Dear All, Today I ran into some strange things concerning homomorphisms between free modules over different rings (QQ vs ZZ). Please geuss which two of the following commands execute without error (using sage 5.0):
(ZZ^2).hom([[2,3],[3,6]],QQ^2) (QQ^2).hom([[1/3,1/4],[1/3,1/6]],ZZ^2) (ZZ^2).hom([[1/3,1/4],[1/3,1/6]],QQ^2) Here are the answers: sage: (ZZ^2).hom([[2,3],[3,6]],QQ^2) #ok Free module morphism defined by the matrix [2 3] [3 6] Domain: Ambient free module of rank 2 over the principal ideal domain Integer Ring Codomain: Vector space of dimension 2 over Rational Field sage: (QQ^2).hom([[1/3,1/4],[1/3,1/6]],ZZ^2) #should not be possible Free module morphism defined by the matrix [1/3 1/4] [1/3 1/6] Domain: Vector space of dimension 2 over Rational Field Codomain: Ambient free module of rank 2 over the principal ideal domain Integer Ring sage: (ZZ^2).hom([[1/3,1/4],[1/3,1/6]],QQ^2) #goes boom with an error you might have excpected in the previous example Traceback (most recent call last): ... TypeError: no conversion of this rational to integer I wonder if anybody ran into similar things. And whether people are working on this. I tried to search trac but the most related ticket I could find is http://trac.sagemath.org/sage_trac/ticket/1947 which it about morphisms between vectorspaces over different fields. -- -- To post to this group, send an email to sage-devel@googlegroups.com To unsubscribe from this group, send an email to sage-devel+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/sage-devel URL: http://www.sagemath.org
