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It is easy to prove that for a an irreducible polynomial $P$ of degree $d$ of $\mathbb F_q[T]$, one can embed $\mathbb F_{q^d}$ in $\mathbb F_q(T)_P$ (the completion of $\mathbb F_q(T)$ at $P$) and there exists an element $\alpha\in\mathbb F_{q^d}$ such that $\mathbb F_q(T)_P=\left\{\sum_{n\ge -m} a_n(T-\alpha)^n\mid \alpha_n\in\mathbb F_{q^d},\,m\in\mathbb Z\right\}$

I am looking for a reference where this kind of things is proved.

Thanks in advance

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  • $\begingroup$ Are you just localizing or are you also $(P)$-adically completing? Otherwise it is hard to make sense of your infinite sum. $\endgroup$
    – Kevin Casto
    Commented Sep 25, 2023 at 5:22
  • $\begingroup$ I complete $(P)$-adically.. I am going to modify my post $\endgroup$
    – joaopa
    Commented Sep 25, 2023 at 6:06
  • $\begingroup$ This kind of thing was proven by Cohen in jstor.org/stable/1990313 though it is rather more general, I think without explicitly containing the precise result, than you are asking about. $\endgroup$
    – Simon Wadsley
    Commented Sep 25, 2023 at 9:43

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Have you tried using Hensel's lemma to show $|T - \alpha|_P < 1$ for some $\alpha$ in $\mathbf F_{q^d}$ and then show $T-\alpha$ is a uniformizer in the completion?

Once you have a uniformizer $t$ in the completion, show $\mathbf F_{q^d}[t]$ is dense in the valuation ring and then that the valuation ring is $\mathbf F_{q^d}[[t]]$ with the $t$-adic topology, so the completion (the fraction field of the valuation ring) is the Laurent series field \mathbf F_{q^d}((t))$. Use $t = T-\alpha$ and then you're done.

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  • $\begingroup$ Thanks for the detailed answer (I knew to prove the assertion) I am looking for a reference only where it is stated explicitly. $\endgroup$
    – joaopa
    Commented Sep 25, 2023 at 6:36

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