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I’m struggling with a following question. Having a polynomial $P(x) = x^{10} + x + 1 \in \mathbb{Q}[x]$ find its Galois group. The question for $x^{3} + x + 1$ is much easier for me, because, all the Galois groups of cubics are subgroups of $S_{3}$ and there are only four cases here $\{e\}$, $\mathbb{Z}/2\mathbb{Z}$, $A_{3}$ and $S_{3}$. Each of them has a particular meaning in terms of roots permutations and there is a discriminant criteria, which allows to classify all Galois groups for cubics.

Unfortunately $S_{10}$ is quite huge and the question of thinking about Galois groups as groups permitting roots is not that obvious here. Is there some more abstract way to construct the Galois group for $x^{n} + x + 1$?

The following is probably an important remark. As $\forall x\in\mathbb{R}:P(x) = x^{10} + x + 1>0$, $P(x)$ $\color{red}{\text{is irreducible both over $\mathbb{Q}$ and $\mathbb{R}$}}$. For arbitrary $n$ it is $\color{red}{\text{at least irreducible}}$ over $\mathbb{Q}$, as both leading coefficient and free term are $1$, but nor $1$ nor $-1$ are roots.

UPD: As it was mentioned in comments, of course, the last statement means only that there is no decomposition to linear polynomials. However for $n = 3$ it shows that it’s irreducible over $\mathbb{Q}$.

Matthew Willow
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    "For arbitrary $n$ it is at least irreducible over $\mathbf Q$, as both leading coefficient and free term are 1, but nor $1$ nor $−1$ are roots." That is a common and really bad mistake: the lack of rational roots only tells you there is no linear factor. It absolutely does NOT tell you the polynomial is irreducible! Consider $(x^2-2)(x^2-3)$: it has no root in $\mathbf Q$ but that doesn't mean it's irreducible over $\mathbf Q$. And your $x^{10}+x+1$ is definitely not irreducible over $\mathbf R$: an irreducible polynomial over $\mathbf R$ has degree at most $2$. – KCd Dec 26 '22 at 22:40
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    The determination of which $x^n+x+1$ are irreducible over $\mathbf Q$ for $n \geq 2$ is given in Example A.2 of https://kconrad.math.uconn.edu/blurbs/ringtheory/irredselmerpoly.pdf. In short: it is irreducible if and only if $n = 2$ or $n \not\equiv 2 \bmod 3$. (For $n \equiv 2 \bmod 3$, $x^2+x+1$ is a factor, so the polynomial is reducible when $n \equiv 2 \bmod 3$ and $n > 2$.) In particular, $x^{10}+x+1$ is irreducible over $\mathbf Q$. But the proof is not as straightorward as checking there are no rational roots. – KCd Dec 26 '22 at 23:00
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    See H. Osada, The Galois groups of the polynomials $x^n + ax^s + b$, II, Tohoku Math. J. 39 (1987), 437-445. By Theorem $1$ and the remark on p. $441$, $x^n + x + 1$ has Galois group $S_n$ as long as $x^n + x + 1$ is irreducible over $\mathbf Q$. – KCd Dec 26 '22 at 23:10
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    A proof that $x^n-x-1$ has Galois group $S_n$ over $\mathbf Q$ is in https://kconrad.math.uconn.edu/blurbs/gradnumthy/galoisselmerpoly.pdf. It needs algebraic number theory in Step $2$. Probably the proof carries over to $x^n+x+1$ when it is irreducible over $\mathbf Q$. I should check that and update the file if it works out. – KCd Dec 26 '22 at 23:13
  • @KCd, oh yes, of course, you’re absolutely right Thank you! – Matthew Willow Dec 26 '22 at 23:43
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    Step 2 in the proof in the link of my previous comment is indeed applicable to $x^n + \delta x + \varepsilon$ where $\delta,\varepsilon \in {\pm 1}$, so if that polynomial is irreducible over $\mathbf Q$ then its Galois group over $\mathbf Q$ is $S_n$. – KCd Dec 27 '22 at 19:41
  • @KCd, I’ve proved that $x^{10} + x + 1$ is irreducible, using https://kconrad.math.uconn.edu/blurbs/ringtheory/irredselmerpoly.pdf, however the second article is much more complicated for me. I'm trying to understand Step 2, but not very successfully. However, it seems to me, that for this particular case there is a simpler proof. If I am to construct such a proof, I will definitely post it here. – Matthew Willow Dec 28 '22 at 01:21
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    See Theorem 2.1 in https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisSnAn.pdf: by factoring $x^{10}+x+1$ modulo suitable primes, show its Galois group over $\mathbf Q$ contains a transposition and a 7-cycle, as that would imply the Galois group is $S_{10}$. – KCd Dec 28 '22 at 01:49
  • @KCd, yeah! i was thinking about something very similar! Thanks! – Matthew Willow Dec 28 '22 at 02:07
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    The factorization of $x^{10}+x+1 \bmod 2$ has irreducible factors of degrees $3$ and $7$, so the Galois group over $\mathbf Q$ has an element $\sigma$ whose permutation on the roots has cycle type $(3,7)$, so $\sigma^3$ is a $7$-cycle in the Galois group. The factorization of $x^{10}+x+1 \bmod 13$ has irreducible factors of degrees $1$, $2$, and $7$, so the Galois group over $\mathbf Q$ has an element $\tau$ whose permutation on the roots has cycle type $(1,2,7)$, so $\tau^7$ is a transposition in the Galois group. – KCd Dec 28 '22 at 02:14
  • @KCd the usage of Dedekind theorem about permutations cycle type is very effective, however its proof (https://kconrad.math.uconn.edu/blurbs/gradnumthy/dedekind-index-thm.pdf) is quite rough. – Matthew Willow Dec 28 '22 at 18:07
  • I know that link very well (guess why), and it's irrelevant since the existence of an element of the Galois group permuting the roots in a certain way is not the theorem that is proved there. What you want is Theorem 4.13 in https://kconrad.math.uconn.edu/blurbs/galoistheory/galoisaspermgp.pdf. There only easy proof of this theorem uses algebraic number theory. While there are proofs that aim to avoid algebraic number theory, I really don't think anyone can understand a proof of the theorem that way. This theorem of Dedekind is much easier to learn how to use than it is to prove. – KCd Dec 28 '22 at 21:37
  • @KCd Is there a reason for not factoring in mod 7 instead of mod 13 where in that case there is a cycle of type (2,8), for showing existence of a transposition in the Galois group – テレビ スクリーン Dec 29 '22 at 15:17
  • @テレビスクリーン if a subgroup of $S_n$ contains a permutation $\sigma$ of cycle type $(2,8)$, why must the subgroup contain a transposition? – KCd Dec 29 '22 at 15:53
  • I see, $\sigma^8$ is not a transposition. Thank you – テレビ スクリーン Dec 29 '22 at 16:20
  • This tred is helpful: https://math.stackexchange.com/questions/836795/proof-of-dedekinds-theorem-on-the-galois-groups-of-rational-polynomials – Matthew Willow Dec 29 '22 at 18:10

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