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I am examining the roots of the equation in $x$, $\sum_{q=0}^{2k-1} (-1)^{q} {2k+1 \choose q+1} x^{2k-q} m^{q}+r=0$ where $m$ and $r$ are positive integers.

I want to know whether the roots of this can always be fully expressed in radicals (which I suspect that it does), and if so, what can be said regarding the discriminants of the roots. I need as much info about these discriminants as possible. Does anyone know how to properly go about this?

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    $\begingroup$ I guess it is common to say "solution" to an equation, vs "zeros/roots" of a polynomial. $\endgroup$
    – YCor
    May 30 at 15:10

2 Answers 2

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Upon dividing the given polynomial by $m^{2k}$ and replacing $x$ with $mx+m$, it assumes the simpler form \begin{equation} s+(x+1)^{2k+1}-x^{2k+1} \end{equation} for some $s$.

Generically its Galois group is the wreath product $C_2^k\rtimes S_k$. This follow from Hilbert's irreducibility theorem together with the fact that its Galois group over $\mathbb C(s)$, where we consider $s$ as a transcendental over the complex numbers, is $C_2^k\rtimes S_k$, so in general the roots cannot be expressed in radicals once $k\ge5$.

Write $n=2k$, set $f(x)=(x+1)^{n+1}-x^{n+1}$ and consider the branched cover $a\mapsto f(a)$. First note that $f(x-1/2)$ is an even function, so the monodromy group $G$ of this cover is a subgroup of $C_2^k\rtimes S_k$.

The critical points of the cover are the roots of $f'(x)$. Let $\gamma$ be such a root, so $(\gamma+1)^n-\gamma^n=0$ and therefore $f(\gamma)=\gamma^n$. The possibilities for $\gamma$ are $\gamma=\frac{1}{\zeta-1}$, where $\zeta$ is an $n$-th root of unity different from $1$. Note that $-\gamma$ corresponds to $1/\zeta$. Looking at absolute values we see that each $f(\gamma_1)=f(\gamma_2)$ if and only if $\gamma_1$ and $\gamma_2$ are complex conjugate. Note that $\gamma=-1/2$ corresponds to $\zeta=-1$. (Recall that $n$ is even.) So we see that the finite generators of the monodromy group of the cover is one transposition and $(n-2)/2$ double transpositions. The double transpositions induce transpositions on the blocks, so the action on the blocks is $S_k$. The transposition fixes all blocks and flips the two elements in one block. This together yields the claimed structure of the group.

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  • $\begingroup$ This is probably a stupid question but wouldn't $s$ just be an integer, so that it is not reasonable to pretend it is an indeterminate/transcendental over $\Bbb C$? $\endgroup$
    – FShrike
    May 31 at 10:49
  • $\begingroup$ @FShrike Let $f(x)$ be as above and $s$ be a transcendental over $\mathbb C$. By Hilbert's Irreducibility Theorem, $\text{Gal}(f(X)-s_0, \mathbb Q)=\text{Gal}(f(X)-s, \mathbb Q(s))$ for "most" rational $s_0$. Furthermore, $\text{Gal}(f(X)-s, \mathbb C(s))\le\text{Gal}(f(X)-s, \mathbb Q(s))$. $\endgroup$
    – Peter Mueller
    May 31 at 11:06
  • $\begingroup$ (I know little, am just a student) so if I understand correctly: you know $\operatorname{Gal}(f(X)-s,\Bbb C(s))\cong C_2^k\rtimes S_k=:K$ for transcendent $s$ and you conclude $\operatorname{Gal}(f(X)-s,\Bbb Q(s))$ contains a copy of $K$. For “most” rational $s_0$ you also know: $G:=\operatorname{Gal}(s_0+X^{n+1}-X^n,\Bbb Q)\cong\operatorname{Gal}(f(X)-s,\Bbb Q(s))$ so you conclude that the group of interest, $G$, contains $K$ most of the time and is consequently not solvable. But for the OP’s specific polynomial that doesn’t prove it because the integer $s_0$ might not be included in ‘most’. $\endgroup$
    – FShrike
    May 31 at 11:34
  • $\begingroup$ I hope that’s right $\endgroup$
    – FShrike
    May 31 at 11:34
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    $\begingroup$ @FShrike Let $f_{m,r}(x)$ be the polynomial in the question, and $t$ be a transcendental over $\mathbb C$. My argument shows the following: Given any rational $m\ne0$, then the Galois group of $f_{m,t}$ over $\mathbb Q(t)$ is not solvable if $k\ge5$. Now by HIT, this Galois group is the same as the one of $f_{m,r}$ over $\mathbb Q$ for infinitely many integers $r$. $\endgroup$
    – Peter Mueller
    May 31 at 15:02
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Q: Can the roots of this sum always be fully expressed in radicals?

A: No. I note that the sum can be carried in closed form: $$r+\sum_{q=0}^{2k-1} (-1)^{q} {2k+1 \choose q+1} x^{2k-q} m^{q}=$$ $$\qquad=r-(-1)^{2 k} m^{2 k}+m^{-1}x^{2 k} \left(m \left(1-\frac{m}{x}\right)^{2 k}-x \left(1-\frac{m}{x}\right)^{2 k}+x\right).$$ The roots are radicals for $k\leq 4$, but for $k=5$, for example, the roots involve the solution of a quintic equation.

Specific example, $r=1,m=1,k=5$:

$$1 - 11 x + 55 x^2 - 165 x^3 + 330 x^4 - 462 x^5 + 462 x^6 - 330 x^7 + 165 x^8 - 55 x^9 + 11 x^{10}$$ does not seem to have roots that can be expressed as radicals, it needs the roots of $1+11x+44x^2+77x^3+55x^4+11x^5$.

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    $\begingroup$ According to Maple, the Galois group of $1+11x+44x^2+77x^3+55x^4+11x^5$ is cyclic of order 5. $\endgroup$
    – Richard Stanley
    May 30 at 15:23

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