Let $k$ be an algebraically closed field and consider the ring $R = k[X_1, \ldots, X_n]$ of polynomials in $n$ variables over $k$.

Is the "general" polynomial in $R$ reducible or irreducible?

The exact meaning of the set of "general" polynomials is left to the pleasure of the reader, but two possibilities that come to mind are "outside a set of measure zero" (if $k = \mathbb C$ and w/r/t the Lebesgue measure) or "in the complement of a countable union of Zariski closed sets".

If $n = 1$ then any polynomial factors as a product of linear ones, so "most" polynomials are reducible. If $n > 1$ the opposite might be true, because by mumbling "Bertini theorem" the general member in a linear system corresponding to an ample divisor might be irreducible; since homogeneous polynomials in many variables are examples of such things (over projective space), there the general (homogeneous) polynomial could be irreducible.

  • 4
    $\begingroup$ Dear Gunnar, It's a bit hard to answer this question as written, because for polynomials to form a finite dimensional space (where measure, or the Zariski topology, make good sense) one has to bound the degree. Otherwise, you have the problem that all degree $\leq d$ polynomials are themselves of measure zero in, and a proper Zariski closed subset of, degree $\leq d+1$ polyomials. Would you be happy for an answer for polynomials of bounded degree (with $d$ arbitrary but fixed)? Regards, $\endgroup$ – Matt E Aug 5 '13 at 0:40
  • $\begingroup$ Dear Matt, Very good remark, thank you. Yes, I would be perfectly happy with an answer for polynomials of bounded degree, which seems to be exactly what Qiaochu has just given. :) $\endgroup$ – Gunnar Þór Magnússon Aug 5 '13 at 1:11
  • $\begingroup$ Dear Gunnar, Just as an aside: this issue that when the degree isn't bounded the space of polynomials is not finite dimensional is why Hilbert schemes are problematic for non-projective varieties. Regards, $\endgroup$ – Matt E Aug 5 '13 at 3:52

I agree with Matt E's comment that we should bound degrees. This makes the argument straightforward. The space of nonzero polynomials in $n$ variables of degree at most $d$ modulo scalar multiplication is a projective space of dimension ${d+n \choose n} - 1$. The subspace of all reducible polynomials is the union of the images in this projective space of $d - 1$ morphisms from products of lower-dimensional projective spaces taking the form

$$(f(x_1, ... x_n), g(x_1, ... x_n)) \mapsto f(x_1, ... x_n) g(x_1, ... x_n)$$

where $f$ has degree at most $d_1 \ge 1$, $g$ has degree at most $d_2 \ge 1$, and $d_1 + d_2 = d$. These products of projective spaces have dimension ${d_1 + n \choose n} + {d_2 + n \choose n} - 2$.

Claim: If $n \ge 2$ this dimension is always strictly less than ${d+n \choose n} - 1$.

Proof. The polynomial $f(d) = {d+n \choose n} - 1$ has no constant term and non-negative coefficients. The claim then follows from the observation that $(d_1 + d_2)^k > d_1^k + d_2^k$ when $k \ge 2, d_1, d_2 \ge 1$. $\Box$

It follows that the generic polynomial (in the complement of a finite union of Zariski closed subsets) of degree at most $d$ is irreducible for $n \ge 2$.

  • $\begingroup$ That was very enjoyable, thank you Qiaochu. $\endgroup$ – Gunnar Þór Magnússon Aug 5 '13 at 1:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.