Two questions about discriminants of polynomials in ℚ[x]

Question

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:

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 $f$ 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.

Answer 1

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.

Answer 2

If you're interested in families of polynomials that don't have Galois group $S_n$ you should look at Noam Elkies' web page: "Trinomials $ax^n+bx+c$ with interesting Galois groups" about families of trinomials with interesting Galois groups, and the references at the bottom of the page which give pointers to papers about other interesting families.

Answer 3

If I plug the following polynomial: 3x^3-x into the formula for the discriminant I think I get 12. The formula for the discriminant of the cubic I am using is here:

http://en.wikipedia.org/wiki/Discriminant

I think b and d being zero cancel out all terms but -4a(c^3) which is 12.

Answer 4

For irreducible cubics, the answer is provided by cyclotomy: cubic extensions with square discriminants are cyclic, hence cyclotomic. Computing the possible discriminants for reducible cubics seems at least feasible.

For general polynomials, I predict trouble ahead. In the 1930s, Udo Wegner tried to prove results in the direction of "if the discriminant of an irreducible polynomial looks like the discriminant of a polynomial with Galois group G, then it must have Galois group G" (I think G was the Galois group of binomial extensions generated by $x^n-a$), but I would not trust any of his results without proving them myself. He wrote one article together with Reichardt, whose results can be trusted:

• H. Reichardt, U. Wegner, Arithmetische Charakterisierung von algebraisch auflösbaren Körpern und Gleichungen von Primzahlgrad, J. Reine Angew. Math. 178 (1937), 1-10, doi:10.1515/crll.1938.178.1

In the second part of their article they deal with the question how certain discriminant divisors can be used to show that the polynomial is not solvable.

The context of your question is not clear to me: if you want polynomials with Galois group different from $S_n$, you can of course take the cyclotomic polynomial or $f(x) = x^n-a$, but you know that. For less trivial examples, check out the book

• Christian U. Jensen, Arne Ledet, and Noriko Yui , Generic polynomials: constructive aspects of the inverse Galois problem, Mathematical Sciences Research Institute Publications 45 (2003), online version at MSRI.