Exercise 8.6.5

Proof.

(a)

Already done in Exercise 5.3.16:

$M=k(t)$ has characteristic $p$ and $f=x^p-x+t\in M[x]$.

$f^{\prime }=-1$, thus $f\wedge f^{\prime }=1$, so $f$ is separable.

As $\alpha$ is a root of $f$, $f(\alpha )={\alpha }^p-\alpha +t=0$, thus

$$\begin{array}{llll}\hfill f(\alpha +1)& ={(\alpha +1)}^p-(\alpha +1)+a& \hfill & \\ \hfill & ={\alpha }^p+1-\alpha -1+a& \hfill & \\ \hfill & =0& \hfill & \end{array}$$

$\alpha +1\in L$ is also a root of $f$.

So $\alpha ,\alpha +1,\dots ,\alpha +p-1$ are roots of $f$. These roots are distinct since $0,1,\dots ,p-1$ are the $p$ distinct elements of the prime subfield of $F$, isomorphic to ${𝔽}_p$, and identified with ${𝔽}_p$.

Thus $f$ is divisible by $(x-\alpha )\cdots (x-\alpha -p+1)$, of degree $p=\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->(f)$. As both polynomials are monic,

$$\begin{array}{lll}\hfill f=(x-\alpha )(x-\alpha -1)\cdots (x-\alpha -p+1)& & \hfill \text{(1)}\end{array}$$

so the roots of $f$ are $\alpha ,\alpha +1,\dots ,\alpha +p-1$, and $L=M(\alpha )$.

Here $\mathbb{Z}∕\mathit{p\mathbb{Z}}={𝔽}_p$ is the prime field of $M$, so $[i]\in {𝔽}_p\subset M$. $\phi$ is well defined: if $\alpha +i=\alpha +j$, then $[i]=[j]$. Moreover $\phi$ is a group homomorphism: if $\sigma ,\tau \in \mathrm{Gal}(L∕M)$, then $\sigma (\alpha )=\alpha +i,\tau (\alpha )=\alpha +j$ for integers $i,j$, and

Hence

Moreover, if $\sigma \in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$, $\sigma (\alpha )=\alpha$. As $L=M(\alpha )$, and $\sigma$ fixes $M$, $\sigma =e$, so $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{e\}$ and $\phi$ is injective. □