I'm trying to find the Galois group of $P(X)=X^5+3X^2+1 \in \mathbb{Q}[X]$, but I have no idea how to proceed.

First, the polynomial is irreducible in $\mathbb{Q}[X]$ (we can reduce modulo $2$ for instance). By checking the extrema, we verify there's only one real root. Let $R$ denote the set of roots. Since the polynomial has real coefficients, taking complex conjugates show the $5$ distinct (since $char\ \mathbb{Q}=0$ and $P$ is irreducible) roots can be written as $R=\{x,\alpha,\bar{\alpha},\beta,\bar{\beta}\}$, with $x\in\mathbb{R}$.

Let $K$ be the splitting field of the polynomial, and $G$ the associated Galois group, with $|G| = n$. Since the polynomial is a quintic, we know that $5| n$ and $n|5!=120$

By Cauchy's Theorem, $G$ contains an element of order $5$ i.e. a $5$-cycle $(12345)$. Complex conjugation defines a double transposition $(23)(45)$ that leaves fixed the real root; this implies, in particular, that $2|n \Rightarrow 10|n$.

This is where I start having doubts. These two elements seem generate the $3$-cycle $(345)$ as follows: $$12345 \rightarrow 23451 \rightarrow 24315 \rightarrow 15243\rightarrow 12534$$ Since a $3$-cycle has order $3$, we now have $3|n \Rightarrow 30|n$.

However, this seems to be wrong according to this list. Where do I go from here?

  • $\begingroup$ Your error is in thinking that you are free to pick the numbers in both the 5-cycle and the product of two 2-cycles as you see fit. If you have $(12345)$ and $(25)(34)$, then you won't get 3-cycles. Instead you get the dihedral group $D_5$ of order ten. So you need to know exactly which root is which in your 5-cycle. $\endgroup$ – Jyrki Lahtonen Sep 10 '16 at 22:52
  • $\begingroup$ Related: math.stackexchange.com/questions/1375747 $\endgroup$ – Watson Dec 26 '16 at 13:07

We know that the transitive automorphism(the one of of order 5) and the complex conjugation (the double transposition) are part of the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$. Unfortunatelly this isn't enough for us to distinguish whether the Galois group is $A_5$ or $S_5$, as the two forementioned automorphisms generate an automorphism group isomorphic to $A_5$, but that doesn't mean that there isn't another automorphism on the roots, which combined with these two automorphisms spans over $S_5$. So as I mentioned combining all possible combinations of $(12345)$ and $(23)(45)$ won't help us much, except leaving us with two candidates $A_5$ or $S_5$.

But thankfully here the Dedekind's Theorem comes in very handy. Factoring the polynomial in $\mathbb{Z}_{19}$ we have that:

$$x^5 + 3x^2 + 1 \equiv (x+2)(x+9)(x+13)(x^2+14x+16) \pmod{19}$$

Therefore by the Dedekind's Theorem the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$ contains an automorphism isomorphic to the cycle of the type $(1,1,1,2)$, which is a transposition. Now using the fact that if a subgroup of $S_n$, $n$ a prime, contains a $n-$cycle and transposition it's actually $S_n$ itself we can conclude that the Galois group of $x^5 + 3x^2 + 1$ over $\mathbb{Q}$ is $S_5$

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    $\begingroup$ The claim "if a subgroup of $S_n$ contains a $n-$cycle and transposition it's actually $S_5$ itself" is true if $n$ is prime, I think. $\endgroup$ – Watson Sep 7 '16 at 12:23
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    $\begingroup$ But for instance the subgroup generated by $(1,3)$ and $(1,2,3,4)$ in $S_4$ is the dihedral group of order $8$ (see math.stackexchange.com/questions/530062). $\endgroup$ – Watson Sep 7 '16 at 12:33
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    $\begingroup$ I agree that $S_n$ is generated by $(1,2)$ and $(1,2,...,n)$. But if we "just rename the roots", then you should have $(1,2)$ and $(1,3,2,4)$ (we must swap $2$ and $3$ everywhere), which is not $S_4$ again. $\endgroup$ – Watson Sep 7 '16 at 12:57
  • $\begingroup$ Stefan, @Watson is right. When $n$ is a prime any 2-cycle $\tau$ and any $n$-cycle $\sigma$ will do. This is because all non-trivial powers $\sigma$ are still $n$-cycles, and we can use the power where the two numbers in $\tau$ are adjacent. Only then will renaming the roots help. When $n$ is not a prime, some power of $\sigma$ will still map one of the numbers in $\tau$ to the other. But, that power may no longer be an $n$-cycle. $\endgroup$ – Jyrki Lahtonen Sep 10 '16 at 23:01
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    $\begingroup$ @JyrkiLahtonen I got it! I just confused the fact that $S_n = \langle (1,2),(1,2,3,...,n) \rangle$ with "any n-cycle and transposition generate $S_n$ for all $n \in \mathbb{N}$", while the latter is true only for prime numbers. $\endgroup$ – Stefan4024 Sep 11 '16 at 0:23

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