scalar in QQ times matrix over ZZ

[René Birkner]

Since we now can join matrices over ZZ and QQ (thanks Dan), wouldn't
it be nice to multiply a number in QQ with a matrix over ZZ, which
returns a matrix over ZZ? So far he "can't promote scalar to ring of
matrix".

[Daniel R. Grayson]

It might be nice for it to return a matrix over QQ (not over ZZ). But
even then, I would hesitate to do this, as a lot of unintended
computation might be incurred in the general case: imagine the matrix
is a map f : M <---- N of R-modules, and the scalar is an element b of
a R-algebra S. What does b*f mean in this case? Probably it should
be a map M**S <--- N**S. If M and N are not free, then computing the
tensor products M**S and N**S could take a lot of time (involving
Groebner basis computations). I suppose we could signal an error if M
and N are not free, hmm. I welcome further discussion.

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[René Birkner]

hmm, you're right I was in my own world (which at that point of time
consisted only of ZZ and QQ...). It should be consistent. But I think
for free modules it should be ok. I would would also like to here more
opinions...