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I'm trying to prove that the following polynomial is not solvable by radicals:

$$p(x) = x^5 - 4x + 2 $$

First, by Eisenstein is irreducible.

(It is not difficult to see that this polynomial has exactly 3 real roots)

How can I proceed?

Thank you!

user1612
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    Isn't the general method that of computing the Galois group of the polynomial and showing it is not solvable as a group? – user99680 Jun 18 '14 at 01:09
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    Do you mean to say that by Eisenstein this is IRREDUCIBLE? – Vladhagen Jun 18 '14 at 01:09
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    One or more of the techniques in http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf might be helpufl. – Qiaochu Yuan Jun 18 '14 at 01:12
  • Where did this example come from? I found it as exactly the nonsolvable quintic mentioned by Rotman in his textbook Galois Theory (2nd edition). See Theorem 75 on page 74. – KCd Jun 18 '14 at 03:22
  • How did you see it has exactly 3 real roots? – user428487 Nov 22 '20 at 07:35
  • @user428487 hand-wavely speaking, start with $x^5-4x$ which clearly has $3$ real roots. By adding $+2$ you move the graph up, but it doesn't reduce the number of the roots, since it's easy to see that the local minimum value is less then $-2$. – S. R Nov 15 '21 at 23:24

3 Answers3

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Roadmap for work below: It will suffice to show that the polynomial has an unsolvable Galois group, namely $S_5$. To do this, we will show that the Galois group, when viewed as a permutation group, has a $5$-cycle and a $2$-cycle; these serve as a generating set for $S_5$.

Let $K$ be the splitting field of $f$. Going forward, it will be helpful to think of the Galois group $\text{Gal}(K/\mathbb{Q}) \subseteq S_5$ as a permutation group acting on the five roots of $f$.

Now, since $f$ is an irreducible quintic, we can adjoin one of its real roots to $\mathbb{Q}$ to yield a degree-$5$ extension of $\mathbb{Q}$, giving the tower of fields $\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset K$. Applying the multiplicativity formula, $|\text{Gal}(K/\mathbb{Q})| = [K:\mathbb{Q}] = 5\cdot[K:\mathbb{Q}(\alpha)]$, so we see that $5$ divides $|\text{Gal}(K/\mathbb{Q})|$. Therefore, by Cauchy's theorem, there must exist an element in $\text{Gal}(K/\mathbb{Q})$ of order $5$. This element is necessarily a $5$-cycle (easy to see if you think about decomposing the element into disjoint cycles; what's the order of such a decomposition in terms of the lengths of the disjoint cycles?).

Moving on, you've already noted that the polynomial has exactly two complex roots which are necessarily complex conjugates, say $a + bi$ and $a-bi$. There exists a $\phi \in \text{Gal}(K/\mathbb{Q})$, namely complex conjugation, wherein $\phi(a+bi) = a-bi$ and fixes the $3$ real roots. In particular, $\phi$ is a $2$-cycle.

Next, it is a theorem that any $2$-cycle together with any $p$-cycle will generate the entire symmetric group $S_p$ for any prime $p$. From this, we can conclude that $\text{Gal}(K/\mathbb{Q}) \cong S_5$.

All that remains is to show that $S_5$ is not a solvable group. $S_5$ cannot be solvable because $A_5$ is its only normal subgroup, and $A_5$ has no normal subgroups, so we cannot construct a chain $\{e\} = G_0 \subset G_1 \subset \cdots \subset G_n = S_5$ such that each $G_{j-1}$ is normal in $G_j$ and $G_{j}/G_{j-1}$ is abelian. Therefore, we can conclude the roots of $f$ are not solvable by radicals.

Edit: $ \ $ SteveD in the comments below has brought my attention to Jordan's theorem, which states that, if a subgroup $H \leq S_n$ is primitive and contains a $p$-cycle for a prime $p< n \! - \! 2$, then $H \cong S_n \text{ or } A_n$. One can also show that transitive groups of prime order are primitive, which would allow us to conclude $\text{Gal}(K/\mathbb{Q}) \cong S_n \text{ or } A_n$ after demonstrating the existence of a $5$-cycle or of a $2$-cycle in this group. From here, we are able to conclude that the polynomial cannot be solvable by radicals since neither $S_n$ nor $A_n$ is solvable.

