$-module.

I think the above is used in a proof that I'm trying to understand. But it's not carried out so I guess it must be really obvious.

Haderlump
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2 Answers2

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If $B=k[x_1,\dots,x_n]$ and $\dim_k(B/A)<\infty$ we get that $x_i$ is integral over $A$ for each $i$.

The powers of the element $\hat x_i\in B/A$ are linearly dependent over $k$, so there exists $N\ge 1$ and $a_1,\dots,a_N\in k$ such that $\hat x_i^N+a_1\hat x_i^{N-1}+\cdots+a_N=0$ in $B/A$, that is, there exists $a\in A$ such that $x_i^N+a_1x_i^{N-1}+\cdots+a_N-a=0$. This shows that $x_i$ is integral over $A$, so the extension $A\subset B$ is finitely generated and integral, and therefore it's finite.

  • Why is $A \subset B$ finitely generated? – Haderlump Jun 24 '13 at 23:49
  • @Haderlump In general, if one has rings $R \subset S \subset T$ and $T$ is finitely generated over $R$ then it is finitely generated over $S$: if you have a polynomial expression with coefficients in $R$ then that's also a polynomial expression with coefficients in $S$. – TTS Jun 25 '13 at 01:54
  • Ah yes, of course.. thank you all! – Haderlump Jun 25 '13 at 06:33
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I think this works. Take $b_1, \dots, b_n$ in $B$ such that their images in $B/A$ span that space over $k$. Then $\{1, b_1, \dots, b_n\}$ spans $B$ over $A$.

TTS
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