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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!

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If you ask Magma to compute the Galois group of a polynomial, then this is basically what it does. For low degree polynomials (It think less than 20 or something like that), it has the right invariants precomputed, for higher degrees it computes them on the fly. –  Alex B. May 24 '12 at 23:59
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1 Answer 1

up vote 10 down vote accepted

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.

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This is a great answer. Thank you Filipe. –  Dave M da C May 28 '12 at 20:02
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