Galois Theory HT25, Quintic equations

By Eisenstein’s Criterion at $p = 3$ together with Gauss’ lemma, $f$ is irreducible over $\mathbb Q$ and it is also separable over $\mathbb Q$. Therefore it has 5 distinct roots in a splitting field $K$.

Since $\mathbb C$ is algebraically closed, we will identify $K$ with a subfield of $\mathbb C$ and will identify

\[G = \text{Gal} _ {\mathbb Q}(f) = \text{Gal}(K/\mathbb Q)\]

with a subgroup of $S _ 5$.

Since:

• $f(-2) = -17 < 0$
• $f(0) = 3 > 0$
• $f(1) = -2 < 0$
• $f(2) = 23 > 0$

It follows that $f$ has a real root in each of $(-2, 0)$, $(0, 1)$ and $(1, 2)$.

Suppose that $f$ has five real roots. Then by the Mean Value Theorem,

\[f' = 5t^4 - 6\]

would have at least four real roots, which is not the case.

Hence $f$ has precisely three real roots. So complex conjugation $c : \mathbb C \to \mathbb C$ preserves $K$ and hence $c \vert _ K \in G$ and $c \vert _ K$ is a transposition: $c$ fixes the three real roots and swaps the two non-real roots of $f$.

As $f$ is irreducible over $\mathbb Q$, $[\mathbb Q(z) : \mathbb Q] = 5$ for any $z \in V(f)$. Hence $5$ divides $[K : \mathbb Q]$, which is equal to $ \vert G \vert $.

By Cauchy’s Theorem, $G$ contains an element $\sigma$ of order $5$ (Cauchy’s theorem says that if $p$ is a prime dividing the order of the group, then the group contains an element of order $p$). Then $\sigma$ is necessarily a $5$-cycle (consider the ways you can shuffle $5$ elements around).

Since $S _ 5$ is generated by a transposition and a $5$-cycle, it follows that $G = S _ 5$.

But $S _ 5$ is not solvable, so no roots of $f$ can lie in a radical extension of $\mathbb Q$.