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I need to find $[L:K]$ and $\textrm{Aut}(L/K)$, where $K := \mathbb{F}_3(t)$ and $L/K$ is the the splitting field for polynomial $f = X^6 - 2tX^3 + 1 \in K[X]$.

What I've done so far / what I know:

  • I know that all the $K$-automorphisms in $\textrm{Aut}(L/K)$ just permute the roots. So if I could figure what the unique roots are, then I'd have a full description of $\textrm{Aut}(L/K)$.

  • $L = K(\alpha_1, \cdots, \alpha_n)$ where the $\alpha_i$'s are the roots of the irreducible divisors of $f$.

  • By showing, via Eisenstein, that $f(X+1)$ is irreducible, I know that $f$, too, is irreducible. Thus, $f$ is the minimal polynomial over $K$ for each of its roots $\alpha_1, \cdots, \alpha_6$. Since the formal derivative $D(f) = 0$, we know that $f$ is not separable and some of those roots are repeated.

  • $L/K$ is a normal extension, so $|\textrm{Aut}(L/K)| = [L:K]_S$, the separability degree. And $[L:K]_S < [L:K]$, since $f$ is not separable.

Not sure where to go with this from here. I'd appreciate some help.

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Let $\rho$ be any root of $f$. Then $\rho^3$ is a root of $Y^2-2tY+1$, whose roots are $t\pm\sqrt{t^2-1}$. Thus you have the two-step inclusion $K\subset K(\sqrt{t^2-1}\,)\subset L$. Perhaps this is enough to answer your questions.

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