# Statistic of cubic Irreducible polynomial with cyclic Galois group

Van der Waerden proved for monic irreducible polynomials $f$ of degree $n$ with bounded height $X$:
$$|\{ f(x)\in \mathbb{Z}[x]: Gal(f)\neq S_n \}|=o(X)$$ Where height of $f$ is defined by maximum of absolute value of its coefficient.

So if we order monic irreducible polynomial of degree $n$ with respect to height, then we expect to obtain symmetric group as its Galois group. Now what if we want to replace height by discriminant.

Do we have any table or any conjecture about this? More precisely is it true:

$$|\{ f\in \mathbb{Z}[x]: monic, irreducible \deg(f)=n, |Disc(f)|\leq X, Gal(f)\neq S_n\}|=o(X)$$

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You might find the slides for Carl Pomerance's talk on counting fields to be relevant: math.dartmouth.edu/~carlp/rademacherlecture4.pdf –  Ben Linowitz Nov 16 '10 at 22:45
There are dozens of recent articles relevant to this question; Bhargava, Belabas, Cohen, ... Perhaps a good place to start is Cohem, Diaz y Diaz, Olivier: "A survey of discriminant counting" in ANTS-V. –  Franz Lemmermeyer Nov 16 '10 at 22:48
@myself: I meant Cohen, not Cohem. As for Pomerances nice notes mentioned by Ben: the Delone-Faddeev correspondence is actually due to F. Levi, Leipziger Berichte 66 (1914), 26-37. Delone and Faddeev credit him for this in the preface to their book, but it seems to have been overlooked. –  Franz Lemmermeyer Nov 16 '10 at 22:57
@Franz: I am a bit familiar with Bhargava results, but here I want to count polynomials instead to count number field, I think this would be not exactly similar to counting number field, since different polynomials can give same number field with the same discriminant. Also for example if one look at quartic number field then contribution of $S_4$ and $D_4$ are linear i.e cx. I expect for polynomials contribution of $D_4$ is negligible. –  M.B Nov 16 '10 at 23:11