A question on faithfully flat extension This question arose while reading page 116 of Red Book by Mumford. 

Let $B$ be a faithfully flat extension of $A$. Can I claim that $b \otimes 1 = 1 \otimes b$ in $B\otimes_A B$ if and only if $b\in A$?

 A: You are asking if the sequence
$0 \to A \to B \to B\otimes_A B$, where the second map is given by $b \mapsto b\otimes 1 - 1 \otimes b$, is exact.
You can test this by tensoring up with $B$ over $A$ (since $B$ is faithfully flat over $A$).  Write $B' = B\otimes B$.  Then extending scalars to $B$ we get
$0 \to B \to B' \to B'\otimes_B B'$ (the same sequence but with $A$ replaced by $B$ and $B$ by $B'$).  (To check this is just a manipulation with tensor products.)  Again, $B'$ is faithfully flat over $B$.
You might think that nothing has improved by doing this, but it has: unlike in the general faithfully flat context, we have a homorphism $B' \to B$ (given by $b_1 \otimes b_2 \mapsto b_1 b_2$); geometrically, Spec $B' \to $ Spec $B$ admits a section.  
(This reduction to the case where the faithfully flat morphism admits a section
is the key technical tool in Grothendieck's flat descent arguments.)
Writing $B' = B \oplus I,$ where $I$ is the kernel of $B' \to B$, we get that
$B'\otimes_B B' = B \oplus I \oplus I \oplus I\otimes_B I$,
and in terms of this isomorphism it is easily seen that the morphism $b' \mapsto b' \otimes 1 - 1 \otimes b'$ is injective when restricted to $I$ (since for $b' \in I$ the two terms lie in the two distinct "$I$ summands" in the direct sum description
of $B'\otimes_B B'$). Thus your question has a positive answer in this base-changed context, and hence in the original context.

The general philosophy behind this kind of argument is that for answering these kinds of questions when $A \to B$ is faithfully flat, it is usually no loss of generality to assume in addition that $A$ is a direct summand of $B$ as an $A$-module.
