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!

  • 3
    $\begingroup$ Raise your $(ab)(\ldots)\cdots$ to an odd power... $\endgroup$ – user641 Mar 30 '12 at 14:27
  • $\begingroup$ for the second statement (but not in detail :) - if you change a polynomial a bit, then its roots move a bit; it works especially well if the derivatives at those roots are non-zero. As complex roots of real polynomial come in conjugate pairs, all the roots will be indeed real. BTW I'd like to know it there is an elementary proof of "(2) implies a permutation $(ab)(\dots)\dots$". The proof I know requires the use of $\mathbb{Q}_3$. $\endgroup$ – user8268 Mar 30 '12 at 15:24
  • 4
    $\begingroup$ @user8268: A theorem of Dedekind implies you can reduce a polynomial mod $p$, and the factorization mod $p$ gives you the cycle type of a permutation in the Galois group (as long as the reduction stays separable). This is theorem 4.37 in Jacobson's Basic Algebra I. $\endgroup$ – user641 Mar 30 '12 at 15:35
  • 2
    $\begingroup$ BTW, the statement "If a [G]alois group contains a transposition, th[e]n it is $S_n$" is clearly false, as any polynomial with Galois group $D_8$ shows. $\endgroup$ – user641 Mar 30 '12 at 15:39
  • $\begingroup$ You're right. There must exist a transposition, a $n-1$-cycle and the galois group have to be a transitive subgroup of $S_n$. Then the galois group is $S_n$. But I can't see, why a $n-1$-cycle exists. $\endgroup$ – David75 Mar 30 '12 at 16:09

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 $D_8$).


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