Restriction of scalars for free modules

asked 2019-09-10 19:56:22 +0200

heluani
304 ●8 ●19

updated 2019-09-10 21:26:48 +0200

Suppose I have a free module M over QQ[x]. How do I obtain the underlying QQ vector space? FreeModule does not have this coercion:

sage: R = PolynomialRing(QQ,'x')
sage: R in CommutativeAlgebras(QQ)
True
sage: M = FreeModule(R,3)
sage: M in VectorSpaces(QQ)
False

And CombinatorialFreeModule only takes already subcategories of QQ-vector spaces:


sage: M = CombinatorialFreeModule(QQ, ['a','b','c'], category=Modules(R))
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
....
ValueError: Subcategory of `Category of vector spaces with basis over Rational Field` required; got `Category of modules over Univariate Polynomial Ring in x over Rational Field`

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answered 2019-09-10 21:53:52 +0200

Emmanuel Charpentier
7057 ●6 ●46 ●137

sage: R = PolynomialRing(QQ,'x')
sage: R in CommutativeAlgebras(QQ)
True
sage: M = FreeModule(R,3)
sage: M.parent()
<class 'sage.modules.free_module.FreeModule_ambient_pid_with_category'>
sage: m=M.random_element()
sage: m.parent()
Ambient free module of rank 3 over the principal ideal domain Univariate Polynomial Ring in x over Rational Field

Does this answer your qestion ?

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Perhaps, I don't understand, which object is an instance of VectorSpaces(QQ).parent_class? or which object is in VectorSpaces(QQ)?

heluani ( 2019-09-11 16:12:25 +0200 )edit

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Asked: 2019-09-10 19:56:22 +0200

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Last updated: Sep 10 '19

Restriction of scalars for free modules edit