# My minimal polynomial over this field of rational functions has two indeterminates, please help me understand my error

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 ?

• I only see a single variable $x$. Remember that $t\in E$, so it is a constant for the purposes of this task. – Jyrki Lahtonen Jul 10 at 3:27
• 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 $F$. – Jyrki Lahtonen Jul 10 at 3:30
• 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 $Aut(E)$ – reuns Jul 10 at 20:47

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 $$F[x]$$.