# Most discriminants are almost squarefree

Write, for $f(x) = x^d + a_2 x^{d-2} + \cdots + a_d\in \mathbb{Z}[x]$, $H(f) := \max(|a_i|^{\frac{1}{i}})$.

Does anyone know of a reference that would allow me to show that the proportion of $f$ with $\deg{f} = d > 4, H(f)\leq X$, and discriminant having a square factor $> X^\delta$ is $O(X^{-c})$ for some $c = c(d,\delta) > 0$?

The proportion of polynomials in $\mathbb{Z}_p[x]$ with discriminant divisible by $p^2$ is $O(p^{-2})$, so that handles primes $p\leq X$. There has been so much work on squarefree discriminants of polynomials that this must be straightforward given the methods, but unfortunately I don't know them.

Thanks much!!

Edit:

I found a paper of Bhargava on the "geometric sieve", and I can almost answer the question by using his methods, except his condition 6 (and subsequent trick to reduce to a higher-codimension variety) doesn't apply. He solves this problem for $d=4$ via something that seems special to that case (he calls it the "embedding sieve" --- it's at the end of the paper). Does anybody know if there is a similar embedding available for higher $d$? I suppose this is now all of a sudden a question in invariant theory, which I essentially know nothing about. Thanks again!

• Mar 9, 2015 at 14:20
• Why is it true that "The proportion of polynomials in $\mathbb{Z}_p[x]\dots$ is $O(p^{-2}?$" Feb 4, 2016 at 16:43
• Let f(a_1,...,a_n) := disc_z(z^n + ... + a_n). Claim: the measure of the set S of (a_1,...,a_n) such that |f(a_1,...,a_n)|\leq p^{-2} is O_n(p^{-2}). Proof: split the set into two: the set T of (a_1,...) where (a_1,...,a_{n-1}) are such that f(a_1,...,a_{n-1},x) is not squarefree as a polynomial in x, and the others. For the others, note that f(a_1,...,a_{n-1},x) has leading term \pm n^n x^{n-1} as a polynomial in x, so that a_n is within p^{-2} of some root of this polynomial (since it's squarefree and we may assume p>n so that p doesn't divide the leading term). Feb 4, 2016 at 19:06
• Hence the measure of S-T is O_n(p^{-2}). But the measure of the set T is zero: indeed, (a_1,...,a_{n-1})\in T if and only if disc_x(f(a_1,...,a_{n-1},x)) = 0, which is a positive-codimension condition since e.g. f(0,...,0,-n,t) has a root at t=-(n-1)\zeta for each (n-1)-st root of unity \zeta, so that disc_x(f(0,...,-n,x))\neq 0. Feb 4, 2016 at 19:06
• (Let me know if I made a mistake in the above! I forget what justification I had in mind originally, or even if I had one in mind at all, since I have an unfortunate tendency to assume things that look plausible to prove when writing in an informal context...) Feb 4, 2016 at 19:12

I am pretty sure that it is unknown how to prove such a theorem. (Edit: See the comment, this may no longer be true!)

By sieve methods, if you could prove your claim for an arbitrarily small value of $\delta$ (and with a larger error term allowed), then you would establish the density of squarefree discriminants in your family (since the small primes are easy to handle).

I presume that this paper by Poonen represents the state of the art. It is conditional on ABC, so if you believe Mochizuki's proof then you're done.

Otherwise, you have to ask that the discriminant polynomial be "special" in some way, and use the fact that it is not just a generic polynomial. As you mentioned one approach leads to invariant theory, representation theory, etc. but as field counting goes our knowledge is so far limited to degrees $\leq 5$.

(Of course if anyone does know how to solve this problem I would be very interested!)

• Jerry Wang recently gave a talk here (Waterloo) where he claimed to have proved the expected counting theorem for square-free discriminants for every degree with Bhargava and Shankar. The trick is to generalize the 'using geometry of numbers to count orbits having bounded invariants' method to cases where the group acting on the space is not reductive. This paper has not yet appeared, even on arxiv, as far as I know. I assume it will be a sequel to this paper: arxiv.org/pdf/1512.03035.pdf Feb 4, 2016 at 14:47
• @StanleyYaoXiao: Cool!! Feb 4, 2016 at 16:16
• Yeah! I actually asked Bhargava about this this past summer and he said he, Arul, and Jerry had proved what I needed. I figured I could just wait til the preprint came out to answer this question since I assumed nobody had noticed it on MathOverflow! But anyway it's an amazing result! Feb 4, 2016 at 19:16