Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

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.

share|improve this answer

$\def\FF{\mathbf{F}_p} \def\FBAR{\overline{\mathbf{F}}_p}$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 $\FF(y)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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