Of course, this depends on the field, and how we measure "how many," but it seems I cannot find an answer to this except over finite fields.

My question specifically is

If we have a field $F = \mathbb{Q}$ or $\mathbb{R}$, how many polynomials of degree $n$ height $\le h$ will be squarefree relative to the number of all polynomials of height $ \le h$?

The height of $$p(x) = c_0 + c_1 x + \dots + c_n x^n$$ is $$\max\{ |c_0|, \dots, |c_n| \},$$ and relative to the number of all polynomials of height $\le h$ means that the proportion taken with respect to an appropriate measure (I am not sure the best way to do this for $\mathbb{Q}$, but for $\mathbb{R}$ this should be a product of the Lebesgue measure of course).

Or really I'm interested in any sort of result in this manner, even if not exactly what I've written here. We know that the proportion of squarefree integers is $6/\pi^2$, so I was hoping for something like this for polynomials since $F[x]$ is also a PID.

  • $\begingroup$ Could also consider $\mathbb{F}_q[x]$ for finite fields $\mathbb{F}_q$ (for which I believe the proportion is $1-1/q$). $\endgroup$ – arctic tern Jun 22 '16 at 1:58
  • $\begingroup$ Heuristically, if you play with various parametrizations, the subspace of nonsquarefree polynomials should have positive codimension, hence "almost all" polynomials are squarefree. $\endgroup$ – arctic tern Jun 22 '16 at 2:10
  • $\begingroup$ @arctictern Could you elaborate on your heuristic to show the nonsquarefree polynomials have positive codimension? $\endgroup$ – Polynomeal Jun 22 '16 at 3:06

Consider the subspace of $\mathbb{R}^n$ consisting of vectors whose coordinates are not all distinct. This is a union of positive codimension subspaces. For example, the subspace of vectors with first two coordinates equal but otherwise pairwise distinct has codimension $1$, just like the line $y=x$ in the plane $\mathbb{R}^2$. Since every monic polynomial of degree $n$ with real zeros is uniquely determined by $n$ zeros counted with multiplicity, we should expect the same of them, and perhaps even for all real coefficient polynomials including those with complex zeros.

Let $\mathbb{R}[x]_n$ the the $(n+1)$-dimensional subspace of polynomials of degree $n$. (It's not a vector subspace because the leading coefficients must be nonzero but whatever.)

For every nonnegative integer solutions to $r+2s=n-1$ there is a map,

$$ \mathbb{R}^\times\times\mathbb{R}^{r-1}\times(\mathbb{C}\setminus\mathbb{R})^s\to\mathbb{R}[x]_n \\ (a,(x_1,\cdots,x_r),(z_1,\cdots,z_s))\mapsto a(x-x_1)^2\prod_{i=2}^r (x-x_r)\prod_{j=1}^s(x-z_j)(x-\overline{z_j})$$

and for every solution to $r+2s=n-2$ there is a similar map


defined similarly but with $(x-z_1)^2(x-\overline{z_1})^2$ instead of $(x-x_1)^2$.

There are only finitely many of the above maps, each one having a domain of dimension $\le n$, and the subspace of $\mathbb{R}[x]_n$ of nonsquarefree polynomials is the union of the images of these maps (since every such polnomial has one repeated either real or complex zero), so this subspace has positive codimension; in particular it has measure zero in any measurable region. Thus, "almost all" polynomials are squarefree.


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