finding $[K(α):K] $, and $\operatorname{Aut}(K(α)/K)$ when $K=F_3(t)$ and $α$ is root of the polynomial $f(x)=x^6−tx^3+1$

Question

in the previous post:$K=F_3(t)$ show that for every root $\alpha$ of the polynomial $f(x)=x^6-tx^3+1$, $K(\alpha)/K$ is a splliting field I proof that $K(\alpha)$ is splitting field over $K$ when $\alpha$ is root of the polynomial $f\in K[x]$ now I need to know what is $[K(\alpha):K]$ and $\operatorname{Aut}(K(\alpha)/K)$

I assume that because $f$ is a 6 degree minimal polynomial in $K$ so $[K(\alpha):K]=6$ and because I have only 2 roots for $f$ so $\operatorname{Aut}(K(\alpha)/K)=(Id, \sigma)\to \sigma (\alpha)=\beta , \sigma (\beta)=\alpha$ when $\alpha,\beta$ are roots of the polynomial $f$. Am I right?

Answer 1

Yes. Your answers are correct. An ultra-detailed solution below just in case you need some of those steps, or want to see alternative arguments.

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Let $g(x)=x^2-tx+1$, so $f(x)=g(x^3)$.

1. We first see that $g(x)$ is irreducible over $K$. This is because (complete the square!) $g(x)=(x+t)^2-(t^2-1)$ has no zeros in $K$. Namely, uniqueness of factorization in the polynomial ring $\Bbb{F}_3[t]$ tells that $t^2-1=(t-1)(t+1)$ is not the square of any rational function in $t$.
2. Let $u$ and $v$ be the zeros of $g(x)$. We then have the relations $u+v=t$, $uv=1$, so the field $L:=K(u)$ is the splitting field of $g(x)$ over $K$ and $[L:K]=2.$
3. We see that $t=u+v=u+1/u$ is actually an element of the the field $\Bbb{F}_3(u)$. It follows that $L=\Bbb{F}_3(u)$.
4. Uniqueness of factorization in the polynomial ring, only this time in $\Bbb{F}_3[u]$, again implies that $u$ cannot be a cube in the field $\Bbb{F}_3(u)$. Therefore it isn't a cube of any element of the field $L$. It follows that the polynomial $h(x)=x^3-u$ is irreducible over $L$.
5. If $\alpha$ is a zero of $h(x)$, then, by irreducibility of $h(x)$, $[L(\alpha):L]=3$. Because we are un characteristic three, we have $$h(x)=(x-\alpha)^3.$$
6. The element $\beta=1/\alpha\in L$ satisfies the equation $$\beta^3=1/\alpha^3=1/u=v,$$ so $$g(x)=(x-\alpha^3)(x-\beta^3)$$ and therefore $$f(x)=g(x^3)=(x^3-\alpha^3)(x^3-\beta^3)=(x-\alpha)^3(x-\beta)^3.$$ This means that $L(\alpha)$ is the splitting field of $f(x)$ over $K$. Because $u=\alpha^3$ we also see that $L(\alpha)=K(\alpha)$.
7. The tower of extensions $K\subset L\subset L(\alpha)$ gives, by multiplicativity of the extension degree, that $[K(\alpha):K]=[L(\alpha):K]=6$. Consequently $f(x)$ is the minimal polynomial of $\alpha$ over $K$. In particular, it is irreducible over $K$.
8. Any $K$-automorphism $\tau$ of $K(\alpha)$ is uniquely determined by $\tau(\alpha)$. As $\alpha$ and $\beta=1/\alpha$ are the only zeros of $f(x)$, there are exactly two automorphisms. The identity mapping, the $K$-automorphism interchanging $\alpha$ and $\beta$.