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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

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    $\begingroup$ Sorry, but what is an AC? $\endgroup$
    – Qiaochu Yuan
    Commented Apr 21, 2011 at 18:29
  • $\begingroup$ algebraically closed, I imagine $\endgroup$
    – JSE
    Commented Apr 21, 2011 at 18:53
  • $\begingroup$ Ah, I misparsed that sentence (missed "field $k$"). $\endgroup$
    – Qiaochu Yuan
    Commented Apr 21, 2011 at 19:04

2 Answers 2

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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.

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  • $\begingroup$ Long, are you saying that it's true for 3 polynomials without a common divisor in $\mb C[x,y,z]$? Thanks for the references -- still looking through them for exactly what I need. $\endgroup$
    – Sue Sierra
    Commented Apr 21, 2011 at 20:41
  • $\begingroup$ Dear Sue, I think we need the ideals gen. by the elements to have height 3, which is stronger than saying they have no common divisor. I have to run now, but will try to locate a precise reference later. $\endgroup$
    – Hailong Dao
    Commented Apr 21, 2011 at 21:48
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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.

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  • $\begingroup$ The answer below makes it clear that there's no hope. Are you remembering anything else that might be useful to understand when this does happen? $\endgroup$
    – Sue Sierra
    Commented Apr 21, 2011 at 20:43
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    $\begingroup$ Sue, definitely Bertini theorems is the right way to go for references. One other reference I found is Exercise 10.4 of Eisenbud's "Commutative Algebra". $\endgroup$
    – Karl Schwede
    Commented Apr 21, 2011 at 21:38

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