If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$.

Asked Jul 31 '18 at 17:34

Active Jul 31 '18 at 17:34

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I found this exercise in Matsumura, Commutative Ring Theory, (Exercise 7.2).

If $A\subset B$ are integral domains with the same field of fractions and $B$ is faithfully flat over $A$, then $A=B$.

Can anyone give me a proof for this statement or tell me where to find it?

asked Jul 31 '18 at 17:34

mathcourse

May be you can use that: Let $A$ be an integral domain and $K$ be its field of fractions. Also let $B$ be a finitely generated $A$-submodule of $K$. Then $B$ is flat iff $B$ is locally free of rank $1$. – xarles Jul 31 '18 at 22:45

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