# Calculating fixed field of a particular action

Let $K = \mathbb F_p(x)$, and let $H = \left\{\begin{pmatrix} d & a \\ 0 & 1 \end{pmatrix} \ \big| \ a \in \mathbb F_p, d \in \mathbb F_p^\times\right\}$ be a group under multiplication which acts on $K$ via $\begin{pmatrix} d & a \\ 0 & 1 \end{pmatrix} x = dx + a$.

How can I find the fixed field of $H$?

If $f = \frac{g(x^p - x)}{h(x^p - x)}$, then it's image under a generic point is $\frac{g(d(x^p-x))}{g(d(x^p-x))}$, which is close but isn't quite there.

Any help would be appreciated. Thanks

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The key identity here is

$$S(x)=(x^p-x)^{p-1}=x^{p-1}(x^{p-1}-1)$$

so that the polynomial $S$ above can be written both as a function if $x^p-x$ and a function of $x^{p-1}$. We show below that the fixed field is ${\mathbb F}_p(S(x))$.

Let $f(x)$ be a rational fraction fixed by $K$. We can write $f(x)=c\frac{u(x)}{v(x)}$ where $c$ is a constant and $u$ and $v$ are coprime unitary polynomials in $x$. By hypothesis, $f$ is fixed by $x \mapsto dx+a$, so $f(dx+a)=f(x)$ and hence $u(dx+a)v(x)=v(dx+a)u(x)$. By Gauss' lemma, $u(x)$ divides $u(dx+a)$.

Let $m$ be the degree of $u$. Then $d^mu(x)$ divides $u(dx+a)$. But since those two polynomials share the same degree and leading term (equal to 1), they must be equal. So $u(dx+a)=d^mu(x)$ for any $d,a$.

Taking $a=0$ and $d$ primitive as in Hurkyl's answer, we see that there is a polynomial $G$ such that $u(x)=G(x^{p-1})$.

Similarly, taking $a=d=1$, we see that $u(x+1)=u(x)$. We deduce that $w=u(x)-u(0)$ is identically zero on ${\mathbb F}_p$, so $w$ is a multiple of

$$\prod_{q\in {\mathbb F}_p}(x-q)=x^{p}-x$$ By induction, it is easy to see then that there is a polynomial $H$ such that $u(x)=H(x^p-x)$.

Therefore $$u(x)=G(x^{p-1})=H(x^p-x)$$

So ${\sf deg}(u)=(p-1){\sf deg}(G)=p{\sf deg}(H)$. So ${\sf deg}(u)$ is a multiple of $p(p-1)$. If $u$ is nonconstant, we can make an euclidian division of $u$ by $S$, and by induction we have $u(x)\in {\mathbb F}_p(S(x))$. Similarly $v(x)\in {\mathbb F}_p(S(x))$, so $f(x) \in {\mathbb F}_p(S(x))$, qed.

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$\def\FF{\mathbf{F}_p} \def\FBAR{\overline{\mathbf{F}}_p}$Consider the Laurent series of an element $f(x) \in \FF(x)$ in the fixed field of $H$:

$$f(x) = \sum_{k=n}^{+\infty} c_k x^k$$

Because $f(x)$ is fixed by the map $x \mapsto dx$, we have

$$\sum_{k=n}^{+\infty} c_k x^k = \sum_{k=n}^{+\infty} c_k d^k x^k$$

Choosing $d$ to be primitive, we see that we can only have $c_k \neq 0$ when $d^k = 1$: i.e. we can write $f(x) = g(x^{p-1})$.

Now that I have time to look at this again, I see Ewan Delanoy has already finished off an approach in this style (and I think the OP already had the ideas I hadn't included here) so I'll show another approach entirely.

Let $K$ be the fixed field. We want to see how $\FF(x)$ could be constructed as a Galois extension of $K$. The obvious candidate for a generator is $x$ and its conjugates. The defining polynomial would have to be

\begin{align*} f(t) &= \prod_{h \in H} (t - h(x)) \\ &= \prod_{d \in \FF^*} \prod_{a \in \FF} (t - (dx+a)) \\ &= \prod_{d \in \FF^*} \prod_{a \in \FF} ((t - dx) - a) \\ &= \prod_{d \in \FF^*} ((t - dx)^p - (t - dx)) \\ &= \prod_{d \in \FF^*} (t^p - t - d(x^p - x)) \\ &= (t^p - t)^{-1} \prod_{d \in \FF} (t^p - t - d(x^p - x)) \\ &= (t^p - t)^{-1} (x^p - x)^p \prod_{d \in \FF} \left( \frac{t^p - t}{x^p - x} - d \right) \\ &= (t^p - t)^{-1} (x^p - x)^p \left( \left(\frac{t^p - t}{x^p - x}\right)^p - \left(\frac{t^p - t}{x^p - x} \right) \right) \\&= \left( (t^p - t)^{p-1} - (x^p - x)^{p-1} \right) \end{align*}

And therefore, $K$ is isomorphic to $\FF(y)$ via $y \mapsto (x^p - x)^{p-1}$, and we have an explicit degree $p(p-1)$ polynomial $f(t) = (t^p - t)^{p-1} - y$ that defines $\FF(x)$ as an extension of $\FF(y)$.

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