Mathematics Asked by An Hoa on December 3, 2020

The exercise on page 30. It says that:

Let $A, B$ be integral domain having the same field of fractions, $B supseteq A$. Prove that $B$ is faithfully flat over $A$ only when $B = A$.

My solution seems to be easy, which makes me doubt its correctness. My idea is to prove that for any $x in B$, $x in A$ so that $B subseteq A$. By assumption that $A, B$ have the same field of fraction, any element of $B$ is of the form $a/b$ for $a, b in A$, $b not= 0$. By faithful flatness, the sequence of $A$-modules

$$0 rightarrow A rightarrow A[X]/(bX – a) rightarrow 0$$

is exact if and only if the sequence of $B$-modules

$$0 rightarrow underbrace{A otimes_A B}_{B} rightarrow A[X]/(bX – a) otimes_A B rightarrow 0$$

is exact. We have

begin{align*}

A[X]/(bX – a) otimes_A B &cong A[X]/(bX – a) otimes_{A[X]} A[X] otimes_A B\

&cong A[X]/(bX – a) otimes_{A[X]} B[X]\

&cong B[X]/(bX – a)\

&cong B &text{ because }a/b in B

end{align*}

So the second sequence is exact: $0 rightarrow B rightarrow B rightarrow 0$. This implies the first sequence is exact and that can only occurs when $a/b in A$.

There is fatal flaw towards the end of your proof when you say $B[X]/(bX - a) cong B$ because $a/b in B$.

For example, let $A$ be a domain and $a in A$. Then $a^2 / a = a in A$ but $A[X]/(aX - a^2)$ is not even a domain, so clearly is not isomorphic to $A$.

A correct solution might look like:

Let $A subseteq B$ and $a/b in B$. Then $(b :_A a) = (b :_A a)B cap A = (bB :_A aB) cap A = B cap A = A$. The first equality uses that $I = IB cap A$ for all $A$-ideals under faithfully flat extension. The second equality is a fundamental property of colon ideals under flat extensions. So $a = bc$ for some $c in A$, i.e. $a/b in A$.

Answered by Badam Baplan on December 3, 2020

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