Application of the Artin-Schreier Theorem

Solution 1:

For (b), suppose that $\gamma\in K$ satisfies $\gamma^p-\gamma=\alpha$, and make a construction that relates $\gamma$ to $\beta$. Once you've related $\gamma$ to $\beta$, you can get a contradiction from the fact that $\beta$ has trace $1$.

For (c), first show that there is an extension of $\sigma$ to an automorphism $\sigma^*$ of $K(\theta)$ such that $\sigma^*(\theta)=\theta+\beta$. Then show that $(\sigma^*)^i$ fixes $\theta$ if and only if $p^m\mid i$. Then deduce that $K(\theta)=F(\theta)$.

Solution 2:

For (b), suppose that $x^p-x-\alpha$ is reducible in $K[x]$. Then, by the Artin-Schreier theorem, it has all of its roots in $K$. Let $\theta$ be such a root. Then $\theta^p-\theta-\alpha=0$ implies $\sigma^p(\theta)-\sigma(\theta)-\sigma(\alpha)=0$. This gives us $$[\sigma(\theta)-\theta]^p-[\sigma(\theta)-\theta]-[\sigma(\alpha)-\alpha]=[\sigma^p(\theta)-\sigma(\theta)-\sigma(\alpha)]-[\theta^p-\theta-\alpha]=0,$$thus $\sigma(\theta)-\theta$ is a root of $x^p-x-(\sigma(\alpha)-\alpha)$. From part (a), we know $\beta$ is a root of $x^p-x-(\sigma(\alpha)-\alpha)$. It's easy to show that $\beta+i$ is also a root for any $1\leq i\leq p-1$, so we must have $\sigma(\theta)-\theta=\beta+i$ for some $i$. Now, since $\theta\in K$, Tr$(\sigma(\theta)-\theta)=0$ and $i\in F$ implies $Tr(i)=0$. Now we have $$0=Tr(\sigma(\theta)-\theta)=Tr(\beta+i)=Tr(\beta)+Tr(i)=Tr(\beta),$$ a contradiction since Tr$(\beta)=1$. This means $x^p-x-\alpha$ is irreducible.

For (c), first note that $\sigma^*(\theta)=\theta+\beta$ implies $$(\sigma^*)^n(\theta)=\theta+\beta+\sigma(\beta)+\cdots+\sigma^n(\beta).$$ That means $$(\sigma^*)^{p^{m-1}}(\theta)=\theta+Tr(\beta)=\theta+1.$$ This implies $(\sigma^*)^{p^m}(\theta)=\theta$, hence the order of $\sigma^*$ is at most $p^m$. Then $H=\langle\sigma^*\rangle$ is a finite group of automorphisms of $K(\theta)$, so by Artin's Theorem, $K(\theta)$ is Galois over the fixed field of $H$, $K(\theta)^H$, and the Galois group is $H$. Now, since $F$ is fixed by $\sigma$, it is fixed by $\sigma^*$, which implies $F\subseteq K(\theta)^H$. Since degrees multiply in towers, we have $$p^m=p\cdot p^{m-1}=[K(\theta):K][K:F]=[K:F]=[K(\theta):K(\theta)^H][K^(\theta)^H:F].$$ This implies $[K(\theta):K(\theta)^H]=p^d\leq p^m$, so the order of $\sigma^*$ is $p^d$. Suppose it is strictly less than $p^m$. Then the order of $\sigma^*$ divides $p^{m-1}$, so $(\sigma^*)^{p^{m-1}}\equiv id$, but we already know $(\sigma^*)^{p^{m-1}}(\theta)=\theta+1$, a contradiction, hence the order must be $p^m$, which gives us $[K(\theta):K(\theta)^H]=p^m$, implying $K(\theta)^H=F$. Thus $K(\theta)$ is Galois and cyclic over $F$, and this implies that any intermediate extension is Galois and cyclic, so $F(\theta)/F$ is Galois and cyclic, as desired. Also, $\langle\sigma^*\big|_{F(\theta)}\rangle\subseteq Gal(F(\theta)/F)$ and $|Gal(F(\theta)/F)|=p^d\leq p^m$. Using the same argument as above in fact gives equality, so we have show that $F(\theta)$ is a Galois, cyclic extension of degree $p^m$.