Proving the roots of a polynomial are a subfield

Question

Define $f(X)=X^{p^r}-X \in \mathbb{F}_p[X]$ where $r\geq 1,\ p$ prime. I'm to show its roots are a subfield of its splitting field $E$ (I have already shown it has $p^r$ distinct roots in $E$).

What I currently have:

$0,1$ are clearly roots of $f$.

Suppose $a,b$ are roots of $f$. Then $(a+b)^{p^r}-(a-b) = a^{p^r}-a+b^{p^r}-b = 0$ so we have closure under addition.

But now, I can't get add./mult. inverses or closed under multiplication! For instance, to try to show existence of additive inverses I tried:

$(-a)^{p^r}-(-a) = (-1)^{p^r}a^{p^r}+a$, and unless $a=0$ or $p^r$ is odd, this isn't going to be zero..

How should I progress with this?

Answer 1

A nonempty subset of a finite group that is closed under the group operation is already a subgroup. That completes the additive part.

For multiplication, assume $a^{p^r}=a$ and $b^{p^r}=b$. Then $(ab)^{p^r}=a^{p^r}b^{p^r}=ab$. Was with addition this together with the existence of one nonzero root) shows that the set of roots (minus $0$) are a group under multiplication.

Answer 2

The splitting field is $\mathbb{F}_{p^r}$. To see this, it will suffice to prove that every element of $\mathbb{F}_{p^r}$ is a root. Note that that the multiplicative subgroup of non-zero elements of $\mathbb{F}_{p^r}$ has order $p^{r}-1$. Therefore, every non-zero-element $a\in\mathbb{F}_{p^r}$ satisfies $a^{p^r-1}=1$. In particular, $a^{p^r}-a=0$. This also holds for $a=0$. So, every element of $\mathbb{F}_{p^r}$ is a root of $x^{p^r}-x$.

Since $\mathbb{F}_{p^r}$ is a subfield of itself, this completes the proof.