Rationality conditions for determining Galois groups

Question

Let $F$ be a field and $h \in F[x]$ be an irreducible, degree $n$ monic polynomial. Let $G$ denote the Galois group of $h$.

It is well known that $G \subset A_n $ if and only if the discriminant of $h$, which we'll denote by $D(h)$, is a square in $F$. We could think of this as being a rationality condition: we are demanding an $F-$rational solution to the equation $y^2 = D(h)$.

My question is if this is always possible for any subgroup $H \subset Sym(n)$. That is, does there exist a polynomial $f\in F[a_0,\ldots,a_{n-1}]$ in the coefficients of $h$ and a polynomial $\phi \in F(y) $ such that $G \subset H$ only if $\phi(y) = f$ has a solution in $F$? Is it possible to make this a sufficient condition also? (I suspect that the answer is yes to the former and no to the latter).

Further, if such a $\phi$ exists, can we control its degree? Is such a condition unique and if there are many, is there a simplest?

Thanks!

Answer 1

More or less, yes. Fix a transitive subgroup $H \subset S_n$. Let $S_n$ act in the usual way in the field $\mathbb{Q}(x_1,\ldots,x_n)$ where the $x_i$ are algebraically independent. Then the fixed field $\mathbb{Q}(x_1,\ldots,x_n)^{S_n} = \mathbb{Q}(a_1,\ldots,a_n)$ where $\prod(X-x_i) = \sum a_iX^{n-i}, a_0=1$. Now, write $\mathbb{Q}(x_1,\ldots,x_n)^{H} = \mathbb{Q}(a_1,\ldots,a_n,z)$, by the primitive element theorem and $z$ satisfies some equation $f(a_1,\ldots,a_n,z)=0$, for some polynomial $f$ whose degree can probably be figured out from the order of $H$. Now, given $\alpha_1,\ldots,\alpha_n \in F$, where $F$ is a field of characteristic zero, the splitting field of $\sum \alpha_iX^{n-i}, \alpha_0=1$ is contained in $H$ if and only if there is $\beta \in F$ with $f(\alpha_1,\ldots,\alpha_n,\beta)=0$. When $H=A_n, f=z^2-$ discriminant.