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Quotients in CombinatorialFreeModule

asked 2020-11-18 12:53:15 +0200

qfaes gravatar image

updated 2020-11-30 11:13:44 +0200

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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Well, CombinatorialFreeModule is for free modules only.

FrédéricC gravatar imageFrédéricC ( 2020-11-19 13:07:12 +0200 )edit
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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 13:26:32 +0200 )edit

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.

qfaes gravatar imageqfaes ( 2020-11-30 10:56:05 +0200 )edit

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answered 2020-11-30 11:13:56 +0200

qfaes gravatar image

It is pointed out by FrédéricC in the comments that the reason for this is simply that the quotient is no more a free module, hence the wrong result. Nevertheless, this could return an error message/warning. I have the feeling that this should be done soon as it is in some Todo list on the Sage documentation : https://doc.sagemath.org/html/en/refe...

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Asked: 2020-11-18 12:53:15 +0200

Seen: 58 times

Last updated: Nov 30 '20