# Proportion of irreducible polynomials $P$ such that $\mathbf Z[X]/(P)$ is the ring of integers of $\mathbf Q[X]/(P)$

I know that number fields have been the object of many statistical experiments. Is there some kind of heuristics for the following?

Fix a degree $d$ and fix a bound $N$ on the coefficients of a monic integral degree $d$ polynomial $P$ (a more natural choice would probably be to bound the discriminant of $P$). Among those such polynomials which are irreducible, what is the proportion of those for which $\mathbf Z[X]/(P)$ is the ring of integers of $\mathbf Q[X]/(P)$?

I wrote a very basic and naive script in pari/gp to test this and it seems to me that the proportion is about 60%. Is anything known or conjectured about this?

At least, is it easy to see that this proportion is non-zero?

What about the similar but different question: for which proportion of $P$ does the ring of integers have a power basis?

• Here is a relevant MO question mathoverflow.net/questions/21267/… which does not answer yours, unfortunately. I wonder if your proportion comes from the local obstructions. May 23, 2014 at 15:02
• This doesn't answer your question, but ... you might be interested in knowing that for rather specific families of polynomials $P$ (rather than all $P$ with bounded height), Bardestani has some density results: arxiv.org/abs/1202.2047 May 23, 2014 at 17:07
• In case it may interess someone: the proportion 60% found is conjectured by Lenstra to be $\frac 6 {\pi^2}$ (cf. the paper of Ash--Brakenhoff--Zarrabi mentionned in David Speyer answer). May 27, 2014 at 12:45

I summarize the first two pages of Kedlaya, A construction of polynomials with squarefree discriminants

When $$P$$ is irreducible and the discriminant $$\Delta(P)$$ is square free, the number field $$\mathbb{Q}[x]/P(x)$$ has ring of integers $$\mathbb{Z}[x]/P(x)$$... When the coefficients of $$P$$ are chosen randomly, this is expected to occur with probability $$\prod_p a_p$$ ... where $$a_p$$ denotes the probability that $$\Delta(P)$$ is not divisible by $$p^2$$. These probabilities have been computed by Brakenhoff:

I'll omit the table, but they are all of the form $$1-O(1/p^2)$$, so the product is nonzero.

Unfortunately, ... it seems quite difficult to prove that a polynomial takes squarefree values with the expected probability ...

Keldaya then summarizes work of Granville and Poonen which, assuming the ABC conjecture, implies that $$\Delta(P)$$ is squarefree with probability $$\prod a_p$$, where the limit is taken over boxes in $$\mathbb{Z}[x]_{\deg n}$$ of a certain shape (roughly, much longer in one direction than the others.).

However, without assuming any conjectures, it is difficult to establish even the existence of infinitely many polynomials of a given degree with squarefree discriminant.

Kedlaya then explains that constructing infinitely many such polynomials is his main result.

In short, there is a good conjecture for the probability of squarefree discriminant, but people can't unconditionally show that it is even positive. Squarefree discriminant is a bit more special than $$\mathbb{Z}[x]/P(x)$$ integrally closed, but I think this is suggestive.

• Thanks for the reference. Note that, however, numerical experiences seem to show that squarefree discriminants explain only half of the proportion obtained (and Stickelberger's relation does not help much). May 24, 2014 at 13:50
• Actually the reference by Ash--Brakenhoff--Zarrabi answers exactly my question. May 27, 2014 at 13:15

The conjecture mentioned in the comments was proven in 2016 by Bhargava--Shankar--Wang:

The density of monic irreducible polynomials $$f$$ with integer coefficients such that $$\mathbb Z[x]/(f(x))$$ is the ring of integers in its fraction field is $$\zeta(2)^{-1} = .607927\ldots\,\,\,$$.

This is Theorem 1.2 of Squarefree values of polynomial discriminants I, https://arxiv.org/abs/1611.09806. An updated version is at http://www.math.uwaterloo.ca/~x46wang/Papers/monicsieve4.pdf.

Note that the density they use is with respect to the height, $$H(x^n+a_1x^{n-1}+\cdots+a_n):=\max(|a_1|,|a_2|^{1/2},\ldots,|a_n|^{1/n}),$$ which is a little different from the one proposed in the question on this page.