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$F$ is a field of characteristic $p$ and $a\neq c^p-c$ for $c\in F$. Then determine the galois group of $x^p-x-a$.

First I showed that this is an irreducible polynomial and has no multiple roots. This will imply that the polynomial had $p$- distinct roots in its splitting field. Hence we can get a $p$-cycle. This means that the Galois group is $S_p$. Is my reasoning correct?

Rutherford Mark
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    Your $c^p - a$ should be $c^p - c.$ Concerning the question, just having a $p$-cycle is hardly sufficient. A separable irreducible cubic has Galois group $A_3$ or $S_3$. Both contain a $3$-cycle. You should think about how the roots are related to each other. If you have one root $\alpha$ then find a formula for the remaining roots in terms of $\alpha$. – KCd Apr 26 '15 at 00:46
  • By https://math.stackexchange.com/questions/81583/how-do-i-prove-that-xp-xa-is-irreducible-in-a-field-with-p-elements-when , the polynomial is irreducible over $\mathbb{F_p}$. Hence, if $\alpha$ is a root, then the map $\sigma$ that sends $\alpha$ to $\alpha+1$ (which is also a root) is an automorphism of the Galois group. Clearly $\sigma$ has order $p$. Moreover, since $\alpha+n$ is a root for all $n\in\mathbb F_p$, the splitting field is of degree $p$. So the galois group is generated by $\sigma$. – 19021605 Jun 28 '25 at 12:54
  • Another argument: the maps $\sigma$ that sends $x$ to $x^p$ is a $\mathbb{F}_p$-automorphism of the splitting field of order $p$. So the Galois group has at least order $p$. Since adjoining one roots give one root, the extension is of degree at most $p$. We have $|\mathrm{Gal}(\mathbb{F}_p(\alpha)/\mathbb{F}_p)|=|\mathbb{F}_p(\alpha):\mathbb{F}_p|=p$ with generator $\sigma$ for the group. – 19021605 Jun 29 '25 at 09:32

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Actually, if $\alpha$ is a root, then so is $\alpha+n$, for all $n\in\mathbb F_p$. Thus, adjoining one root gives the entire splitting field, and thus the splitting field is an extension of degree $p$. The Galois Group is thus isomorphic to $C_p$.

Nishant
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