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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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    I guess you mean "ring of quotients" and not "quotient field"? It is not usually going to be a field, you know...2012-09-01
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    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 $K(A)$.2012-09-01
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    A quotient field is a field. It is also called a field of fraction.2012-09-01
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    http://en.wikipedia.org/wiki/Field_of_fractions2012-09-01
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    Thanks Henri. I understand we are done if there is such a map. But how we rewrite $a'/a\in K(A)$ when $a\in A\setminus B$?2012-09-01
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    @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 $\mathbb{Z}$.2012-09-01
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    @Tom in case you don't trust wikipedia, suppose your ring $A$ has zero divisors. How could it be a subset of a field? This is why it has to be at least a domain.2012-09-01
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    Hurkyl rschwieb Thank you for comments. You are right.2012-09-01
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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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    Thanks. I understood the first half. But why is $S^{-1}B$ a field?($S:=A \setminus 0$)2012-09-01
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    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 .2012-09-01
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    Great! I understood all now! Thank you so much!2012-09-01
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    Ah, I'm very happy about that, Tom!2012-09-01
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    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-0.2012-09-01
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    Maybe I'm confused, but does your "useful lemma" imply that $K[x]$ is a field?2017-12-01
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    @Daniel: yes, for every $r\in R$ it is true that $K[r]$ is a field.2017-12-01
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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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    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.2012-09-01
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    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 $S^{-1}B$.2012-09-01
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    @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.2012-09-01
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    @GeorgesElencwajg Thanks for your kind comments. I upvoted your answer as well!! Speaking about base change, a few days ago I posted this question [here](http://math.stackexchange.com/questions/188791/defining-the-lie-bracket-on-mathfrakg-otimes-bbbr-bbbc/188797#188797) on complexifications. Now someone asked me a few days ago if the bracket is well-defined, and the only answer I could come up with is because you have a tensor product of vector spaces and so nothing strange will happen. Do you have anything to add to this? Thanks.2012-09-01
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    Benja, I understand the two statement but why do the two imply $S^{-1}B$ is a field?2012-09-01
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    Dear Benja, no, I have nothing to add: both the question and the answer are fine (as are the other answers).2012-09-01
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    (1)$S^{-1}A$ is integral over $S^{-1}B$. (2) $S^{-1}A$ is a field. These two are ok for me.2012-09-01
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    Thank you Georges. You said $S^{-1}B$ is a field quickly and then implied $S^{-1}A$ is a field. But for me, the second is obvious but the first is not obvious.2012-09-01
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    No, Benja. The appendix does not answer mine. What I want to know is not SA but SB2012-09-01
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    @GeorgesElencwajg I would like to say thanks to you, I remember you posting an answer to one of my questions on $S^{-1}A$ - modules! You explicit showed how a certain scalar jumped over to the other side of an elementary tensor and that for me was FrEAkY.2012-09-01
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    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!2012-09-01
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    Benja. thank you for your kind answer.2012-09-01