Exercise 6.2.5

Proof.

(a)

$L=F(\alpha )$ and $\alpha$ has for minimal polynomial $f=x^p-x+a$, thus $[L:F]=p$.

By Exercice 5.3.16, we know that $L=F(\alpha )=F(\alpha ,\alpha +1,\dots ,\alpha +p-1)$ is the splitting field of

$$f=x^p-x-a=(x-\alpha )(x-\alpha -1)\cdots (x-\alpha -p+1).$$

Therefore $F(\alpha )$ is the splitting field of a separable polynomial $f\in F[x]$, and by theorem 6.2.1

$$|\mathrm{Gal}(L∕F)|=[L:F]=p.$$

Every group of order $p$, where $p$ is prime, is cyclic and isomorphic to $\mathbb{Z}∕\mathit{p\mathbb{Z}}$ :

$$\mathrm{Gal}(L∕F)\simeq \mathbb{Z}∕\mathit{p\mathbb{Z}}.$$

(b)

$F\subset L$ is by part (a) a normal extension, and $f\in F[x]$ is irreducible over $F$ by hypothesis. The roots of $f$ in L are $\alpha ,\alpha +1,\dots ,\alpha +p-1$. By Proposition 5.1.8 , there exists a field isomorphism ${\sigma }_i:L\to L$ which is identity on $F$ and which takes $\alpha$ on $\alpha +i,i\in {𝔽}_p$. Then ${\sigma }_i\in \mathrm{Gal}(L∕F),\sigma (\alpha )=\alpha +i$. As $L=F(\alpha )$, $\sigma$ is uniquely determined by the image of $\alpha$.

Conclusion: $\alpha$ being a fixed root of $f$, and $i\in {𝔽}_p$, there exists a unique ${\sigma }_i\in \mathrm{Gal}(L∕F)$ such that ${\sigma }_i(\alpha )=\alpha +i$.

(c)

Let

$$\phi \{\begin{array}{ccc}\hfill \mathrm{Gal}(L∕F)\hfill & \hfill \to \hfill & \hfill {𝔽}_p\hfill \\ \hfill \sigma \hfill & \hfill \mapsto \hfill & \hfill \sigma (\alpha )-\alpha \hfill \end{array}$$

$\text{∙}$ For all $\sigma \in \mathrm{Gal}(L∕F)$, $\phi (\sigma )\in {𝔽}_p$ since $\sigma (\alpha )$ is a root of $f$, so $\sigma (\alpha )-\alpha =i\in {𝔽}_p$.

$\text{∙}$ $\phi$ is bijective by part(b), since for all $i\in {𝔽}_p$, there exists a unique $\sigma \in \mathrm{Gal}(L∕F)$ such that $\phi (\sigma )=\sigma (\alpha )-\alpha =i$.

$\text{∙}$ $\phi$ is a group homomorphism : if $\sigma ,\tau \in \mathrm{Gal}(L∕F)$, and $\phi (\sigma )=i,\phi (\tau )=j$, then $\sigma (\alpha )=\alpha +i,\tau (\alpha )=\alpha +j(i,j\in {𝔽}_p)$.

$(\sigma \circ \tau )(\alpha )=\sigma (\alpha +j)=\sigma (\alpha )+\sigma (j)=(\alpha +i)+j=\alpha +(i+j)$ ( $\sigma (j)=j$ since $\sigma$ is identity on $F$, a fortiori on ${𝔽}_p\subset F$).

As $(\sigma \circ \tau )(\alpha )=\alpha +(i+j)$, $\phi (\sigma \circ \tau )=i+j=\phi (\sigma )+\phi (\tau )$.

So $\phi :\mathrm{Gal}(L∕F)\to {𝔽}_p$ is a group isomorphism.