Are polynomials with non-($S_n$ or $A_n$) Galois groups discrete?

There are various results, beginning with van der Waerden, on the asymptotic number of integral polynomials with bounded coefficients, whose Galois group is $S_n$. Some of these results also include $A_n$ in the number of polynomials. In particular, Rainer Dietmann (On the distribution of Galois groups, 2012) shows that the number of polynomials of degree $n$ and height $\leq H$ which do not have Galois group $S_n$ or $A_n$ to be $\ll H^{n-1+e(n)+\varepsilon}$ for a function $e(n)$ which rapidly tends to zero.

On the other hand, Joos Heintz (On polynomials with symmetric Galois group which are easy to compute, 1986) shows that the polynomials over a Hilbertian field $k$, considered as points in $k^{n+1}$, which have Galois group $S_n$ are dense.

Is it known or suspected that the polynomials (over $\mathbb{Z}$, $\mathbb{Q}$, or number fields) with a given Galois group $G \not= S_n,A_n$ are discrete? Is the answer different if instead one takes all polynomials with Galois group different from $S_n,A_n$ or if $A_n$ is dropped from the excluded groups?

EDIT: Certainly it is true over $\mathbb{Z}$ since $\mathbb{Z}^{n+1}$ is discrete. However, I wonder how close two polynomials with Galois group $G \not= S_n,A_n$, or even two distinct groups $\not= S_n,A_n$, can get.

• For a dense subset of $a\in\mathbb{Q}$, $x^n-a$ has Galois group $\mathbb{Z}/n\rtimes(\mathbb{Z}/n)^*$. – MTyson Aug 10 '17 at 18:14
• Surely the question makes no sense in $\mathbb Q$: replacing $X$ with $\frac {n-1}nX$ yields a nearby polynomial with the same Galois group. – Anthony Quas Aug 10 '17 at 18:20
• @AnthonyQuas But under that substitution, the higher power coefficients will tend to 0. – Jacob Bond Aug 10 '17 at 18:29
• I find the question of discreteness in this context to be ambiguous. I gather the you are parameterizing polynomials of degree $n$ with $\mathbb Q$ coefficients by their vector of coefficients in $\mathbb Q^n$. That's fine, but to talk about discreteness, you need a topology, and there really isn't a natural topology on $\mathbb Q$. It appears maybe you're embedding $\mathbb Q$ in $\mathbb R$ and using the induced real topology, but you could as well embed in $\mathbb Q_p$ and use the $p$-adic topology. It actually makes more sense to embed $\mathbb Q$ In the adeles ... – Joe Silverman Aug 10 '17 at 22:23
• ... $\mathbb A_{\mathbb Q}$, but of course, $\mathbb Q$ is discrete in the adeles. This is why results along these lines are often phrased instead in terms of some sort of density, for example the density obtained by ordering the rational numbers by height (as in Dietmann's paper). And as Anthony Quas noted, since you get the same Galois group for $f(x)$ and $f(ax+b)$, and indeed even for $f((ax+b)/(cx+d))$ as long as $ad-bc\ne0$, any topology on $\mathbb Q^n$ that would be useful for this question must be invariant for these transformations. And $\mathbb Q^n/PGL_2(\mathbb Q)$ is a messy set! – Joe Silverman Aug 10 '17 at 22:27