Radical Solvable quintic polynomial.

Question

I am trying to solve the following exercise:

Let $k$ be a field of characteristic 0, and let $f(x)\in k[x]$ be a polynomial of degree 5 with splitting field $E/k$. Prove that $f(x)$ is solvable by radicals if and only if $[E:k]<60$.

What I already know:

1. Since $\operatorname{char}k=0$, it follows that $f(x)$ is solvable by radicals if and only if $G=\text{Gal}(E/k)$ is solvable.
2. Since $\operatorname{char}k=0$, every irreducible polynomial in $k[x]$ is separable, so that $f(x)$ is also separable. Therefore, $E/k$ is Galois. and $|G|=[E:k]$.
3. $|A_5|=60$
4. $\text{Gal}(E/k)$ can be embedded in $S_5$.
5. $A_5$ is simple and thus $S_5$ is not solvable.

Any help? I've thought about it for days. BTW: I have not learned discriminant yet. I only know the Fundamental Theorem of Galois Theory and other basic Galois Theory.

Answer 1

Let us add the restriction that $f$ is irreducible, since that is the hardest case to consider.

The forward direction is simple: if $[E:K] \geq 60$, then Lagrange's theorem would imply that $Aut(K/\mathbb{Q}) \cong A_5$ or $S_5$, which is no good as you've noted. So let's focus on the converse.

If $[E:K] < 60$, then $Aut(K/\mathbb{Q})$ is a subgroup$S_5$, and in particular, Lagrange's theorem would require $|Aut(K/\mathbb{Q})|$ to be a divisor of $120$. So the automorphism group has possible orders $2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40$. This may seem like a lot, so let's eliminate some.

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 $|Aut(K/\mathbb{Q})| = [K:\mathbb{Q}] = 5\cdot[K:\mathbb{Q}[\alpha]]$, then we know that $5$ divides $|Aut(K/\mathbb{Q})|$. Therefore, we eliminate a bunch of possiblities from our original list, and the remaining possibilities are $5, 10, 15, 20, 30, 40$.

Next, all abelian groups are solvable. Can you prove that all groups of order $5$ and of order $15$ are cyclic (thus abelian)?

Finally, all we have left are groups of order $10$, $20$, $30$, and $40$. For each case, you can use Sylow theory to arrive at the appropriate chain of subgroups you need to show solvability.

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Obviously, you will need to add some stuff to this argument to finish up. For one, what if $f$ were reducible?