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When is the following true and why?

If a commutative ring $A$ is a finitely generated module over a commutative ring $B$, then its quotient field $K(A)$ is a finitely generated module over $K(B)$.

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    This doesn't work for ring extensions of _finite type_, however (see [here](https://mathoverflow.net/q/188377)).2017-10-17

2 Answers 2

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It is always true that $K(A)$ is finitely generated as a module over $K(B)$ .

Since you mention quotient fields, we may assume that $A$ is a domain.
We may also assume that $B\subset A$ and thus that $B$ is a domain too: if this is not the case and you have an algebra $\phi:B'\to A$, just consider $A$ as a $B=\phi(B')$-algebra.

With these preliminaries out of the way, let $S=B\setminus \lbrace 0\rbrace$.
Then (by base change or by hand) $S^{-1}A$ is a finitely generated $S^{-1}B$-module .
Since $S^{-1}B$ is a field and $S^{-1}A$ is a domain, $S^{-1}A$ is a field too by the Useful Lemma below.
Finally,we have $A\subset S^{-1}A\subset K(A)$ and since $S^{-1}A$ is already a field, we have $S^{-1}A=K(A)$ : we have proved that $K(A)=S^{-1}A$ is a finitely generated module (=finite-dimensional vector space ) over the field $K(B)=S^{-1}B$.

Useful Lemma:
If the commutative domain $R$ is a finite-dimensional algebra over a field $K$, then $R$ is a field.
Proof:
Let $0\neq r\in R$. The multiplication map $m_r:R\to R:x\mapsto rx$ is an injective endomorphism because $R$ is a domain, hence it is surjective by elementary linear algebra .
Thus there exists $s\in R$ with $m_r(s)=rs=1$ and thus $r$ is invertible: $r^{-1}=s$.

NB I have simplified my original proof by invoking the Useful Lemma instead of integral extensions.

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    @GeorgesElencwajg Sorry again, but as I understand, if $x$ is an indeterminate, then the useful lemma implies that the polynomial ring $K[x]$, is a field. Am I wrong?2017-12-02
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I would just like to add to Georges' answer. He mentions that if $A$ is finitely generated as a $B$ - module, then $S^{-1}A$ will be finitely generated as an $S^{-1}B$ module where $S = A - \{0\}$. I would like to mention that this result can be proved as follows:

Recall we have the $S^{-1}B$ - module isomorphism $S^{-1}A \cong S^{-1}B \otimes_B A$ such that every element in the tensor product can be written as an elementary tensor of the form $\frac{1}{s} \otimes a$ for some $s\in S$ and $a \in B$. But then $A$ was finitely generated as a $B$ - module so we can write $a = \sum_{i=1}^n a_ib_i$ for some $b_i \in B$ and $a_i's $ the generators of $A$. Hence we can write $a$ as

\begin{eqnarray*} \frac{1}{s} \otimes a &=& \frac{1}{s} \otimes \sum_{i=1}^n a_ib_i \\ &=& \sum_{i=1}^n \frac{b_i}{s} \otimes a_i \\ &=& \sum_{i=1}^n \frac{b_i}{s} \left(\frac{1}{1} \otimes a_i\right) \end{eqnarray*}

showing that $S^{-1}A$ is finitely generated as an $S^{-1}B$ - module.

$\hspace{6in} \square$

Appended for the OP: Suppose we have integral domains $A,B$ with $A \subseteq B$ and $B$ integral over $A$. Then $A$ being a field implies that $B$ is a field.

Proof: Take any $y \in B$ and by assumption $y$ satisfies an integral dependence relation

$y^n + a_{n-1}y^n + \ldots a_0 = 0$

with the $a_i \in A$. Then you can rewrite this as

$(y^{n-1} + a_{n-1}y^{n-2} + \ldots + a_1)y = -a_0.$

Now $a_0 \neq 0$ for otherwise this contradicts $B$ being an integral domain. It follows that we can divide through by $-a_0$ to get that

$(\text{some stuff})y = 1$

where the stuff is the inverse of $y$, consequently $B$ is a field.

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    $B$enja. thank you for your kind answer.2012-09-01