Ring Homomorphism of a Ring with Characteristic 2

Question

The book I'm reading makes the following claim:

Let $R$ be a commutative ring of characteristic $2$. Then the mapping $a \rightarrow a^2$ is a ring homomorphism from $R$ to $R$

I tried verifying this but I get stuck really quick: $\phi(a+b) = a^2 + b^2 + 2ab$. How do I use what I know about the characteristic of the ring to simplify this? (Similarly with the product verification)

Answer 1

If the ring $R$ has characteristic 2, this means that $r + r = 0_{R}$ for all elements $r \in R$.

Thus $\phi(a+b) = a^{2} + b^{2} + ab + ab = a^{2} + b^{2} = \phi(a) + \phi(b)$ since $2ab = ab + ab = 0$ as $ab \in R$.

In the case of products, $\phi(ab) = (ab)^{2} = a^{2}b^{2} $ by commutativity, since $R$ is commutative $abab = aabb$. Hence $\phi(ab) = a^{2}b^{2} = \phi(a) \phi(b)$.

Answer 2

Note that characteristic two means $a+a=0$ for every $a\in R$

Then $\phi(a+b)=a^2 + b^2 + 2ab=a^2 +b^2=\phi(a)+\phi(b)$ since $2ab=0$

With the product we have $\phi(ab)=abab=a^2b^2=\phi(a) \phi(b) $ since R is commutative.

Now we have $\phi $ preserves the operations in R.