irreducibility of generic linear combination of polynomials? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:12:29Z http://mathoverflow.net/feeds/question/62557 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62557/irreducibility-of-generic-linear-combination-of-polynomials irreducibility of generic linear combination of polynomials? Sue Sierra 2011-04-21T18:15:13Z 2011-04-22T01:08:38Z <p>I would be shocked if the following were not true, but I can't seem to see a proof.</p> <p>Claim:</p> <p>Let $R$ be an integral domain containing an AC (uncountable if you wish) field $k$. Let $a, b \in R$, and suppose that the ideal $(a, b)$ is height 2. </p> <p>Then for general $\alpha, \beta \in k$, the element $\alpha a + \beta b$ is irreducible. </p> <p>Thanks!</p> <p>Sue </p> http://mathoverflow.net/questions/62557/irreducibility-of-generic-linear-combination-of-polynomials/62558#62558 Answer by Karl Schwede for irreducibility of generic linear combination of polynomials? Karl Schwede 2011-04-21T18:29:32Z 2011-04-21T21:35:10Z <p>Sue, this shouldn't work in characteristic $p > 0$. For example, consider $k[x,y]$ where $k$ is an uncountable perfect field of characteristic $p > 0$. Choose $a = x^p$, $b = y^p$. Then $\alpha a + \beta b$ is always reducible, it has a $p$th root. </p> <p>In characteristic zero however, at least in the geometric setting (finite type over an algebraically closed field), this should be basically a version of Bertini's theorem in a form close to the one in Remark 7.9.1 in Chapter III of Hartshorne (you probably already knew about that). </p> <p><strong>EDIT:</strong> As Long points out, this doesn't even work (with two variables). For another reference to track down, a (global) special case of the statement for 3 or more terms is Exercise 11.3 in Chapter III of Hartshorne. </p> http://mathoverflow.net/questions/62557/irreducibility-of-generic-linear-combination-of-polynomials/62566#62566 Answer by Hailong Dao for irreducibility of generic linear combination of polynomials? Hailong Dao 2011-04-21T19:33:05Z 2011-04-22T01:08:38Z <p>This is not true even over $\mathbb C$. Take $\mathbb C[x,y]$ and $x^2, y^2$. You need general combination of a regular sequence of length at least $3$. Search for "local Bertini theorem" and "Flenner". </p> <p>ADDED: the relevant reference is Satz 4.9 and 4.10 (<em>Die Sätze von Bertini für lokale Ringe</em> by H. Flenner, Mathematische Annalen, (299), 1977). This works for $n$ elements such that the ideal generated by them has height at least $3$ (over a infinite field of char. $0$). There is no hope in char. $p>0$ no matters how many elements you pick, since one can expand Karl's example. </p> <p>The height at least $3$ condition can't be weakened (think about $x^2, xy, y^2$ over $\mathbb C$). However, in the case of $2$ elements, if you assume $a,b$ are irreducible to begin with, then I would guess what you want has a much better chance. </p>