Kaj Hansen
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    Your second paragraph is a bit misleading; it is false in general that a transitive subgroup of $S_n$ necessarily contains an $n$-cycle. For example, $V_4$ is a transitive subgroup of $S_4$ not containing a $4$-cycle. (It happens to be true for $n = 5$, though. I'm not sure how general this is.) And in your fourth paragraph I believe you need $p$ to be a prime. – Qiaochu Yuan Jun 18 '14 at 02:49
  • @QiaochuYuan, thanks for pointing that out. What a terrible argument in retrospect. I think I've patched it up. – Kaj Hansen Jun 18 '14 at 02:59
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    @QiaochuYuan: when $p$ is prime then every transitive subgroup $G$ of $S_p$ contains a $p$-cycle: the stabilizer subgroup of, say, 1 will have index $p$ in $G$, so $p$ divides $|G|$ and thus $G$ has an element of order $p$. Being a subgroup of $S_p$, an element of order $p$ is a $p$-cycle because $p$ is prime. – KCd Jun 18 '14 at 03:14
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    @QiaochuYuan: $p$ is not required prime, but there is a condition on the coprimality of the gap between the positions in the transposition and the length of the big cycle. Since 5 is prime, this is automatic here. See Theorem 2.8 at http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf Corollary 2.10 is the restriction you mention. – Eric Towers Jun 18 '14 at 04:15
  • Interesting contribution @EricTowers. It's a shame that a question with this response is not its own thread. – Kaj Hansen Jun 18 '14 at 05:26
  • @EricTowers: There is not enough control on the gap in the $2$-cycle in this type of application to Galois theory in order to use that result in group theory to say that together with an $n$-cycle they generate $S_n$ when $n$ is not prime. – KCd Jun 18 '14 at 23:23
  • I think the key concept here is primitivity. A primitive subgroup that contains a transposition is automatically all of $S_n$ (this is usually called Jordan's theorem). And transitive subgroups of $S_p$ are always primitive. – Steve D Apr 24 '17 at 22:11
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    You're right @SteveD, see the link below. Very cool! I was, until now, unaware of that theorem. That could've shortened my post significantly; I'll make an edit to mention that. https://en.wikipedia.org/wiki/Jordan%27s_theorem_(symmetric_group) – Kaj Hansen Apr 25 '17 at 00:02
  • Yes it is a cool theorem! Note that $f$ being irreducible already implies the Galois group is transitive. Thus showing a transposition is enough to get your end result. This allows you to generalize that any irreducible polynomial of prime degree, with exactly two complex roots, has full Galois group. But I still like your argument better, because it doesn't use such serious machinery. – Steve D Apr 25 '17 at 03:48
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Here's a general theorem which fits your problem perfectly:

If $f$ is an irreducible polynomial of prime degree $p$ with rational coefficients and exactly two non-real roots, then the Galois group of $f$ is the full symmetric group $S_p$. [Wikipedia]

In your case, $p=5$ and $S_5$ is not solvable.

lhf
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Since most polynomials $f$ of degree $n$ have Galois group $G=S_n$, it is desirable to have an easy algorithm to verify this. Here it is: Choose primes $p$ until each of the below three things happen:

  1. The polynomial $f$ factors modulo $p$ into product of linears and one irreducible quadratic. (All the linears must be distinct). This will ensure you have a transposition in the Galois group.
  2. The polynomial $f$ factors modulo $p$ into one linear and one irreducible of degree $n-1$. This will ensure that you have an $n-1$ cycle in the Galois group.
  3. The polynomial $f$ is irreducible modulo $p$. This will ensure that you have an $n$ cycle in the Galois group.

(We assume $n>2$, for $n=2$ the first step suffices, and for $n=1$ there is nothing to be done.) From 3, the Galois group $G$ is transitive, from 2 it is doubly transitive, hence from 1 it contains all transposition and so equals $S_n$.

This will always work if the Galois group is $S_n$ due to chebotarev's theorem; and it is supposed to kick-in pretty fast (that is $p$ is supposed to be relatively small in terms of $n$ and the size of the coefficients) but not a lot is rigorously known on how fast.

This algorithm may be made much more efficient, but the goal was to keep it simple.

Lior B-S
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