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That those vectors are not equal is truly disturbing. This should either raise an exception about not having compatible parents or should be True. I'll let the coercion guru's argue about that. :)
Fixed the Z/8Z case. As for Z/7Z and Q, they are incomparable, which by convention means == returns False. (If it gave an error,we would have to re-think nonsense like "some string" != random_matrix(ZZ, 3) returning False).
I think this bit of code should not produce an exception. The vectors should both be coerced to belong to Z8!^3 and compared.
Note that a similar thing seems to work in other cases (because 7 is prime and Z7 is a field?).
This may or may not be related, but combining QQ and Z7 produces some wrong results:
That those vectors are not equal is truly disturbing. This should either raise an exception about not having compatible parents or should be True. I'll let the coercion guru's argue about that. :)
Component: algebra
Author: Robert Bradshaw
Reviewer: Mike Hansen
Merged: sage-4.3.1.rc1
Issue created by migration from https://trac.sagemath.org/ticket/3638
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