# Can sextic resolvent of an irreducible quintic have repeated root?

Consider formal variables $x_1,\cdots,x_5$. Denote $$\theta_1 = x_1^2 x_2 x_5 + x_1^2 x_3 x_4 + x_2^2 x_1 x_3 + x_2^2 x_4 x_5 + x_3^2 x_1 x_5 + x_3^2 x_2 x_4 + x_4^2 x_1 x_2 + x_4^2 x_3 x_5 + x_5^2 x_1 x_4 + x_5^2 x_2 x_3$$

The stabilizer of $\theta_1$ under the action of $S_5$ is a group $M$, which has order $20$ (isomorphic to the general affine group over $\mathbb{F}_5$). The orbit consists of six elements, denote them by $\{\theta_1, \cdots, \theta_6\}$. Any conjugates of $M$ in $S_5$ is the stabilizer of some $\theta_i$ (because $M$ is self-normalizing in $S_5$).

When $x_i$ are subsituted as roots of an irreducible quintic $f$ (say over a field $k$ with characteristic $0$), denote the polynomial $f_{20}(x):=(x-\theta_1)\cdots(x-\theta_6)$, evidently it is in $k[x]$.

In this well-known paper, it is stated that $f(x)$ is solvable by radical iff $f_{20}(x)$ has a root in $k$.

Assuming $f_{20}(x)$ has distinct roots, this equivalence is easy to establish.

Indeed, if Galois group is $A_5$ or $S_5$, since they act transitively on $\{\theta_1, \cdots, \theta_6\}$, if one roots is in $k$, then all roots are the same, contradiction. On the other hand, if Galois group is conjugate to $M, D_5$ or $C_5$, then it is contained in the stabilizer of some $\theta_i$, this $\theta_i$ must be in $k$.

Now there is a subtleties lurking in the background, what if $f_{20}(x)$ has repeated root? In this case, one implication still holds: that is

• Galois group is $D_5, C_5$ or $M$ $\implies$ $f_{20}(x)$ has a root in $k$

However, I am troubled in establishing the reverse. If Galois group is $A_5$ or $S_5$, then either $f_{20}(x)$ has no root in $k$, or it has and all roots are the same. I could not remove the latter possibility. So my question is

Given an irreducible quintic, can $f_{20}(x)$ have repeated root? What if we impose an additional assumption that the Galois group of $f$ is $A_5$ or $S_5$?

I tried to factorize the difference $\theta_i - \theta_j$, but it turn out cannot be factorized. The paper I mentioned above completely ignores this issue. Maybe this is really something trivial, but I cannot figure this out.

Thank you for your time.

• If the Galois group contains $A_5$, and $f_{20}$ has repeated roots, then all roots of $f_{20}$ are identical. You may be able to find some other linear combination of the $\theta_i$ that must vanish and is easier to factor? – Lord Shark the Unknown Mar 5 '18 at 11:27

Lemma If a degree five polynomial $f(x)$ over $K$ of characteristic zero has Galois group containing $A_5$, then the sextic resolvent is separable.

Proof Suppose otherwise. Since $A_5$ will act transitively on the $\theta_i$, the only way the $\theta_i$ can not all be distinct is if they are all equal and fixed by $G$. In particular, we may assume that $\theta:=\theta_1 \in K$. Since $G$ contains $A_5$ which contains $(12)(34)$, we find that $$0 = (12)(34) \theta - \theta = (x_2 - x_3)(x_1 - x_4)(-x_2 x_3 + x_1 x_4 - x_1 x_5 + x_2 x_5 + x_3 x_5 - x_4 x_5).$$ Irreducible polynomials over characteristic zero fields are separable, and so the roots $x_i$ are distinct. Hence, if $$\psi := -x_2 x_3 + x_1 x_4 - x_1 x_5 + x_2 x_5 + x_3 x_5 - x_4 x_5,$$ then $\psi = 0$. Since $A_5 \subset G$ contains $(123)$, we similarly have $$0 = (123) \psi - \psi = (x_1 - x_2)(x_3 + x_4 - 2 x_5),$$ so, as before, $\phi:=x_3 + x_4 - 2 x_5 = 0$. Finally, $$0 = (12)(35)\psi - \psi = 3(x_5 - x_3),$$ contradicting the fact the $x_i$ are distinct, and we are done. $\square$

We also have:

Lemma If a degree five polynomial $f(x)$ over $K$ of characteristic zero is irreducible, then the sextic resolvent is separable.

Proof We have $5 | |G|$ by the Orbit-Stabilizer theorem. Without loss of generality, we may assume $g = (1,2,3,4,5) \in G$. The element $g$ sends (after making a choice of ordering): $$\theta_2 \mapsto \theta_3 \mapsto \theta_4 \mapsto \theta_5 \mapsto \theta_6 \mapsto \theta_2.$$ It follows easily that the sextic resolvent can only have repeated roots if $$\theta_2 = \theta_3 = \theta_4 = \theta_5 = \theta_6.$$ Let us fix our choices by saying that $\theta_2 = (12) \theta_1$. We find that $$(x_5 - x_3)(x_1 - x_4)(\theta_2 - \theta_3) - (x_5 - x_1)(x_3 - x_4)(\theta_6 - \theta_4)$$ $$= (x_1 + x_3 - 2 x_2)(x_1 - x_4)(x_4 - x_3)(x_1 - x_5)(x_3 - x_5)(x_4 - x_5).$$ In particular, since $x_i \ne x_j$ for $i \ne j$, we have that $$x_1 + x_3 = 2 x_2,$$ and by applying $g$ we find that $x_2 + x_4 = 2 x_3$, $x_3 + x_5 = 2 x_4$, $x_4 + x_1 = 2 x_5$, and $x_5 + x_2 = 2 x_1$. These together imply that all the $x_i$ are equal, a contradiction. $\square$

It's easy to produce examples of reducible polynomials with non-separable sextic resolvent, for example by taking all the roots $x_i$ to be equal. However:

Remark If a degree five polynomial $f(x)$ over $K$ of characteristic zero is separable, then the sextic resolvent need not be separable.

For example, the polynomial $x^5+x^4+x^3+x^2+x$ is separable with Galois group $\mathbf{Z}/4 \mathbf{Z}$, but the resolvent sextic is $$(x+1)^2 (x^4 + 4 x^3 + 16 x^2 + 49 x + 61).$$ More generally, $x^5 + p x^4 + q x^3 + r x^2 + (r/p)^2 x$ has resolvent sextic divisible by

$$\left( x - \frac{(pq-2r)r}{p^2} \right)^2$$

The general condition for the sextic resolvent of $x^5 + p x^4 + q x^3 + r x^2 + s x + t$ to have repeated roots is a degree $30$ polynomial on $p$, $q$, $r$, $s$, and $t$, where the implied degrees of these terms are $1$, $2$, $3$, $4$, and $5$ respectively. I omit it, but note that it has exactly $335$ non-zero terms.

• Very nice solution! Your method seems can be generalized to any resolvent of any irreducible degree $n$ polynomial. If the roots of resolvent are distinct, then we get some information on the Galois group. – pisco Mar 9 '18 at 1:30