Supppse $B$ is a faithfully flat $A$-algebra, and $b$ an element in $B$. Does $1 \otimes_A b= b\otimes_A 1$ in $B\otimes_A B$ imply $b \in A$?
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Yes, $A \to B \rightrightarrows B \otimes_A B$ is exact. This is well-known. It can be proven for example as follows: The sequence is split-exact after tensoring with $B$ over $A$. But $B$ is faithfully flat, so this proves exactness of the original sequence.
Martin Brandenburg
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You beat me to the answer! Just an additional remark: these are the first terms of the Amitsur complex of $B$ over $A$, and Martin's argument proves more generally that the whole Amitsur complex is exact. – Dan Petersen Nov 24 '13 at 09:38
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@Dan: Yes, exactly. – Martin Brandenburg Nov 24 '13 at 09:39