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?

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. 


Yes, it is always possible. First note that it suffices to construct a totally complex Galois extension $K/\mathbb{Q}$ of degree $2n$ with Galois group $S_{2n}$. By the primitive element theorem, this extension is of the form $\mathbb{Q}[t]/(f(t))$ for some irreducible polynomial $f$, the minimal polynomial of an algebraic number $\alpha \in K$. Then there exists $n \in \mathbb{Z}^+$ such that $n \alpha$ is an algebraic integer: take the minimal polynomial of that algebraic integer: it generates the same field extension. To construct the desired extension $K$, in turn it suffices to find an irreducible polynomial with $\mathbb{Q}$coefficients with no real roots and whose Galois group is the largest possible $S_{2n}$. This is possible by a weak approximation / Krasner's Lemma argument. I will just sketch it for now; I can fill in more details if needed. The idea is to find a finite set of primes $p$ and degree $2n$ polynomials $f_p$ such that the Galois group of $f_p$, as a group of permutations on the roots of $f_p$, is of a certain form (e.g. contains a specific transposition). Also let $f_{\infty}$ be any degree $2n$ polynomial over $\mathbb{R}$ without real roots. Then by Krasner's Lemma, there exists a polynomial $f$ which is sufficiently $p$adically close to each $f_p$ and to $f_{\infty}$ to have the same local behavior: in particular, to factor the same way over $\mathbb{Q}_p$ and over $\mathbb{R}$ and to generate the same local Galois groups. Then, by identifying the local Galois groups with decomposition groups at $p$ (of unramified extensions), if one has enough primes so as to get permutations of every possible cycle type, then the global Galois group of $f$ certainly must be $S_{2n}$. Indeed, to see this we use the following result from lecture notes of Keith Conrad (and Bertrand's postulate!): http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf Theorem: For $n \geq 2$, a transitive subgroup of $S_n$ which contains a transposition and a $p$cycle for some prime $p > \frac{n}{2}$ is $S_n$. The condition at infinity means that $\mathbb{Q}[t]/(f(t))$ is totally complex, hence so is its splitting field. To ensure that $f$ is irreducible, we may apply Krasner's Lemma again and take its coefficients sufficiently close to those of an irreducible degree $2n$ polynomial over $\mathbb{Q}_p$ (for a different $p$ from those used thus far) so as to be irreducible over $\mathbb{Q}_p$, which implies irreducibility over $\mathbb{Q}$. This can in principle be made explicit, but I might search the literature for a known classical family of polynomials doing what you want before I tried to carry out this construction explicitly. 

