Let $f(X)$ be an irreducible polynomial of degree 5 with coefficents in the field of rational numbers $\mathbb{Q}$. Assume that $f$ has at least one non-real root in the complex field $\mathbb{C}$. Assume further that the discriminant of $f$ is a square in $\mathbb{Q}$.

Let $r$ be a root of $f$, and let $K$ be the field $\mathbb{Q}(r)$, so that $f$ factors in $K[X]$ as $$f = (X − r)g$$ with $g$ of degree 4. Prove that $f$ is solvable by radicals if and only if $g$ is reducible in $K[X]$.

My thoughts:

I know that since the discriminant of $f$ is a square in $\mathbb{Q}$, $G\subset A_5,$ and since $f$ is irreducible, $G$ must be transitive. So we must have that either $G\cong A_5$ or $G\cong D_5$ (the dihedral group of order 10). Moreover, since char $\mathbb{Q}=0,$ we know that $f$ is solvable by radicals if and only if $G$ is solvable.

Since $A_5$ is simple and non-abelian, $A_5$ is not solvable. On the other hand, $D_5$ is solvable: $D_5$ has an element $r$ with $r^5=e$, and $[D_5 : \left<r\right>]=2,$ so $\left<r\right>\triangleleft D_5.$ Now

$$D_5\triangleright\left<r\right>\triangleright\{e\}.$$ So it suffices to show that $g$ is reducible in $K[X]$ if and only if $G\cong D_5.$

If $f$ has exactly 2 non-real roots, then the only nontrivial $\mathbb{Q}$-automorphism is complex conjugation, so $G$ contains a transposition, which implies $G\cong S_5$. Thus, $f$ must have exactly one real root and 4 non-real roots.

This is where I'm stuck. How can we relate the reducibility of $g$ in $K[X]$ to the Galois group of $f$?

  • $\begingroup$ If the only nontrivial automorphism is conjugation then the Galois group has order $2$ and certainly cannot be $S_5$. $\endgroup$ – Matt Samuel Jun 26 '16 at 22:11
  • $\begingroup$ Perhaps you mean that complex conjugation is the only nontrivial automorphism of $K$ that fixes its real subfields? $\endgroup$ – Servaes Jun 26 '16 at 22:22

HINT: The subgroup of $G$ fixing $r$ acts transitively on the remaining roots iff $g$ is irreducible.

  • $\begingroup$ I know if $g$ were an irreducible quadratic on its own, then it's Galois group would be isomorphic to a transitive subgroup of $S_4$, none of which are subgroups of $D_5$. How do I account for the fact that the subgroup is fixing $r$? $\endgroup$ – user346096 Jun 27 '16 at 22:13
  • $\begingroup$ I'm not sure what you're asking; you seem to have proved one implication, and your argument can be reversed. If $G\cong A_5$ then for any root, the subgroup fixing that root acts transitively on the remaining roots. This implies $g$ is irreducible. $\endgroup$ – Servaes Jun 29 '16 at 11:44

Another way to think the problem is the following.

First, show that $A_5$ cannot have a subgroup of size $20.(1)$

If such subgroup $H$ existed, $H$ would be a subgroup of $A_5$ of index $3$, which cannot be normal, why would this imply $A_5$ to have a normal subgroup of index $2$?; consider the action of $G$ on the cosets of $H$.

Let $F$ be the splitting field of the polynomial $f$. Then $[F:\Bbb Q]=|Gal(F/\Bbb Q)|$.$(2)$

If $g$ is irreducible in $K[X]$, we get $20\mid[F:\Bbb Q]$. By $(2)$ we have $[F:\Bbb Q]\mid |A_5|$, because of $(1)$ we must have that $Gal(F/\Bbb Q)=A_5$(why?), hence $f$ is not soluble by radicals.

If $g$ is reducible in $K[X]$, there are two cases:

  • Suppose $g=h_1h_2$, where $h_1$ and $h_2$ are irreducible over $K[X]$ both of degree $2$. Then we must have that either $[F:\Bbb Q]=20$ or $[F:\Bbb Q]=10$(why?), by what we saw above the first case cannot happen, so that $|Gal(F/\Bbb Q)|=10$, and any group of this size must be soluble; use Sylow's third theorem.
  • If $g=h_1h_2$, with $h_1$ irreducible of degree $3$, let $\gamma$ be a root of $h_1$. We claim that $F=K[\gamma]$. Otherwise, we'd get $[F:\Bbb Q]=30$(why?), so that $|Gal(F/\Bbb Q)|=30$, however as $A_5$ is simple of size $60$, it cannot have subgroups of size $30$. Thus $[F:\Bbb Q]=15$, i.e., $|Gal(F/\Bbb Q)|=15$, hence $Gal(F/\Bbb Q)$ is soluble.
  • $\begingroup$ If $g$ is irreducible in $K[X]$, the do we get $20\,|\, [F:\mathbb{Q}]$? $\endgroup$ – user346096 Aug 8 '16 at 19:36
  • $\begingroup$ @user346096 As $f$ is irreducible over $\Bbb Q$ we have $[K:\Bbb Q]=5$. Now if $ \alpha$ is a root of $g$, then if $g$ is irreducible $[K(\alpha):K]=4$, thus $[K(\alpha):\Bbb Q]=20$, but $[K(\alpha):\Bbb Q]\mid [F:\Bbb Q]$. $\endgroup$ – Camilo Arosemena-Serrato Aug 8 '16 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.