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Let $k$ be a field, $ A \subseteq B$ finitely generated $k$-algebras, with $B$ integral over $A$. Is it true that $B$ is a finite $A$-module?

In the case where $A=k[x]$, $x$ transcendental over $k$, and $B$ is the integral closure of $A$ in a finite extension $L$ of the field of fractions $K=k(x)$ this follows from the properties of Dedekind domains (and/or function fields). What about the more general cases?

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2 Answers 2

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Yes. Note that because $B$ is a finitely generated $k$-algebra, $B$ is also a finitely generated $A$-algebra (as $k$ is a field, the homomorphism $k \to A$ is injective, thus wlog $k \subset A$.)

We need two lemmas which are not difficult to prove.

Lemma 1 If $A$ is a commutative ring and $x$ is integral over $A$, then $A[x]$ is a finitely generated $A$-module

Proof: Let $x^n+a_{n-1}x^{n-1}+\dots+a_0=0$, then it is easy to see that $1,x, \dots x^{n-1}$ generates $A[x]$ as an $A$-module.

Lemma 2 If $B$ is a $A$-algebra such that $B$ is a finite $A$-module and $M$ is a finite $B$-module, then $M$ is a finite $A$-module (by restriction of scalars).

Proof: Let $b_1, \dots b_n$ be a $A$-generating system of $B$ and $m_1, \dots m_k$ be a $B$-generating system of $M$. Then obviously $(b_i m_j)_{1 \leq i \leq n, 1 \leq j \leq k}$ is an $A$-generating system of $M$.

Now we have the following proposition (note that no hypothesis on any ring being an algebra over a field is needed.)

If $B$ is a finitely generated integral $A$-algebra, then $B$ is a finite $A$-module

Proof: We proceed by induction of the number of generators $n$ of $B$ as an $A$-algebra. Lemma 1 takes care of the case $n=1$.

For the induction step, assume that $B$ is generated by $x_1, \dots x_{n+1}$ as an $A$-algebra. By the induction hypothesis, the $A$-algebra $B'$ generated by $x_1, \dots x_n$ is a finite $A$-module. But we have $B=B'[x_{n+1}]$ such that by lemma 1, $B$ is a finite $B'$-module. Thus by lemma 2, $B$ is a finite $A$-module.

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Right, if $B$ is a finitely generated $k$-algebra then it is a fortiori a finitely generated $A$-algebra*. Since $B$ is finitely generated and integral over $A$, it is finite over $A$ (see for example Lang's Algebra, Proposition VII.1.2).

*To be finitely generated over $k$ means there are elements $x_1,\dots,x_n$ such that every element of $B$ is a polynomial in $x_1,\dots,x_n$ with coefficients in $k$. Since $k \subseteq A$, a polynomial with coefficients in $k$ is also a polynomial with coefficients in $A$. So every element of $B$ can be written as a polynomial in $x_1,\dots,x_n$ with coefficients in $A$.

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