# Proving that a polynomial is not solvable by radicals.

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!

• 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 1:09
• Do you mean to say that by Eisenstein this is IRREDUCIBLE? – Vladhagen Jun 18 '14 at 1:09
• One or more of the techniques in math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf might be helpufl. – Qiaochu Yuan Jun 18 '14 at 1:12
• 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 3:22
• How did you see it has exactly 3 real roots? – user428487 Nov 22 '20 at 7:35

## 3 Answers

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.

• 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 2:49
• @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 2:59
• @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 3:14
• @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 math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf Corollary 2.10 is the restriction you mention. – Eric Towers Jun 18 '14 at 4:15
• 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. en.wikipedia.org/wiki/Jordan%27s_theorem_(symmetric_group) – Kaj Hansen Apr 25 '17 at 0:02

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.

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.