Does the Galois group of a quintic with a single real root contain $\mathbb{A}_5$?

Question

I'm quite stuck on this question from Stewart's Galois Theory (Exercise 15.10): if an irreducible quintic $f(t)\in \mathbb{Q}[t]$ has one real root and two complex conjugate pairs as roots, does the Galois group contain the alternating group $\mathbb{A}_5$?

Let $G$ be the Galois group of $f(t)$. As $f(t)$ is irreducible, $G$ is isomorphic to a subgroup of the symmetric group $\mathbb{S}_5$, and the extension degree of the splitting field over $\mathbb{Q}$ is divisible by 5. We can use Cauchy's Theorem to say that $G$ has a 5-cycle, and we should have disjoint 2-cycles corresponding to the complex conjugate roots, but I'm not sure how to show whether $\mathbb{A}_5$ is contained in $G$ based on this. Based on the answer, can we say that if $G\leq \mathbb{S}_5$ contains $\mathbb{A}_5$, then $G$ is not soluble, as $\mathbb{A}_5$ is not soluble?

Answer 1

Given the quintic polynomial with one root, if its discriminant is a perfect square then its Galois group, G is A5. This is because only even permutations are possible with a discriminant that is a perfect square. The 3-cycles generate A5. For example, consider the following polynomial: x^5+x^4-2x^2-2x-2. There is one real root between x=1 and x=2. The discriminant is 18,496=136^2. G=A5 has the following elements: 20 3-cycles, 24 5-cycles, 15 2x2-cycles and the identity for a total of 60 elements.

1. Conrad K., "Galois Groups as Permutation Groups", A Galois group is shown how it is related to a group of permutations.
2. Conrad K., "Recognizing Galois Groups Sn and An", This paper gives a description of the Galois groups or subgroups of the symmetric group Sn and alternating group An.
3. Conrad K., "Generating Sets", This paper shows how a group can be created with a set of elements.

4. Schwartz A., Pohst M. and Diaz Y Diaz F., "A Table of Quintic Number Fields", Math. of Comp., Vol. 63,No. 207, July 1994, pp. 361-376.

Answer 2

No. For $a \in \Bbb Z$ for which $f(t) := t^5 + a$ is irreducible over $\Bbb Z$, the Galois group $\operatorname{Gal}(f)$ is the Frobenius group $F_5 = \operatorname{Aff}(\Bbb F_5) = \Bbb F_5 \rightthreetimes \Bbb F_5^*$ of order $20$, and in particular it cannot contain $A_5$. (For that matter, $F_5 \not< A_5$.)

If $g$ is a quintic irreducible over $\Bbb Z$ with exactly $1$ real root, complex conjugation exchanges the members of both conjugate pairs of roots, so $\operatorname{Gal}(g)$ contains a double transposition, but there need not be an element of the Galois group that exchanges the members of one conjugate pair but fixes the other complex roots, i.e., a single transposition.

In fact, the double transposition is the square of a $4$-cycle, and $F_5$ is generated by such a $4$-cycle and a $5$-cycle.

Remark If $g(t)$ is a quintic irreducible over $\Bbb Z$ with exactly $3$ real roots (e.g., $t^5 - 3 t + 1$), then complex conjugation does determine a transposition. But since $g$ is irreducible it contains a $5$-cycle, and a $2$-cycle and a $5$-cycle in $S_5$ together generate all of $S_5$, so for such $g$, $\operatorname{Gal}(g) \cong S_5$.