Suppose $f \in \mathbb{Q}[x]$ is monic, with roots $\alpha_1,\dotsc,\alpha_n$. Define the discriminant of $f$ to be the number $ \Delta = \prod_{i<j} (\alpha_i – \alpha_j)^2$. Let $D(f) = \sqrt{\Delta} = \prod_{i<j} (\alpha_i – \alpha_j)$ be the square root of the discriminant. Here are the questions:
- 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,
- 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 $f$ 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).
EDIT (transplanted from a post below by LSpice): Kristal, yes, yes, sorry. There are at least 2 mistakes in the original post:
- I forgot to mention that the polynomial $f$ is assumed monic
- The sentence "But it seems clear that there is no cubic having discriminant $2^{2} \times 3$" should say "… no cubic having $D = \sqrt{\Delta} = 2^{2} \times 3$". I reach this conclusion just by looking at the roots as points on a line and the numbers $(\alpha_i – \alpha_j)$ as 'oriented distances between the points'.
For your example, $3x^3 – x = 3x(x^2 – 1/3) = 3x(x – 1/\sqrt{3})(x + 1/\sqrt{3})$ so that $\Delta = 3^{2(3-1)} [(0 + 1/\sqrt{3})(0 – 1/\sqrt{3})(1/\sqrt{3} + 1/\sqrt{3})]^2 = 3^4 \dot (1/3)^2 \dot (2/\sqrt{3})^2 = 12$, just as you claimed. But my question is about the square root of the discriminant, which is the interesting quantity for me because the Galois group of $f$ is a subgroup of $A_n$ if and only if $D(f)$ is rational!
It appears I can't edit my original post (because I didn't register to make it). Thank you all very much for the references, especially David Brown.
Best Answer
You might enjoy reading Kiran Kedlaya's paper Unramified alternating extensions of quadratic fields on constructing $A_n$ extensions of $\mathbb{Q}$ with specified discriminant, and following the references there in. The main point of the paper is to construct polynomials with squarefree discriminant, subject to various conditions.