# Galois group of $x^5-x-1$ over $\Bbb Q$ using integral extension ring theory

Let $$f= x^5 − x − 1$$ and $$L$$ be the splitting field of $$f(x)$$ over $$\Bbb Q$$. Suppose $$B$$ is an integral closure of $$\Bbb Z$$ in $$L$$ and $$P$$ is a maximal ideal of $$B$$ such that $$P \cap \Bbb Z = 2\Bbb Z$$. I am trying to compute the Galois group of $$f$$, i.e., the group $$Gal(L/\Bbb Q)$$, using the following theorem:

Theorem. Let $$A$$ be an entire ring, integraly closed in its quotient field $$K$$. Let $$L$$ be a finite Galois extension of $$K$$. Let $$K(\alpha)$$, where $$\alpha$$ is integral over $$A$$, and let $$f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$$ be the irreducible polynomial of $$\alpha$$ over $$K$$ with $$a_i\in A$$. Let $$\mathfrak{p}$$ be a maximal ideal in $$A$$, let $$\mathfrak{B}$$ be a prime ideal of the integral closure $$B$$ of $$A$$ in $$L$$, $$\mathfrak{B}$$ lying over $$\mathfrak{p}$$. Let $$\bar{f}$$ be the reduced polynomial with coefficients in $$A/\mathfrak{p}$$. Let $$G_{\mathfrak{B}}$$ be the decomposition group. If $$\bar{f}$$ has no multiple roots, then the map $$\sigma \mapsto \bar{\sigma}$$ is an isomorphism of $$G_{\mathfrak{B}}$$ on the Galois group of $$\bar{f}$$ over $$A/\mathfrak{p}$$.

I see that the theorem can be applied with $$A=\Bbb Z, K=\Bbb Q, L=L, \mathfrak{p}=2\Bbb Z$$, and $$\mathfrak{B}=P$$, but I can't see how to compute $$G_\mathfrak{B}$$ or the Galois group of $$\bar{f}$$ over $$A/\mathfrak{p}=\Bbb Z/2\Bbb Z$$. I see that $$\bar{f}$$ is a product of two irreducible polynomials in $$\Bbb Z/2\Bbb Z[x]$$, of degrees $$2$$ and $$3$$. Also how can I deduce the Galois group $$Gal(L/\Bbb Q)$$ by the theorem?

The fact that $$f$$ decomposes that way when reduced modulo 2 tells you that the Galois group $$G$$ of $$f$$, which is a subgroup of $$S_5$$ since $$\deg f=5$$, contains an element $$\sigma$$ that decomposes as a product of a 2-cycle and a 3-cycle. Hence, $$G$$ contains a transposition, namely $$\sigma^3$$, and a 3-cycle, namely $$\sigma^2$$. I'm not sure whether this is already enough to conclude that $$G=S_5$$, but $$f$$ is irreducible mod 3, and therefore $$G$$ also contain a 5-cycle. Since $$S_p$$ is generated by a $$p$$-cycle and a transposition for every prime $$p$$, you get that $$G=S_5$$.