5

Consider $f = x^5 -11x + 1 \in \mathbb{Q}[x]$. I want to prove that its not solvable by radicals. I know that its solvable by radicals iff its galois group is solvable. My attempt was first to use the following:

For any prime $p$ not dividing the discriminant of $f \in \mathbb{Z}[x]$ , the galois group of $f$ over $\mathbb{Q}$ contains an element with cyclic decomposition $(n_1,..,n_k)$ where $n_1,...,n_k$ are the degrees of the irreducible factors of $f$ reduced mod $p$.

Then, I could use this to determine the galois group of $f$. However, the discriminant proved to be super hard to calculate (wolfram alpha works but its not intended to be used). So I am thinking that I got the wrong approach here. Any other hints?

Robin
  • 3,227
LocationMap2
  • 169

3 Answers3

4

It is true that this polynomial has Galois group $S_5$. The discriminant is $-41225931 = -1 \cdot 3^2 \cdot 1409 \cdot 3251$.

But there is an easier way to show that the Galois group is $S_5$ using a related idea to what you wrote.

Suppose $f \in \mathbb{Z}[x]$ is irreducible of degree $n$. For any prime $p$ not dividing the leading coefficient of $f$ and for which $f\pmod p$ has no repeated factor, one can write $$ f(x) = f_1(x) \cdots f_r(x) \bmod p, $$ where each $f_i$ is irreducible mod $p$. Then there is an element in the Galois group of $f$ with cycle type $(\deg f_1) \cdots (\deg f_r)$.

This is often called "a result of Dedekind". See this other question and its answer for a bit more.

Computing the discriminant is annoying, but factoring over $\mathbb{F}_p$ is a computationally friendly, finite task. One can check that mod $5$, the polynomial is irreducible (and thus there is a $5$-cycle in the Galois group). And mod $23$, the polynomial factors as $$ f = (x + 9)(x + 10)(x + 12)(x^2 + 15x + 22). $$ Thus the Galois group contains a transposition.

As any $5$-cycle and any transposition generates $S_5$, we find that the Galois group is necessarily $S_5$.

user26857
  • 52,094
davidlowryduda
  • 91,687
  • I have not seen the result anywhere on factoring over $\mathbb{F}_p$ anywherem so I have a little hard time seeing what you are doing. If you dont mind, could you elaborate on it or maybe you know a reference in dummit&foote? – LocationMap2 Jan 02 '23 at 22:05
  • I dont see how you can skip the condition on the discriminant. What if your chosen $p$ divides it? I.e 5 or 23.. – LocationMap2 Jan 02 '23 at 22:18
  • 1
    Not having repeated roots mod $p$ is equivalent to having discriminant not divisible by $p$. Morally, this is because the discriminant is a product of differences of roots, and repeated roots cause this product to be $0$ (and this holds over $\mathbb{F}_p$ as well). – davidlowryduda Jan 03 '23 at 00:10
  • Actually, computing the discriminant for a trinomial is fairly easy, I just didnt notice I had a trinomial. After that, my initial idea basically becomes yours. However, I dont see two things: Why is the polynomial irreducible mod 5 and why is $x^2 + 15x + 22$ irreducible mod 23? – LocationMap2 Jan 03 '23 at 00:12
  • 1
    To see that $x^2 + 15x + 22$ is irreducible, it suffices to check that it has no roots (as if it factors, then it has either one repeated root or two distinct roots). More generally, there are only finitely many polynomials of a fixed degree over a fixed finite field $\mathbb{F}_p$ and it suffices to check that none of them are divisors. For degree $5$, it's actually more computationally efficient to generate all irreducible quadratic polynomials and check their divisibility: a degree $5$ poly that factors will either have a linear factor (and thus a root) or a quadratic factor. – davidlowryduda Jan 03 '23 at 03:04
3

A slightly more elementary version of the other excellent answer is the following.

By Gauss' Lemma, if $f$ is reducible in $\mathbb{Q}[x]$, then it is reducible in $\mathbb{Z}[x]$, and thus (its reduction modulo $p$) is reducible in any $\mathbb{F}_{p}[x]$, for $p$ a prime.

Now, as stated in the other answer, $f$ is irreducible in $\mathbb{F}_{5}[x]$.

Thus $f$ is irreducible in $\mathbb{Q}[x]$. Thus $5$ divides the order of the Galois group, which thus contains an element of order $5$.

A study of the graph of $f$ shows that it has exactly three real roots. Thus complex conjugation induces a transposition in the Galois group of $f$, which exchanges the two non-real roots.

Now it is not difficult to see that if $p$ is a prime, and $G$ is a subgroup of the symmetric group $S_{p}$ on $p$ symbols which contains an element of order $p$ (i.e. a $p$-cycle) and a transposition, then $G = S_{p}$.

Andreas Caranti
  • 68,873
1

Count the real roots. Descartes' Rule of Signs allows up to two positive roots, and these actually exist because the polynomial is positive at $x=0$ and as $x\to+\infty$, but negative at $x=1$. Then the Rule of Signs indicates exactly one negative root. Total: three real roots and thus one pair of complex conjugate roots.

Then the Galois group has to have a $5$-cycle (from the prime degree of the irreducible polynomial) and a pairwise transposition (from the single pair of complex conjugate roots). The only group within $S_5$ that has both features is all of $S_5$. Conclude from there.

Oscar Lanzi
  • 39,403