Quotients in CombinatorialFreeModule

asked 2020-11-18 05:53:15 -0600

qfaes gravatar image

updated 2020-11-19 11:12:06 -0600

slelievre gravatar image

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.

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Comments

Well, CombinatorialFreeModule is for free modules only.

FrédéricC gravatar imageFrédéricC ( 2020-11-19 06:07:12 -0600 )edit

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 :(

tmonteil gravatar imagetmonteil ( 2020-11-19 06:26:32 -0600 )edit