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Let $A \subseteq B$ be integral domains with the same field of fractions. Assume that $A \to B$ is faithfully flat. Why do we have $A=B$?

This is an exercise in Matsumura's book. Here is my idea: If $b \in B$, consider $I = \{a \in A : ab \in A\}$. This is an ideal of $A$. By asumption $I \neq 0$, and our goal is to show that $I=A$. It suffices to prove $IB=B$. But how can we achieve this?

Martin Brandenburg
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    This is just a hunch, but if they are integral domains with the same field of fractions, wouldn't $B$ be a localization of $A$? – Arthur Sep 03 '13 at 11:28
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    @Arthur: No. This holds in some special cases (for example when $A$ is a PID), but it fails in general. Birational morphisms can be quite complicated. – Martin Brandenburg Sep 03 '13 at 13:12
  • Perhaps Arthur means localisation in the broad sense of $A[S^{-1}]$ for some multiplicatively-closed set $S$. Surely this is true, by taking $S$ to be the set of units in $B$? – Zhen Lin Sep 03 '13 at 13:18
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    @ZhenLin No: http://math.stackexchange.com/a/287259/38268 –  Sep 03 '13 at 13:21
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    Hmmm, pity. Otherwise we could just base change to $B$ and use faithful-flatness to deduce that $\operatorname{coker} f = 0$. – Zhen Lin Sep 03 '13 at 13:22
  • @MartinBrandenburg Wait if $A,B$ have the same fraction field can't we embed $B$ in the fraction field of $A$ and say any $b \in b$ is of the form $a_1/a_2$ where $a_i \in A$? –  Sep 03 '13 at 14:08
  • @user38268: Yes but this doesn't mean that $B$ is a localization of $A$. – Martin Brandenburg Sep 03 '13 at 14:31

2 Answers2

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We can show $I = A$ as follows, the inclusion $I \subseteq A$ being clear. Choose $a \in A$ and write $b = a_1/a_2$. Then $aba_2 = a\cdot a_1 \in A$ and so $a\cdot a_1 \in a a_2 B \cap A = aa_2 A$ where the last equality comes from faithful flatness. It follows that $aba_2 \in aa_2A$ and so $aba_2 = a a_2 a'$ for some $a' \in A$. Thus $ab = a a' \in A$ and so indeed $a \in I$. Thus $I = A$.

Martin Brandenburg
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    Of course it suffices to take $a=1$. So the proof is just: If $b=a_1/a_2$, then $a_1 \in a_2 B \cap A = a_2 A$, hence $b \in A$. I should have seen this! – Martin Brandenburg Sep 03 '13 at 21:13
  • Here is way $aB\cap A=aA$. In general we have $IB\cap A=I$ for $I$ an ideal of $A$. The $\supseteq$ inclusion is easy so we can consider the quotient $(IB\cap A)/I$ and fits this in the exact sequence of $A$-modules $$0\rightarrow (IB\cap A)/I\rightarrow A/I \rightarrow B/IB$$ If we tensor this by $B$ we get $(IB\cap A/I)\otimes_A B=0$ and hence $(IB\cap A)/I=0$, that is, $IB\cap A=I$. – nowhere dense Apr 16 '20 at 12:07
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As pointed out in the comments, the problem is easy if we know that $B$ is a localisation of $A$, because then the codiagonal $\nabla : B \otimes_A B \to B$ is an isomorphism, and so we can (co)base change $A \to B$ along itself to deduce that $A \to B$ is an isomorphism. (This is where we use the fact that $A \to B$ is faithfully flat.) But in fact $\nabla : B \otimes_A B \to B$ is an isomorphism if and only if $A \to B$ is an epimorphism, and this is certainly true if $B$ is flat and embeds in $\operatorname{Frac} A$ as an $A$-algebra. (See comments below.)

Zhen Lin
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  • Ok. But for the last claim we also need that $B$ is flat over $A$, right? – Martin Brandenburg Sep 03 '13 at 14:34
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    Ah, sorry, I was thinking that the usual proof that $A \to \operatorname{Frac} A$ is an epimorphism would go through here as well, but it seems a more subtle approach is needed. Hmmm... – Zhen Lin Sep 03 '13 at 14:40
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    Yes. I meant that $B \otimes_A B \to Q(A) \otimes_A Q(A)$ is injective (flatness), hence $b \otimes 1 = 1 \otimes b$ holds in $B \otimes_A B$. – Martin Brandenburg Sep 03 '13 at 15:05