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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.

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

How can I proceed?

Thank you!

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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, which serves 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, the polynomial has 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 all primes $p$. 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 the only normal subgroup of $S_5$ and because $A_5$ contains no normal subgroups, so we cannot construct a chain of groups $\{e\} \subset G_1 \subset G_2 \subset \cdots \subset G_n = S_5$ 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 $A_n$ are solvable.

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.

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