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I was looking over an old exam paper and I came across a question which confused me, it says:

Let $E = GF(7)(t)$ be the field of rational functions in the indeterminate $t$ over $GF(7)$. Define $σ,τ ∈ Aut(E)$ by $σ(t) = 2t$ and $τ(t) = 1/ t.$ Set $G = \langle σ,τ\rangle$ and $F = Fix(G)$.

Find the minimal polynomial of $t$ over $F$

Well I know that $\sigma(t)$ will be a root to some min. polynomial which would lead me to believe that $2t,4t$ and $t$ were roots but then we'd get the min polynomial $x^3-7x^2t+7xt^2-t^3=x^3-t^3$

But then our minimal polynomial has two indeterminates which seems meaningless to me , could anyone please shed some light on what I'm misunderstanding ?

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    $\begingroup$ I only see a single variable $x$. Remember that $t\in E$, so it is a constant for the purposes of this task. $\endgroup$ – Jyrki Lahtonen Jul 10 at 3:27
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    $\begingroup$ But, it seems to me that $|G|=6$. Clearly $E=F(t)$, so the minimal polynomial should have degree six also. Observe that $t^3\notin F$ for $\tau(t^3)=1/t^3\neq t^3$. This means that the coefficients of your polynomial are not in $F$. $\endgroup$ – Jyrki Lahtonen Jul 10 at 3:30
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    $\begingroup$ Your $F= E^G$ minimal polynomial is $\prod_{g \in G} (x-g(t)) \in E^G[x]$ where $G= \langle \sigma,\tau\rangle \cong S_3$ is your order $6$ finite subgroup of $Aut(E)$ $\endgroup$ – reuns Jul 10 at 20:47
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As Jyrki mentioned it is not hard to see that $G$ has order $6$. Indeed we have that $\sigma^3 = \tau^2 = 1$. Moreover $\tau^{-1} \sigma \tau = \sigma^2$ and hence we have that $G$ is the dihedral group of 6 elements.

Furthermore, as Jyrki also noted we have that $t$ is a primitive element of the extension $F \subset E$. Hence we have that $\sigma(t),\sigma^2(t), \tau(t),\sigma\tau(t),\sigma^2\tau(t)$ are roots of the minimal polynomial of $t$, too. In particular it is given by:

$$(x-t)(x-2t)(x-4t)\left(x - \frac 1t\right)\left(x - \frac 2t\right)\left(x - \frac 4t\right) =$$ $$(x^3 - 7tx^2 + 14t^2x - 8t^3)\left(x^3 - \frac{7x^2}{t} + \frac{14x}{t^2} - \frac{8}{t^3}\right) =$$ $$(x^3 - t^3)\left(x^3 - \frac{1}{t^3}\right) = x^6 - \left(t^3 + \frac 1{t^3}\right)x^3 + 1$$

You can indeed note that this is a polynomial in $F[x]$.

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