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.

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    $\begingroup$ 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

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