Question about Zariski density and polynomials with full Galois group Let $A_3\subseteq {\mathbb Q}^4$ be the sets of all $q=(q_3,q_2,q_1,q_0)$ such that $P_q=X^4+q_3X^3+q_2X^2+q_1X+q_0$ has no rational root. Let $A_2 \subseteq A_3$ be the subset of all $q$'s such that $P$ is irreducible over $\mathbb Q$. Let $A_1 \subseteq A_2$ be the sub-subset of all $q$'s such that $P$ has Galois group $S_4$ over $\mathbb Q$.
My questions : Is $A_1$ Zariski-dense in $A_2$ ? Is $A_2$ Zariski-dense in $A_3$ ?
My thoughts : $B=\cap_{r\in {\mathbb Q}} \lbrace q \ | \ P_q(r)\neq 0 \rbrace$ is a countable intersection of open sets, so it is probably not open or closed.
 A: The Hilbert irreducibility theorem says that $A_1$ is already Zariski dense in all of $\mathbb Q^4$, so the answer to both of your questions is yes.
The Hilbert irreducibility theorem says that $A_1$ is already Zariski dense in all of $\mathbb Q^4$, so the answer to both of your questions is yes.
First, one can see that the Galois group of your polynomial over the field $K=\mathbb Q(q_0,q_1,q_2,q_3)$ is $S_4$. A lazy way to see this is to observe that the Galois group is at most that large and that there are specializations which have Galois group $S_4$ (of course this is ``historically'' a backward approach).
Next, let $\lambda$ be a primitive element for the splitting field of $f(x)$ over $K$, which we may take to be a $K$-linear combination of the roots of $f(x)$, and $g(x)$ the minimal polynomial of $\lambda$ over $K$. Tweaking the linear combination if necessary, we may take $\lambda$ to be integral over $\mathbb Q[q_0,q_1,q_2,q_3]$ so that $g(x)$ is in $\mathbb Q[q_0,q_1,q_2,q_3][x]$, making it a polynomial. By construction, $g$ is irreducible in $K[x]$. By the HIT the set of specializations which preserve the irreducibility of $g$ is Zariski dense. All we need to do is check that these also give rise to specializations of $f(x)$ which have Galois group $S_4$. 
This is easy: the Galois group of $f(x)$ (specialized somewhere) embeds in $S_4$. On the other hand, the specialization of $\lambda$ is still a linear combination of roots of $f(x)$ and so the splitting field of $f(x)$ contains it. But $\lambda$ has degree $4!$ over $\mathbb Q$ by our choice of specialization and so the degree of the splitting field must be as large as possible, meaning we have the desired Galois group.
I suppose to be more a little careful with the specialization (and how it extends to $\lambda$) one would want to talk about a prime of the integral closure of $\mathbb Q[q_0,q_1,q_3,q_4]$ in $K(\lambda)$ lying over the ideal $(q_0-a_0,q_1-a_1,q_2-a_2,q_3-a_3)$. The HIT is a statement about decomposition groups in this setting. If $\lambda$ were not chosen to be integral one would have to localize to resolve that issue, and then use the version of HIT which gives specializations which don't vanish on specified polynomials in order to avoid trouble with the denominators. 
