# 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. – Felipe Voloch May 23 '14 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 – so-called friend Don May 23 '14 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). – Oblomov May 27 '14 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). – Oblomov May 24 '14 at 13:50
• Actually the reference by Ash--Brakenhoff--Zarrabi answers exactly my question. – Oblomov May 27 '14 at 13:15