Examining the Galois group of an quintic irreducible polynomial over $\Bbb Q$

Question

Suppose $f$ is a monic irreducible quintic polynomial in $\Bbb Q[x]$ (necessarily separable), and let $F$ be the splitting field of $f$ over $\Bbb Q$ in $\Bbb C$. Let $u_1,\dots,u_5$ denote the distinct roots of $f$ in $F$. Suppose moreover that the following holds:

• (a) $F=\Bbb Q(u_1,u_2)$,
• (b) $u_i\notin \Bbb Q(u_1)$ for $i>1$,
• (c) $u_2,u_3,u_4,u_5$ are conjugates over $\Bbb Q(u_1)$, and
• (d) $u_i-u_j$ are pairwise distinct for all choices of $i,j$.

I want to show that there exists a $\Bbb Q$-automorphism $\sigma$ of $F$ such that $\sigma(u_1)=u_3$ and $\sigma(u_2)=u_5$, and compute the degree of $u_1+u_2$ over $\Bbb Q$.

It is obvious that $[\Bbb Q(u_1):\Bbb Q]=\deg f=5$. (c) implies that the degree of $u_2$ over $\Bbb Q(u_1)$ is $4$, so (b) implies that $[F:\Bbb Q]=20$. I got stuck here. Should I have to examine transitive subgroups of $S_5$ of order $20$?

Answer 1

Good job figuring out that $[F:\Bbb{Q}]=20$. Let's write $G=Gal(F/\Bbb{Q})$ and $K=\Bbb{Q}(u_1)$, $H=Gal(F/K)$.

Roadmap/extended hint:

• How many elements are there in the subgroup $H$?
• For all $\sigma\in G$ why is it that $\sigma(u_1)=u_1$ if and only if $\sigma\in H$?
• As $f$ is irreducible, $G$ acts on the roots transitively, so there exists an automorphism $\alpha\in G$ such that $\alpha(u_1)=u_3$. Let $S$ be the set of such automorphisms $\alpha$. How many elements are there in $S$? Hint: if $\alpha_1,\alpha_2\in S$ then where will $\alpha_1^{-1}\alpha_2$ map $u_1$?
• If $\alpha_1,\alpha_2\in S, \alpha_1\neq\alpha_2$, why does it follow that $\alpha_1(u_2)\neq\alpha_2(u_2)$?
• List the possibilities of $\alpha(u_2), \alpha\in S$.
• Is it possible that $u_i+u_j=u_k+u_\ell$ non-trivially?
• How many distinct numbers $u_i+u_j, i\neq j$ are there?
• Why are they all Galois conjugates of each other (= in the same $G$-orbit)?