Given $n \in \mathbf{N}$,is always possible to construct a monic polynomial in $\mathbf{Z}[x]$ of degree $2n$, whose roots are in $\mathbf{C} \setminus \mathbf{R}$ and whose Galois group over $\mathbf{Q}$ is $S\_{2n}$?
I have an approximate idea of how to solve the problem for the Galois group (I immagine something related to the Hilbert irreducibility theorem), but I have no idea for the condition on the roots.
Furthermore, is it possible to give an explicit example?



A special integral polynomial
An easy way to ensure that a polynomial $g$ of degree $m$ over $\mathbf{Z}$ has
Galois group $S\_m$ is to take primes $p\_1$, $p\_2$ and $p\_3$
with $g$ irreducible modulo $p\_1$, a linear times an irreducible
modulo $p\_2$ and a bunch of distinct linears times an
irreducible quadratic modulo $p\_3$. Then the Galois group
must be doubly transitive and have a transposition, so it's $S\_m$.


Now take $m=2n$ and a polynomial $f$ over $\mathbf{Q}$ with no real
roots (e.g. $(x^2+1)^n$). Replacing the coefficients of $f$ by
close rationals won't create any real roots. So replace the $x^k$
coefficient of $f$ by a sufficient close rational $a\_k/b\_k$
where $a\_k$ and $b\_k$ are congruent modulo $p\_1 p\_2 p\_3$
respectively to the $x^k$ coefficient of $g$ and to $1$. Then
the new polynomial has rational coefficients, no real roots
and Galois group $S\_{2n}$. You can easily convert it to one
with these properties and integer coefficients should you wish.


