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 $a/b \in A$.

  • $\begingroup$ Looks good to me. $\endgroup$ – Eric Wofsey Oct 19 '15 at 2:19

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