20120120, 21:42  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
6,379 Posts 
Theorems about ideals
Suppose I have an ideal of Q[x,y,z,t] generated by f1 and f2, each homogenous of degree 2.
Is there a theorem that, if a polynomial G of degree d is in the ideal, it can be written as q1*f1 + q2*f2 with deg(q1), deg(q2) bounded? I'm not at all sure even where to start looking for such a thing. 
20120120, 22:11  #2  
Apr 2010
224_{8} Posts 
Quote:


20120120, 22:29  #3 
(loop (#_fork))
Feb 2006
Cambridge, England
14353_{8} Posts 
That's what the software I'm using at the moment (sage) is doing; it works perfectly well and determines q1 and q2 happily, but I don't know how to get a degree bound out of the process. The problem is that sage leaks memory at a spectacular rate, and I want to do tests as to whether a fixed G is in the ideals generated by a few billion different pairs f1, f2.
Last fiddled with by fivemack on 20120120 at 22:31 
20120120, 23:33  #4  
Apr 2010
2^{2}·37 Posts 
Quote:


20120121, 00:09  #5 
May 2003
7×13×17 Posts 
fivemack,
That should follow directly from the fact that and are homogeneous. Given , write it as , where is the homogeneous part of in degree . If and is the homogeneous part of in degree then . (To see this, just look at all of the possible terms of a given degree.) In particular, you can choose the degrees on the 's to be bounded by . Last fiddled with by ZetaFlux on 20120121 at 00:09 
20120121, 08:06  #6  
Apr 2010
94_{16} Posts 
Quote:


20120121, 08:51  #7  
Apr 2010
224_{8} Posts 
Quote:
Last fiddled with by ccorn on 20120121 at 08:52 

20120121, 10:21  #8  
Apr 2010
94_{16} Posts 
Quote:
Last fiddled with by ccorn on 20120121 at 10:23 Reason: Replaced "terms" by "parts" where appropriate 

20120121, 11:37  #9 
(loop (#_fork))
Feb 2006
Cambridge, England
6,379 Posts 
Thanks very much. I feel silly now, I should have been able to construct zetaflux's argument on my own ...

20120121, 15:30  #10 
May 2003
1547_{10} Posts 
Don't feel bad, this is one of those things that just appears easy in hindsight. My area of expertise is ring theory, and quite a few of my papers specifically work with homogeneous ideals to construct examples and counterexamples. So I'm just very familiar with this kind of argument.
Last fiddled with by ZetaFlux on 20120121 at 15:31 
20120122, 11:01  #11 
Apr 2010
2^{2}·37 Posts 
Thanks to ZetaFlux, the original question has been answered.
HoweverFivemack: Is your G a homogenized polynomial? If so, assume t is the projection variable. Then G and G*t^k (for some natural number k) are equivalent in the sense that their affine (unhomogenized) counterparts are equal. And it may well happen that G is not in the ideal generated by f1, f2 whereas G*t^k is (from some k onward). Example: f1(x,y,z,t) = y^2 + t^2 f2(x,y,z,t) = xy  xt + yt G(x,y,z,t) = xy  y^2 + xt All these are degree 2, but you need degree1 factors to combine f1, f2 to G*t: q1(x,y,z,t) = x q2(x,y,z,t) = y I note this rather for myself in order to keep in mind why homogenizing does not magically ease all membership questions. Last fiddled with by ccorn on 20120122 at 11:08 
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