My question is about the realization of the symmetric group $S_n$ as a galois group of a real and normal field extension $K/\mathbb Q$. As I read, such a field $K$ can be obtained as the splitting field of a polynomial
$$f(x)=\alpha_0 + \alpha_1 x + … + \alpha_{n-1}x^{n-1}+ x^n \in \mathbb Q[x] $$
in which all the coefficients $\alpha_i$ have to be integral for the prime numbers 2 and 3.
Furthermore, the following properties have to be satisfied:
(1) $\ f \text{ mod }3$ is irreducible
(2) $\ f \equiv (x^2+x+1)\cdot f_1 \cdots f_k \text{ mod }2$
where $f_i \in \mathbb Z / 2 \mathbb Z[x]$ are different and irreducible polynomials with odd degree.
(3) $f$ have to be near at $(x-1)\cdot (x-2) \cdots (x-n) = \beta_0 + \beta_1 x + … + \beta_{n-1}x^{n-1}+ x^n$,
that does mean $|\alpha_i – \beta_i|< \varepsilon$ for all $i$ and $\varepsilon$ should be chosen so small, that there are $n$ real roots of $f$.
Now, it is claimed, that because of (1) the galois group of $f$ over $\mathbb Q$ is a transitive subgroup of $\mathbb S_n$. I think, this is true, because of the fact, that $f$ is also irreducible in $\mathbb Q[x]$ and therefore the galois group acts transitively on the roots of $f$.
Ok, now comes an assertion that I don't understand: Because of (2) the galois group of $f$ over $\mathbb Q$ contains a transposition. Please could you explain that to me? I think, (2) shows, that there is a permutation of the form $(ab)(…)…(…)$. But how to continue the argument?
If the galois group contains a transposition, than it is $S_n$ (this is clear and well-known).
And the second assertion I don't understand well: "Because of (3) the splitting field $K$ of $f$ is real." How to explain this in detail?
Thanks in advance!
Best Answer
As mentioned in the comments, $x^4+7x+14$ satisfies properties (1) and (2), but does not have Galois group $S_4$ (it is actually $D_8$).