irreducibility of generic linear combination of polynomials? I would be shocked if the following were not true, but I can't seem to see a proof.
Claim:
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.  
Then for general $\alpha, \beta \in k$, the element $\alpha a + \beta b$ is irreducible. 
Thanks!
Sue 
 A: 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.  
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).  
EDIT: 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.  
A: 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". 
ADDED: the relevant reference is Satz 4.9 and 4.10 (Die Sätze von Bertini für lokale Ringe 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. 
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. 
