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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!

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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 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

1 Answer 1

up vote 10 down vote accepted

Motivation for work below: We want to show that the polynomial has Galois group $S_5$, which is an unsolvable group. To do this, it will suffice to show that the Galois group, when viewed as a permutation group, has a $5$-cycle and a $2$-cycle.

Let $K$ be the splitting field of $f$. It will be helpful to view the Galois group $G(K/\mathbb{Q}) \subseteq S_5$ as a permutation group acting on the five roots of $f$.

Now, given that $f$ is an irreducible quintic, imagine we adjoin one of its real roots to $\mathbb{Q}$ to produce a degree-$5$ extension. This yields the tower of fields $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset K$. Since $|G(K/\mathbb{Q})| = [K:\mathbb{Q}] = 5\cdot[K:\mathbb{Q}[\alpha]]$, then we know that $5$ divides $|G(K/\mathbb{Q})|$. By Cauchy's theorem, there must exist an element in $G(K/\mathbb{Q})$ of order $5$. In other words, $G(K/\mathbb{Q})$ must contain a $5$-cycle.

Moving on, since the polynomial has two complex roots, say $a + bi$ and $a-bi$, then there exists a $\phi \in G(K/\mathbb{Q})$ such that $\phi(a+bi) = a-bi$ 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 all primes $p$. From this, we can conclude that $G(K/\mathbb{Q}) \cong S_5$.

All that remains is to show that $S_5$ is not a solvable group. This is because $A_5$ is the only normal subgroup of $S_5$, and since $A_5$ contains no normal subgroups, then we cannot construct a chain of groups $\{e\} \subset G_1 \subset G_2 \subset ... \subset S_5$ such that each $G_{j-1}$ is normal in $G_j$ and $G_{j}/G_{j-1}$ is abelian.

Since the Galois group of $f$ is not solvable, then we can finally conclude that the roots of $f$ are not solvable by radicals.

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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 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
Interesting contribution @EricTowers. It's a shame that a question with this response is not its own thread. –  Kaj Hansen Jun 18 '14 at 5:26

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