Exercise 11.1.1

Proof. Let $q=p^n$, where $p$ is prime. There exists an extension ${𝔽}_p\subset L$ such that $x^q-x$ splits completely over $L$. Write

We show that $F$ is a subfield of $L$, with $q$ elements.

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$1\in F$, since $1^q=1$.

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As the characteristic of $L$ is $p$, if $\alpha ,\beta \in F$,

$${(\alpha +\beta )}^{p^n}={\alpha }^{p^n}+{\beta }^{p^n}=\alpha +\beta ,$$

therefore $\alpha +\beta \in F$.

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If $p$ is odd, ${(-\alpha )}^{p^n}=-{\alpha }^{p^n}=-\alpha$, so $-\alpha \in F$, and if $p=2$, $-\alpha =\alpha \in F$.

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${(\mathit{\alpha \beta })}^q={\alpha }^q{\beta }^q=\mathit{\alpha \beta }$, so $\mathit{\alpha \beta }\in F$.

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If $\alpha \in F,\alpha \ne 0$, ${(1∕\alpha )}^q=1∕{\alpha }^q=1∕\alpha$, therefore $1∕\alpha \in F$.

Since the derivative of $f(x)=x^q-x$ is $-1$, then $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f,f^{\prime })=1$. Therefore $f(x)$ has $q$ distinct roots in $L$, therefore $|F|=q=p^n$. □