A counter counterexample example
Taken from Jordan Ellenberg's blog:
``The counterexample Max countered is the central simple algebra A over the
field Q(t) obtained as the tensor product of the two generalized quaternion
algebras (17,t) and (13,6(t-1)(t-11)). This algebra has index 4, which is to
say (if I understand correctly) that its Brauer class isn't the cup product of
two elements in Q(t)^*. On the other hand, it turns out that A \otimes
\mathbf{Q}_v has index 1 or 2 - that is, it's either trivial or a cup product
- for all places v of Q.
This seems like an example of a situation where there's no Hasse principle;
A fails to be a cup product despite the fact that A is a cup product over every
completion of Q. But the truth, as Max explained, is more complicated.''