ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 30 Nov 2020 11:13:56 +0100Quotients in CombinatorialFreeModulehttps://ask.sagemath.org/question/54272/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.Wed, 18 Nov 2020 12:53:15 +0100https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/Comment by FrédéricC for <p>The program </p>
<pre><code>sage: T = CombinatorialFreeModule(ZZ, 'x')
sage: Q = T.quotient_module([2*T.monomial('x')])
sage: Q
Free module generated by {} over Integer Ring
</code></pre>
<p>says that Q is a trivial Z-module, which does not make sense. </p>
<p>If CombinatorialFreeModule handle only quotients over Q, then it should not return a Z-module when quotienting over Z...</p>
<p>Am i missing something here ? The documentation on CombinatorialFreeModule is incomplete, and using FreeModule would make a lot of my previous programs useless.</p>
https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54298#post-id-54298Well, `CombinatorialFreeModule` is for **free** modules only.Thu, 19 Nov 2020 13:07:12 +0100https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54298#post-id-54298Comment by tmonteil for <p>The program </p>
<pre><code>sage: T = CombinatorialFreeModule(ZZ, 'x')
sage: Q = T.quotient_module([2*T.monomial('x')])
sage: Q
Free module generated by {} over Integer Ring
</code></pre>
<p>says that Q is a trivial Z-module, which does not make sense. </p>
<p>If CombinatorialFreeModule handle only quotients over Q, then it should not return a Z-module when quotienting over Z...</p>
<p>Am i missing something here ? The documentation on CombinatorialFreeModule is incomplete, and using FreeModule would make a lot of my previous programs useless.</p>
https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54299#post-id-54299Well, 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 :(Thu, 19 Nov 2020 13:26:32 +0100https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54299#post-id-54299Comment by qfaes for <p>The program </p>
<pre><code>sage: T = CombinatorialFreeModule(ZZ, 'x')
sage: Q = T.quotient_module([2*T.monomial('x')])
sage: Q
Free module generated by {} over Integer Ring
</code></pre>
<p>says that Q is a trivial Z-module, which does not make sense. </p>
<p>If CombinatorialFreeModule handle only quotients over Q, then it should not return a Z-module when quotienting over Z...</p>
<p>Am i missing something here ? The documentation on CombinatorialFreeModule is incomplete, and using FreeModule would make a lot of my previous programs useless.</p>
https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54456#post-id-54456Thanks 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.Mon, 30 Nov 2020 10:56:05 +0100https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?comment=54456#post-id-54456Answer by qfaes for <p>The program </p>
<pre><code>sage: T = CombinatorialFreeModule(ZZ, 'x')
sage: Q = T.quotient_module([2*T.monomial('x')])
sage: Q
Free module generated by {} over Integer Ring
</code></pre>
<p>says that Q is a trivial Z-module, which does not make sense. </p>
<p>If CombinatorialFreeModule handle only quotients over Q, then it should not return a Z-module when quotienting over Z...</p>
<p>Am i missing something here ? The documentation on CombinatorialFreeModule is incomplete, and using FreeModule would make a lot of my previous programs useless.</p>
https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?answer=54458#post-id-54458It 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/reference/modules/sage/modules/tutorial_free_modules.htmlMon, 30 Nov 2020 11:13:56 +0100https://ask.sagemath.org/question/54272/quotients-in-combinatorialfreemodule/?answer=54458#post-id-54458