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 $K(B)$.

  • $\begingroup$ I guess you mean "ring of quotients" and not "quotient field"? It is not usually going to be a field, you know... $\endgroup$ – rschwieb Sep 1 '12 at 11:59
  • $\begingroup$ Try to write any fraction in $K(A)$ as a fraction with denominator from $B$ and deduce that there is an isomorphism of $K(B)$-modules from $K(B)\otimes_B A$ to $K(A)$. $\endgroup$ – HenrikRueping Sep 1 '12 at 12:13
  • $\begingroup$ A quotient field is a field. It is also called a field of fraction. $\endgroup$ – Tom Sep 1 '12 at 12:14
  • $\begingroup$ en.wikipedia.org/wiki/Field_of_fractions $\endgroup$ – Tom Sep 1 '12 at 12:14
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    $\begingroup$ @Tom: Note the first sentence of your wiki link says "... of an integral domain ...". In general the ring of fractions is not a field. Additionally, "quotient field" is a risky name, because, e.g., many would use that phrase to describe the relationship of $\mathbb{F}_p$ (the finite field of $p$ elements) to $\mathbb{Z}$: it is a field which is a quotient of $\mathbb{Z}$. $\endgroup$ – Hurkyl Sep 1 '12 at 17:05

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.
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.

  • $\begingroup$ Thanks. I understood the first half. But why is $S^{-1}B$ a field?($S:=A \setminus 0$) $\endgroup$ – Tom Sep 1 '12 at 13:12
  • $\begingroup$ Dear Tom, sorry, this was a typo: I meant $S=B\setminus \lbrace 0\rbrace$. I have edited my answer accordingly. Thanks for reading my post so attentively . $\endgroup$ – Georges Elencwajg Sep 1 '12 at 14:11
  • $\begingroup$ Great! I understood all now! Thank you so much! $\endgroup$ – Tom Sep 1 '12 at 14:26
  • $\begingroup$ Ah, I'm very happy about that, Tom! $\endgroup$ – Georges Elencwajg Sep 1 '12 at 14:29
  • $\begingroup$ Thank you. Your new proof is more elementary. You helped me so much. Before I posted my question, I tried some textbooks including Eisenbud but no books contained this statement explicitly. Before asking, I couldn't take denominators S:=B-0, I always kept S:=A-0. $\endgroup$ – Tom Sep 1 '12 at 23:57

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.

  • $\begingroup$ Yes, this is what I meant by "base change" but it is nice to have it spelled out for those not yet familiar with this powerful technique: I have just upvoted BenjaLim for his calculation. $\endgroup$ – Georges Elencwajg Sep 1 '12 at 12:38
  • $\begingroup$ Thanks Benja. I understood all of what you said. The only thing left is that $S^{-1}B=K(B)$ when $S:=A\setminus 0$ and $K(B):=T^{-1}B$ with $T=B\setminus 0$. Why $S^{-1}B$ is a field?(in George's message) $b/a$ has an inverse in $S^{-1}B$. $\endgroup$ – Tom Sep 1 '12 at 12:45
  • $\begingroup$ @Tom Well as Georges has mentioned, finitely generated implies that $S^{-1}A$ is integral over $S^{-1}B$, and so $S^{-1}A$ a field implies that $S^{-1}B$ is a field. $\endgroup$ – user38268 Sep 1 '12 at 12:47
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    $\begingroup$ Dear Tom, there was a confusing typo in my answer: it is corrected now (cf. my comment answering yours below my answer) By the way, you should write your comments only below the relevant answer, in this case mine: BenjaLim is not responsible for my silly typo! $\endgroup$ – Georges Elencwajg Sep 1 '12 at 14:16
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    $\begingroup$ Benja. thank you for your kind answer. $\endgroup$ – Tom Sep 1 '12 at 14:47

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