Quotients in CombinatorialFreeModule
The program
sage: T = CombinatorialFreeModule(ZZ, 'x')
sage: Q = T.quotient_module([2*T.monomial('x')])
sage: Q
Free module generated by {} over Integer Ring
says that Q is a trivial Z-module, which does not make sense.
If CombinatorialFreeModule handle only quotients over Q, then it should not return a Z-module when quotienting over Z...
Am i missing something here ? The documentation on CombinatorialFreeModule is incomplete, and using FreeModule would make a lot of my previous programs useless.
Well,
CombinatorialFreeModuleis for free modules only.Well, free structures exist mainly through their quotients, they get their universality (and legitimacy) because they have something to lift ;) Unfortunately, i also encountered similar problems with free algebras :(
Thanks for the answears.
This is infortunate indeed. My question seems stupid now because FrédéricC answear clearly explains why it does not work, but i still think Sage should give a warning/error message when doing this kind of quotients, as i lost some time understanding why my computations were wrong.