# A homomorphism on a nontrivial commutative ring with trivial unit group

Suppose $$R \neq \{0\}$$ is commutative and satisfies $$R^\times = \{1\}$$. I've shown that this implies that $$\operatorname{char}R=2$$ (by showing that $$-1 = 1$$). Now consider the homomorphism $$f \in \operatorname{End}R$$ given by $$f(r) = r^2$$. I'm trying to show that $$f$$ is injective. To this end, I realize that as $$f$$ is in particular a group homomorphism, the statement $$\ker(f) = \{0\}$$ would imply this. In other words, we need to show that $$r^2 = 0 \Rightarrow r = 0$$. I'm not sure how to go about this. I'd also like to know whether this map is generally surjective.

Let $$r\in R$$ be such that $$f(r)=0$$. Then $$r^2=0$$ and hence $$(r+1)^2=r^2+2r+1=1.$$ This shows that $$r+1$$ is a unit, so $$r+1=1$$ and hence $$r=0$$. This shows that $$f$$ is injective.
To see that it is not surjective in general, consider the ring $$R=\Bbb{F}_2[X]$$.