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 ?

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $F[x]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.