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3 fixed discriminant 12 error; added 11 characters in body

Suppose $f \in \mathbb{Q}[x]$ has is monic, with roots $\alpha_{1},\dots,\alpha_{n}$. Define the discriminant of $f$ to be the number $\Delta = \Pi_{i<j} (\alpha_{i} - \alpha_{j})^2$. Let $D(f) = \sqrt{\Delta} = \Pi_{i<j} (\alpha_{i} - \alpha_{j})$ be the square root of the discriminant. Here are the questions:

1) Let $p/q \in \mathbb{Q}$ be fixed. What can be said about the set $\{ f \in \mathbb{Q}[x] : D(f) = p/q \}$? For example, if we fix $p$ to be some prime, then the family of quadratics having this prime as the discriminant are just translates of the polynomial $x(x-p)$ by any rational number. Of course, the whole family has the same (trivial) Galois group. What can be said about the Galois group of families of polynomials of greater degree with fixed discriminant? The Galois groups will be subgroups of $A_{n}$, but is there any reason to expect any 'stability'?

and less interestingly,

2) What rational numbers $p/q \in \mathbb{Q}$ have the property that $p/q = D(f)$ for some $f$ as above, with the degree of $f$ fixed to be some natural number $n$? For example, every rational number is the discriminant of some quadratic. But it seems clear that there is no cubic having discriminant $f$ with $D(f) = 2 \times 2 \times 3$. What can be said in general?

The context of this is I'm trying to generate some families of polynomials that don't have Galois group $S_{n}$ and my simple-minded idea is to use the discriminant to 'pick them out' (it won't do to just generate polynomials randomly, as the set of polynomials with Galois group $S_{n}$ is thick in the set of all polynomials).

Thank you!

2 deleted 10 characters in body; added 10 characters in body

Suppose $f \in \mathbb{Q}[x]$ has roots $\alpha_{1},\dots,\alpha_{n}$. Define the discriminant of $f$ to be the number $\Delta = \Pi_{i<j} (\alpha_{i} - \alpha_{j})^2$. Let $D(f) = \sqrt{\Delta} = \Pi_{i<j} (\alpha_{i} - \alpha_{j})$ be the square root of the discriminant. Here are the questions:

1) Let $p/q \in \mathbb{Q}$ be fixed. What can be said about the set $\{ f \in \mathbb{Q}[x] : D(f) = p/q \}$? For example, if we fix $p/q = n$ p$to be some integerprime, then the family of quadratics having this integer prime as the discriminant are just translates of the polynomial$x(x-n)$x(x-p)$ by any rational number. Of course, the whole family has the same (trivial) Galois group. What can be said about the Galois group of families of polynomials of greater degree with fixed discriminant? The Galois groups will be subgroups of $A_{n}$, but is there any reason to expect any 'stability'?

and less interestingly,

2) What rational numbers $p/q \in \mathbb{Q}$ have the property that $p/q = D(f)$ for some $f$ as above, with the degree of $f$ fixed to be some natural number $n$? For example, every rational number is the discriminant of some quadratic. But it seems clear that there is no cubic having discriminant $2 \times 2 \times 3$. What can be said in general?

The context of this is I'm trying to generate some families of polynomials that don't have Galois group $S_{n}$ and my simple-minded idea is to use the discriminant to 'pick them out' (it won't do to just generate polynomials randomly, as the set of polynomials with Galois group $S_{n}$ is thick in the set of all polynomials).

Thank you!

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