Characteristic and homomorphism from $\mathbb Z$ to $R$

Question

I got asked the following question: Let $R$ be a ring with characteristic $2$. Is $R \cong \mathbb{Z}/(2)$? The way I thought of it is the following. Let $\Phi: \mathbb{Z} \rightarrow R , z \rightarrow 1 + \ldots + 1$ ($z$ times), if $z > 0, z \rightarrow 0$, if $z = 0$ and $z \rightarrow - (1+ \ldots + 1)$ ($z$ times). Then $\textrm{ker}(\phi) = (\textrm{char}(R))_\mathbb{Z}$. My question is, is $\phi$ surjective? In other words, does one get every element of $R$ just by adding $1$ over and over? This might be a fairly trivial question, however, I never really thought about it. This would then imply, that $R \cong \mathbb{Z}/(2)_\mathbb{Z}$ wouldn't it? Thank you in advance for help!

Answer 1

No, $\phi$ does not need to be surjective. A simple example is $R = (\mathbb{Z}/2)^n$. You could also consider a matrix ring over $(\mathbb{Z}/2)^n$ and any subring of $\operatorname{Mat}_m((\mathbb{Z}/2)^n)$ will also work.

Also a finite field of order $2^n$ will be characteristic $2$ and, by extension, any $\mathbb{F}_{2^n}$-algebra will be characteristic $2$. For example $$ \mathbb{F}_{16}[x,y,z]/(x - z^2, y - z^3). $$