# 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,

`CombinatorialFreeModule`

is forfreemodules 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.