Is vectorspace trivial under these conditions?

Question

Let $R$ be a ring.

Looking for a left $R$-module free over abelian group $A$, I arrived at $\left|R\right|\otimes A$ with $r.\left(s\otimes a\right)=rs\otimes a$ where $\left|R\right|$ denotes the underlying abelian group of $R$.

Investigating the special case $R=\mathbb{F}_{2}$ I suddenly realized that in that case: $r\otimes\left(b+b\right)=r\otimes b+r\otimes b=\left(r+r\right)\otimes b=0\otimes b=0$.

Can I conclude that a $\mathbb{F}_{2}$-vectorspace, free over an abelian group, is trivial whenever each element $a$ of the group can be written as $a=b+b$? Or did I make some mistake?

Answer 1

Any module over any ring (with identity) of characterisic $2$ is going to exhibit the behavior you're talking about.

In a module $M$ over such a ring, free or not, $m+m=1\cdot m+1\cdot m=(1+1)\cdot m=0\cdot m=0$.

So if you have any module $M$ over a ring with characteristic $2$, no nonzero element is going to be expressible as $m'+m'$ for some other $m'$ in the module.

Answer 2

If you have an action of such a group on a $\mathbb{F}_2$ vector space then for each $v \in V$ you have $av=(b+b)v=bv+bv=0$. This does not directly comes from the "tensor product" case but it is still easy.