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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.

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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 $R=\Bbb{F}_2[X]$.

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