I have the following question, if $ \varphi: \mathbb{F}_{p^2} \longrightarrow \mathbb{F}_{p^2} $ defined as $\varphi(x) \mapsto x^p$ is an automorphism, then how do I show it is a homomorphism in the first step? (I have a problem in showing this is an additive homomorphism).
Now what I did was:
$$\varphi(a+b) = (a+b)^{p} = \binom{p}{0}a^p + \binom{p}{1}a^{p-1}b+...+\binom{p}{p-1}ab^{p-1}+\binom{p}{p}b^p$$
So now if I can show that $p^2 | \binom{p}{r}$ for all $1 \le r \le p-1$ then I will obtain that $\varphi(a+b) = a^p+b^p =\varphi (a)+\varphi(b).$ But how do I show the above claim?
I read about a Kummer's version of Lucas theorem on the internet, but I don't get how to use that? Is that needed to prove the above claim or something else?
Any hints? Also, I cannot find a duplicate if one exists, then please point out. Thanks.
It's false that $p^2 | \binom{p}{r}$ for any $r$, but this is unnecessary, since $\mathbb{F}_{p^2}$ has characteristic $p$, not $p^2$.
That is, all you need is that $p | \binom{p}{r}$ for $1 \leq r \leq p-1$, which is true.