Let $F$ be a field. I'd like to show that all $F\times F$-modules are locally free. I think I've done that, but I'm confused by some material I found about this. Could you please verify my proof?
The prime ideals of $F\times F$ are $I=F\times\{0\}$ and $J=\{0\}\times F$.
We have a ring homomorphism $F\times F\to F$ given by $(a,b)\mapsto b$. It sends all elements outside $I$ to units. So we have a ring homomorphism $f\colon(F\times F)_I\to F$ given by $(a,b)/(c,d)\mapsto b/d$. Clearly $f((0,b)/(0,1))=b$ and so $f$ is surjective. But $f$ is also injective: If $f((a,b)/(c,d))=0/1$ then $b/d=0/1$, and so $b=0$. But then $(0,1)\cdot(a,b)=(0,0)$ and thus $(a,b)/(c,d)=(0,0)/(1,1)$ because $(0,1)$ is not in $I$.
So $(F\times F)_I\cong F$ as rings. Similarly $(F\times F)_J\cong F$ as rings. So the localization of $F\times F$ at every prime ideal is a field. So if $M$ is an $F\times F$-module then $M_I$ and $M_J$ are vector spaces over $(F\times F)_I$ and $(F\times F)_J$, respectively, and vector spaces free modules! Thus $M$ is locally free.
Then the localized module is a module over a field, a.k.a. a vector space, hence free over the localization.
Products of fields (even infinite products) are all von Neumann regular.
– rschwieb Oct 30 '23 at 22:08