for arbitrary $y∈\Bbb{F}_q((X))^{sep}$, there exists $c∈\Bbb{F}_q((X))$ and $a∈\Bbb{F}_q((X))^{sep}$ such that $y＝ca^q$

Question

$\Bbb{F}_q((X))^{sep}$ be separable closure of $\Bbb{F}_q((X))$.

I want to prove, for arbitrary $y∈\Bbb{F}_q((X))^{sep}$, there exists $c∈\Bbb{F}_q((X))$ and $a∈\Bbb{F}_q((X))^{sep}$ such that $y＝ca^q$.

How can I prove this ?

Firstly I need to specify what $c$ we take, but I'm stupid about this.Thank you for your help.

Answer 1

Let $K = \Bbb{F}_q((X))$ and $y$ separable over $K$.

• What you asked is wrong, try with $q= 2$ and $y= X^{1/3}+X^{2/3}$ then $K( (y/c)^{1/q})/K(y)$ is purely inseparable for all $c\in K^*$, so $a= (y/c)^{1/q}$ can't be separable over $K$.

• What is correct is that $K(y)=K(y^q)$, as otherwise $K(y)/K$ would contain the purely inseparable extension $K(y)/K(y^q)$ so that $K(y)/K$ wouldn't be separable.

So this proves that there exist $a_j\in K$ such that $$y=\sum_{j=0}^{n-1} a_j y^{qj},\qquad n = [K(y):K]$$

This doesn't tell us a way to construct the $a_j$. To do so: write the $(y^q)^0,\ldots,(y^q)^{n-1}$ in the $y^0,\ldots,y^{n-1}$ basis (using the $K$-minimal polynomial of $y$), then invert the obtained $GL_n(K)$ matrix.