### 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.

It is pointed out by FrédéricC below 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/reference/modules/sage/modules/tutorial_free_modules.html