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! 



Rationality conditions for determining Galois groups
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.


