I've been asked to determine the Galois Group of $x^5+ 5x^3 + 5x + 1$. This is what I know so far.

1) The polynomial is irreducible.

2) Its discriminant is $78125=5^7$

Since the discriminant is not a square, the Galois group should be $S_5$ or $F_{20}$. How can I know? In case it was $F_{20}$, is it solvable?

Thanks in advance

  • $\begingroup$ What methods have you learned for computing Galois groups already? (By the way, a computer algebra system will quickly tell you what the answer is, but that is not the same as figuring out an argument that proves the result independently of computers.) $\endgroup$ – KCd Jan 10 '14 at 22:50
  • $\begingroup$ I know the general strategies to compute Galois groups of cubic and quartic polynomials, and its definition as the group of Q-endomorphisms of the splitting field. Nothing for quintics, though $\endgroup$ – Pesi442 Jan 10 '14 at 22:55
  • $\begingroup$ IF it has exactly two nonreal roots, then the group is $S_5$, so first I'd check to see how many nonreal roots it has. That much is an exercise in 1st year Calculus. $\endgroup$ – Gerry Myerson Jan 10 '14 at 22:57
  • $\begingroup$ Already checked, it has 4 non real roots. $\endgroup$ – Pesi442 Jan 10 '14 at 22:59
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    $\begingroup$ Do you know Dedekind's theorem about the factorization of $f(x) \bmod p$ into irreducibles implying the existence of particular cycle types in the Galois group, e.g., mod 2 the polynomial is a linear times a quartic, so the Galois group over $\mathbf Q$ has an automorphism that acts on the roots as a permutation of cycle type $(1,4)$? $\endgroup$ – KCd Jan 10 '14 at 23:02

The paper Solving Solvable Quintics - D. S. Dummit shows the construction of a resolvent $f_{20}$ for the quintic such that $f_{20}$ has a rational root iff the Galois group is contained in $F_{20}$.

Let $x_i$ be roots of the general quintic.

Let $$\theta = x_{0} x_{1}^{2} x_{2} + x_{0}^{2} x_{2} x_{3} + x_{1} x_{2}^{2} x_{3} + x_{0} x_{1} x_{3}^{2} + x_{0}^{2} x_{1} x_{4} + x_{0} x_{2}^{2} x_{4} + x_{1}^{2} x_{3} x_{4} + x_{2} x_{3}^{2} x_{4} + x_{1} x_{2} x_{4}^{2} + x_{0} x_{3} x_{4}^{2}$$

Then we define $f_{20}(x) = \prod_{\sigma} (x - \sigma \theta)$ where the permutations $\sigma \in S_5$ act by permuting the variables $x_{i}$ of $\theta$. The product is taken over the whole orbit. This orbit has size 6. Note that $5!/6 = 20$.

Now $f_{20}$ is invariant under conjugation therefore it lies in the base field. Using symmetic polynomial calculations you may recover (the very long) formula (2) from the paper. The resolvent for our polynomial in particular is:

$$f_{20}(x) = x^{6} + 40 \, x^{5} + 250 \, x^{4} - 4375 \, x^{3} - 21875 \, x^{2} + 25000 \, x - 1187500$$

and $(x-10)|f_{20}(x)$ so combining this with your observation that the discriminant is nonsquare we find that the Galois group of this polynomial is $F_{20}$.

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    $\begingroup$ Nice to have this explained. Hope it won't offend you if I tell that I like Oscar Lanzi's take even more. $\endgroup$ – Jyrki Lahtonen Jul 20 '20 at 11:31
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    $\begingroup$ Not at all, I am very grateful that it was posted so I can see such an efficient solution! $\endgroup$ – river Jul 20 '20 at 11:37
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    $\begingroup$ One way to seek the rational root more efficiently: render $x=5u$. Then all terms are divisible by $5^6=15625$ and a rational root for $u$ would have to divide $1187500/15625=76$. $\endgroup$ – Oscar Lanzi Jul 20 '20 at 12:38

There is no doubt that this is solvable. Render $x=r-1/r$ and show that $r^5-1/r^5=-1$. Thus $r^5=(-1\pm\sqrt5)/2$, etc, and the Galois group can't be larger than $F_{20}$.

  • $\begingroup$ This is a fantastic solution! How would one come up with substitutions like $r - 1/r$? $\endgroup$ – river Jul 20 '20 at 10:23
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    $\begingroup$ I happened to spot it from attempting to find geometric fifth root or angle wuintisection solutions in the past (that failed). If we put $x=2\sinh u$ into the equation the terms containing $u$ match with the five-times argument formula for hyperbolic sine, thus $\sinh 5u=-1/2$. This then implies that $2\sinh u =r-1/r$ implies $\sinh 5u=(-1/2)$. $\endgroup$ – Oscar Lanzi Jul 20 '20 at 10:27

We can use the Magma online calculator to find the Galois group of the equation $f(x)= x^5 + 5 x^3 + 5 x + 1$ in the field of rational numbers.

> P<x> := PolynomialRing(IntegerRing()); 
> f :=  x^5 + 5 * x^3 + 5 * x + 1; 
>G:=GaloisGroup (f);
>IsCyclic (G);
>IsSolvable (G);

We can get that the Galois group of this equation in the rational number field has order 20 and is a solvable group:

enter image description here

This result is for reference only, and cannot replace theoretical analysis.


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