What carries over?
As Peter pointed out, a submodule of a free $mathbb{Z}$-module though
free need not have a complement. Indeed each submodule of a free
$mathbb{Z}$-module is free, but a quotient module need not be, for instance
$mathbb{Z}/2mathbb{Z}$. Also a $mathbb{Z}$-module is free if and oly if
it is projective; this entails that a kernel of a map of free modules
does have a complement.
The set $mathrm{Hom}(F,G)$ for free $mathbb{Z}$-modules need not be free.
If $F$ is free of countably infinite rank and $G=mathbb{Z}$, then
$mathrm{Hom}(F,G)congprod_{j=1}^inftymathbb{Z}$ which remarkably
is not free over $mathbb{Z}$. But $Fotimes G$ is free for free $F$ and $G$.
No comments:
Post a Comment