## Density of Irreducible Polynomials in $\mathbb{Z}[x]$

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Recently I was thinking about some questions concerning $\mathbb{Z}[x]$ and realized that they might be a bit easier if I knew the relative densities of reducible polynomials.

Let $P_d$ denote the set of all elements of $\mathbb{Z}[x]$ of degree $\leq d$. My basic question is:

Question: What fraction of elements of $P_d$ factor in $\mathbb{Z}[x]$?

The fraction $f_d$ of elements of $P_d$ which factor satisfies $f_d\geq 1-\zeta(d+1)^{-1}$ (with $\zeta$ the Riemann Zeta Function) since this is the count of elements which factor as a constant times an element of $P_d$. When $d = 0,1$ one obviously has equality, but what is not clear to me is whether $f_d = 1-\zeta(d+1)^{-1}$ holds for all $d$.

I thought about the analogous problem in $\mathbb{F}_p[x]$ (where one ignores constant factors), but in this case as $d\rightarrow\infty$ one has $f_d\rightarrow 1$. If this behavior carries over to $\mathbb{Z}[x]$ then as $d$ grows large, almost every polynomial of degree $d$ factors which seems absurd; therefore looking at this question in $\mathbb{F}_p[x]$ doesn't seem to help much.

So, letting $f_d = 1-\zeta(d+1)^{-1}+\varepsilon(d)$, I am wondering if the correction term $\varepsilon$ equals zero or, if not, what is $\varepsilon(d)$ and how would one derive it?

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Perhaps I am missing something, but how do you define the 'fraction' (as P_d is infinite)? – quid Mar 30 2011 at 18:42
The conventional way to define the fraction is as the limit of the fraction of reducible elements in a ball centered at the origin as the radius of the ball tends to infinity. – ARupinski Mar 30 2011 at 18:45

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Actually the conventional way would be to use the height of a polynomial. I.e. fix a bound for coefficients, allowing you to count how many there are; at least that is the naive form of height. Why not look at monic polynomials first? – Charles Matthews Mar 30 2011 at 18:57
I deleted a comment asking 'which norm' as CM's comment basically answers this. – quid Mar 30 2011 at 19:00

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See also this question: mathoverflow.net/questions/58397/… – Xandi Tuni Mar 30 2011 at 21:03
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## 3 Answers

I restore the following in clarified form as an over-sized comment; in a temporary (at least I hope it was only temporary) state of confusion I posted it as an answer, which it is not (I was not careful regarding the different notions of ir/reducibility, sorry about that).

The density of integral polynomials of fixed degree that are reducible as rational polynomials, i.e. that are the product of two non-constant integral polynomials, is $0$. (This contains the statement for monic ones as a special case.)

More precisley, the order of the numer of such polynomials of height at most $t$ is `$t^d$` for $d \ge 3$.

That is, it is known that for $d \ge 3$ there exists a constant `$C_d$` such that if `$R_d(t)$` denotes the number of reducible polynomials with height at most $t$ and degree $d$ then `$$ t^d \le |R_d(t)| \le C_d t^d $$`

For $d= 2$ one has an additional logarithmic factor; the order in this case is `$ t^2 log t $`.

This and related ressults are proved by G. Kuba in 'On the distribution of reducible polynomials, Math. Slovaca 59 (2009), no. 3, 349–356.'

(The novelty of the paper is the quality of the estimate; the density result is much older, an upper bound of the form `$t^d (log t)^2$` seems to be mentioned in Polya and Szego.)

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Looks like exactly what I expected would be the case. Thanks. – ARupinski Mar 31 2011 at 4:23

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Take a look at the book: An introduction to sieve methods and their applications By Alina Cojocaru, Maruti Ram Murty

In section 4.3 the Turan Sieve is used to prove that the probability of a random polynomial with integer coefficients is irreducible is 1. It is available online at Google books.

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Although that section only deals with monic polynomials with positive coefficients, it seems that the basic argument should modify to be useable for arbitrary polynomials and with a little work ought to be able to verify that $\varepsilon(d) = 0$ for all $d$. I'll have to look at that a bit. – ARupinski Mar 30 2011 at 21:21

Actually, a lot more is true: a random polynomial has Galois group the full symmetric group (a result of van der Waerden) and this is true under various restrictions, and with various error terms. For a summary and some related results, see

Igor Rivin Walks on groups, counting reducible matrices, polynomials, and surface and free group automorphisms. Duke Math. J. 142 (2008), no. 2, 353–379

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Thanks for that reference, I will have a look at it as well. – ARupinski Mar 31 2011 at 16:00
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