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I am reading this article and it talks about how polynomial modulo $p$ could help determine the Galois group. I am trying it out on the polynomial $f = x^5 - x -1$, which is irreducible over $\mathbb{Q}$.

I reduced $f$ to $f$ modulo 2: $\bar{f} = x^5 + x + 1$. I showed that $\bar{f}$ has no root in $\mathbb{Z}_2$ but factors as irreducible polynomials $\bar{f}=(x^2 + x + 1)(x^3 + x^2 + 1)$. According to the reading, $\text{Gal}(f)$ should contain element $(12)(345)$.

My question is, can $\text{Gal}(f)$ contain a $5$-cycle? I know that $\text{Gal}(f)$ cannot be $S_5$ or $A_5$ because the order of Galois group is $5$. It cannot be a subgroup of $A_5$ since $(345)$ is not even. But these two arguments cannot prove that we cannot find $5$-cycle. I am pretty bad with permutation group so I'm assuming that I'm overlooking something really simple...

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  • $\begingroup$ A polynomial of the form $x^p-x+a$ with $a\not=0$ mod $p$ is an Artin Schreier polynomial, and these have long been known to be irreducible mod $p$... If we know a bit of algebraic number theory, then we might know that this entails that the Galois group includes a $5$-cycle. $\endgroup$ – paul garrett Jul 30 '16 at 22:45
  • $\begingroup$ @paulgarrett I am confused because $(12)(345)$ is in Galois group. But $\langle (12)(345)\rangle$ actually has $6$ elements. Why is that? I thought the order of Galois group is $5$... $\endgroup$ – 3x89g2 Jul 30 '16 at 23:25
  • $\begingroup$ Why do you believe the order of the Galois group is 5? $\endgroup$ – quid Jul 30 '16 at 23:31
  • $\begingroup$ @quid proofwiki.org/wiki/… $\endgroup$ – 3x89g2 Jul 30 '16 at 23:35
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    $\begingroup$ @quid Oh I see where I got wrong. Somehow I thought that degree of extension always equal to degree of irreducible polynomial, which is not the case. I was thinking of simple ones like $x^2 - 2$ but actually $x^3 - 2$ is a counterexample to what I thought. Its splitting field has degree $6$, not $3$... $\endgroup$ – 3x89g2 Jul 31 '16 at 1:43
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The Galois group of an irreducible polynomial over $\mathbb{Q}$ is transitive on the roots. Hence the Galois group $G$ of $x^{5}-x-1$ has order divisible by $5$. Any finite group whose order is divisible by a prime $p$ contains an element of order $p$ (Cauchy's theorem). All elements of order $5$ in $S_{5}$ are $5$-cycles so $G$ certainly contains a $5$-cycle.

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  • $\begingroup$ I actually kept multiplying $(12)(345)$ by itself and I got $\{ (12)(345), (354), (12), (345), (12)(354), e\}$. Where did I go wrong? $\endgroup$ – 3x89g2 Jul 30 '16 at 22:59
  • $\begingroup$ Because that element is not a $5$-cycle, does not mean that there is no $5$-cycle in the Galois group. $\endgroup$ – Geoff Robinson Jul 30 '16 at 23:02
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If the Galois group is of order 5 then it certainly is a cyclic group, by the fact that any group of prime order is cyclic.

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