# Doubtful solution to an exercise on faithful flatness in Matsumura's Commutative Algebra

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 $a/b \in A$.

• Looks good to me. Commented Oct 19, 2015 at 2:19

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 $$A$$.
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. $$a/b \in A$$.