Let $R$ be a ring of prime characteristic, $p > 0$ and $x,y \in R$. Show that:
$$(x+y)^p = x^p + y^p$$
To recall definition; a characteristic of $R$ is $n\in\mathbb N \cup \{0\}$, such that $\forall a \in R, na = 0.$ If such a number does not exist then $n=0$.
Any help is much appreciated.
Consider using the binomial theorem.
First, make sure you understand why the statement of this theorem holds over any commutative ring with unity, then apply it to the expression $(x + y)^p$. Since you are in characteristic $p$, any element of the form $nx$ for $n \in \mathbb{N}$, with $p | n$ will be equal to $0$ (make sure you can explain this step too!). Finally, you need to use the fact that $p | \binom{p}{i}$ for all $1 \leq i \leq p - 1$. This is not too difficult to prove, as long as you're satisfied using combinatorial arguments to conclude that $\binom{a}{b} \in \mathbb{N}$ for all $a, b \in \mathbb{N}$, $b \leq a$. If you're ok with all of that, then just think about the prime factorizations of the numerator and denominator in $\binom{p}{i}$.
Hope this helps!
