Re: module homomorphisms

On Sat, 27 Oct 2007 13:44:57 -0700, Keith Ramsay <kramsay@xxxxxxx>
wrote:


On Oct 27, 12:47 pm, quasi <qu...@xxxxxxxx> wrote:
|I'll let you work out the analogue of the above strategy for the
|general case, m > n, where n is an arbitrary positive integer, but
|hopefully, the underlying idea has been made clear.
|
|Also, I hope to have planted the idea (assuming sci.math was not
|available, or even if it was), that you should have at least tried
the
|simplest case (m=2, n=1).

Don't assume that the original poster didn't do so. I'm not
finding it so easy to generalize your hint. It's easy to
find a nonzero solution (x,y) of ax+by=0 when a and b are
not both zero. The next step would be to try to find a
nonzero simultaneous solution (x,y,z) to ax+by+cz=0 and
dx+ey+fz=0, but that's a different story.

If we let (x,y,z)=(bf-cd, cd-af, ae-bd) (minors of the
matrix), then ax+by+cz=dx+ey+fz=0. Then we have to deal
with the case where bf=cd, cd=af, and ae=bd. So let (x,y,z)
=(-b,a,0), (-c,0,a), or (0,-c,b) as needed to be nonzero
in those cases.

In general, you can assume m=n+1 by restricting to the
submodule of rank n+1 of the domain. The minors give you
a candidate solution. Then if they're all 0, you need
some further argument.

I could be wrong but I'm not convinced this is the easiest
way to go, starting from what you have in Atiyah and
McDonald.

You're right, it's not immediately clear how to generalize from the
case n=1. In fact, there's probably a slicker way, using more advanced
machinery. Still, I sense that the argument I sketched for the case
n=1 captures the underlying key idea. Then again, maybe it's not that
simple -- I'll think about it.

quasi
.