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I would like to have some explanation for the following statement

Let $K$ be an algebraically closed field of characteristic $p>0$, and $K((t))$, the field of Laurent series with coefficients in $K$. The Galois group of the polynomial $X^{p^n}-X=t^{-1}$ is isomorphic to the additive group of $F_{p^n}$, i.e. to $(\mathbb{Z}/p\mathbb{Z})^n$.

Another question:

Are there some extensions with Galois group isomorphic to $(\mathbb{Z}/p^n\mathbb{Z})$ with $n>1$

Josh fisher
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Say that $\alpha$ is a root of your polynomial $X^{p^n}-X-t^{-1}=0.$ Then it is obvious that if $a \in \mathbb{F}_{p^n},$ that $\alpha+a$ is a root as well, since $(\alpha+a)^{p^n}= \alpha^{p^n}+a^{p^n} = \alpha^{p^n}+a.$ So the Galois group is as claimed.

There are finite extensions with Galois groups isomorphic to $\mathbb{Z}/p^n\mathbb{Z}$ with $n>1.$ This can be done using the Witt polynomials, see for example: Cyclic Artin-Schreier-Witt extension of order $p^2$ .

Dedalus
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    Everything you say is, of course, correct. But the answer is missing the piece that the given polynomial is irreducible. That follows from Artin-Schreier theory if we can show that the polynomial has no roots in $K((t))$. Mind you, I'm sure you know how to handle that, but I'm not sure all the readers are aware of all the subtleties. For example, if we replace $t^{-1}$ with $t$, then the series $u=-(t+t^{p^n}+t^{2p^n}+\cdots)\in K((t))$ is a zero of $X^{p^n}-X-t$. – Jyrki Lahtonen Aug 04 '19 at 10:19
  • Thank you for reminding me and posting this comment. You are indeed correct about this point and it is worth pointing out. – Dedalus Aug 04 '19 at 10:49
  • @Jyrki Lahtonen how one can show that the polynomial is irreducible? – Josh fisher Aug 10 '19 at 17:21
  • @Joshfisher This is not hard at all. Have you used the valuation associated to your field, and the non-archimedean property of this valuation? The irreducibility follows immediately from this. – Dedalus Aug 10 '19 at 18:14
  • @Dedalus Do you mean to use the fact that $v_{\mathfrak{P}}(X^{p^n} - X) = v_{\mathfrak{P}}(T^{-1}) = e(\mathfrak{P}/T)$ ? – Josh fisher Aug 10 '19 at 18:20
  • @Joshfisher That's the idea. Let $\mathfrak{P}$ be a place above $(t)$. If $u$ is any zero, it follows that $p^n\nu_{\mathfrak P}(u)=-e(\mathfrak{P}|(t))$ and so forth. – Jyrki Lahtonen Aug 10 '19 at 19:04