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.

  • 1
    $\begingroup$ I think the simplest approach would be to consider the equivalent (I think) question about the Galois group: show that a subgroup of $S_5$ is solvable iff its order is less than $60$. $\endgroup$ – Jack M Jul 11 '14 at 0:16
  • $\begingroup$ @JackM I already noticed that and list it in "What I already know", but I really have no idea how to prove it. $\endgroup$ – YYF Jul 11 '14 at 0:18

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.

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

  • $\begingroup$ I know that Sylow theory is really powerful, but I have not learned it yet. So, do you have a way to bypass it? $\endgroup$ – YYF Jul 11 '14 at 1:11
  • $\begingroup$ Hmmm. Yuck. Showing normality would be a pain without it. Cauchy's theorem could get you somewhere, especially for the order $10$ case. Also, you can do a little bit more chiseling by proving, for example, that $S_5$ contains no subgroup of order $40$. $\endgroup$ – Kaj Hansen Jul 11 '14 at 1:49
  • $\begingroup$ Nor does it have elements of order 15, so it won't have subgroups of order 15 either (as you observed those are necessarily cyclic). $\endgroup$ – Jyrki Lahtonen Jul 11 '14 at 8:43

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