Galois group of quintic. 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?
 A: 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 $f (\bmod p)$ 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 $S_5$.
A: 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 $G = S_{p}$.
