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We consider $$x^5 - x + 5 \in \mathbb{Q}[x].$$

I'm pretty sure that this polynomial is NOT solvable by radicals since its Galois group is isomorphic to $S_5$ (see my question from yesterday).

However, my teacher is pretty sure that it IS solvable by radicals, and he proved it by seeing that its Galois group is isomorphic to $D_5$ (and I think it's not).

He considers the roots of the polynomial $\alpha_1, \alpha_2, \overline{\alpha_2}, \alpha_3, \overline{\alpha_3}$, with $\alpha_1 \in \mathbb{R}, \alpha_2, \alpha_3 \in \mathbb{C}\setminus\mathbb{R}$. These roots form a regular pentagon, and since two pairs of roots are conjugate, we can consider the rotation $a = (1, 2, 3, 4, 5)$ and the permutation $b = (2, 5)(3, 4)$, which generate $$D_5 = \left\langle a, b \mid a^5 = 1 = b^2, ba = a^{-1}b \right\rangle.$$

So then he saw that $D_5$ is solvable by radicals and deduced that the polynomial is solvable by radicals too.

I don't really understand at all what he does, so I am not able to say what is wrong with it… Could you help me? Thanks!

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    The Galois group certainly contains a $5$-cycle, and complex conjugation induces a permutation of consisting of two transpositions, like $(2,5)(3,4)$, but that does not prove that these are the only types of permutations, it proves only that the order of the Galois group is at least $10$. In fact it is the whole of $S_5$. – Derek Holt Dec 14 '23 at 15:19
  • @DerekHolt That's right!!! But how could we prove that there are more types of permutations? –  Dec 14 '23 at 15:39
  • Compare with the answer here. – Dietrich Burde Dec 14 '23 at 15:52
  • @DietrichBurde Thank you so much! –  Dec 14 '23 at 16:08
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    The roots do not form a regular pentagon. A polynomial whose roots form a regular pentagon with one real root and the others complex conjugates would be $$(x-a)^5 - b^5 = x^{5}-5 a ,x^{4}+10 a^{2} x^{3}-10 a^{3} x^{2}+5 a^{4} x -a^{5}-b^{5}$$ – Robert Israel Dec 14 '23 at 16:18
  • @JeanMarie Yes you're right, I'm editing it! –  Dec 14 '23 at 16:26
  • @JeanMarie Oooh thank you!! :D –  Dec 14 '23 at 16:28
  • @RobertIsrael I don't know what theorem is used to justify it but if you write the polynomial in Wolfram Alpha then you'll see that the roots form a regular pentagon. –  Dec 14 '23 at 16:32
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    Reducing modulo 2 gives a product of degree 2 and 3. By a theorem of Dedekind, the Galois group thus must contain an element that is a produict of a 2-cycle and 3-cycle. – ahulpke Dec 14 '23 at 17:58
  • @ahulpke ... under the condition that the discriminant of the polynomial is not dividable by 2 (= odd) which is the case. – Jean Marie Dec 15 '23 at 00:25
  • Another case where one may jump to incorrect conclusions about the shape of the set of roots, if one doesn't do the arithmetic carefully enough (as in Jean Marie's nice answer) :-) – Jyrki Lahtonen Dec 15 '23 at 04:30
  • @JeanMarie About ahulpke's argument: The discriminant condition is not necessary, when, after reduction modulo a prime, there are no repeated factors. I think those two conditions are actually equivalent, but need to check :-) – Jyrki Lahtonen Dec 15 '23 at 04:32

1 Answers1

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No, the pentagon formed by the roots of

$$x^5-x+5=0 \tag{1}$$

is neither regular nor centered in $0$ (one can be fooled by the fact that the axes haven't necessarily the same length if it is not explicitly asked).

enter image description here

One can check it on the approximate values of the roots, which are :

$$\begin{cases}r_1&\approx&-1.4519\\r_2&\approx&-0.36823+1.3592i\\r_3&=&\overline{r_2}\\r_4&\approx&1.0942 + 0.73443i\\r_5&=&\overline{r_4}\end{cases},$$

giving very different sidelengths, for example $\approx 1.73233$ between $r_1$ and $r_2$, and $\approx 2.18836$ between $r_4$ and $\overline{r_4}$.

It is not centered in $0$ because the norms of the $r_k, k=1,\cdots 5$ are :

$$\approx 1.4519, 1.4082, 1.4082, 1.3178, 1.3178$$

Edit : A rigorous proof, by contradiction, can be given. Let us assume that the pentagon is regular and centered in the origin : this would mean that the roots, now with indices $k=0,\cdots 4$ would be :

$$r_k=re^{ik \tfrac{2 \pi}{5}}$$

where $r=r_0\ne 0$ is the unique real root.

Their product, using one of the Viète's formulas, would give rise to :

$$r^5\underbrace{e^{i (0+1+2+3+4) \tfrac{2\pi}{5}}}_{= \ 1}=-5\tag{2}$$

Plugging (2) into the following relationship (true because $r$ is a root of equation (1)) :

$$r^5-r+5=0$$

leads to $r=0$ which is impossible.

As I know you are interested by Sage, here is the little program which has produced in particular the graphical representation :

CC=ComplexField(20)
p=x^5-x+5;
s=p.roots(ring=CC);
L=[];M=[];
for k in range(len(s)):
    m=s[k][0];
    L.append([m.real(),m.imag()]);
    M.append(abs(m));
print(L);
print(M);
Q=points(L,size=50);
show(Q,aspect_ratio=1);
Jean Marie
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  • Thank you again! It is very interesting to me that the roots do not form a regular pentagon, but how could we show it without approximations? Is there a way? –  Dec 14 '23 at 20:06
  • I just added a rigorous proof under the form of an Edit to my answer. – Jean Marie Dec 14 '23 at 21:51
  • And for the case of a regular pentagon not necessarily centred at $0$, see my comment to the OP. – Robert Israel Dec 15 '23 at 02:48
  • @JeanMarie You're amazing! Thank you!! –  Dec 15 '23 at 06:19
  • @RobertIsrael Thanks!! Do you know where can I find a proof of that result? I'm really interested. –  Dec 15 '23 at 06:20
  • @galoisman The roots of $z^n - 1$ are by definition the $n$'th roots of unity, which are $\exp(2 \pi i k/n)$ for $k = 0, 1, \ldots, n-1$ and form the vertices of a regular $n$-gon centred at $0$. Scale and translate... – Robert Israel Dec 15 '23 at 15:04
  • @RobertIsrael But why is $(x - a)^5 - b^5$ the unique polynomial with that property? By the way, what are $a, b$ exactly? –  Dec 15 '23 at 17:38
  • A monic polynomial is uniquely determined by its roots. – Robert Israel Dec 17 '23 at 05:18