If $R$ is a non-trivial ring with $1$, and $f : R \to R, f(x) = x^2$ is a ring homomorphism, then I want to prove that $1+x \in R^{\ast}$ for all $ x \in \ker f.$ First, I have shown that $R$ has characteristic $2$:
$$ f(1) = 1^2 = 1$$ $$ f(x+y) = (x+y)^2 = x^2+2xy +y^2 = f(x) + f(y) = x^2 + y^2$$ $$f(xy)= x^2y^2=f(x)f(y).$$
Because $$x^2+2xy +y^2 = x^2 + y^2, $$ $$2xy=0 \ \forall \ x,y \in R \ \text{and} \ 2xy =0 \in R.$$
So, $$f(2xy)=(2xy)^2=2^2x^2y^2=f(0)=0.$$
Finally, $$ \ker(f) = \{ 2xy \in R \ | \ x,y \in R \} = \{2xy \in R \ | \ xy \in R \},$$ and $\ker(f) = (2).$
However, I am at a complete loss how to proceed from here in proving that $1+x \in R^{\ast}$ for all $ x \in \ker f.$
Any help would be greatly appreciated.
You don't actually need to prove that $R$ has characteristic 2. The fact that $1 + r$ is invertible for $r\in \ker(f)$ comes directly from the calculation $$(1+r)(1-r) = 1 - r^2 = 1 - f(r) = 1$$ This not only shows that $1+r$ has an inverse, but also gives you an explicit formula for the inverse.