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Old 2012-01-20, 21:42   #1
fivemack
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Default 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.
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Old 2012-01-20, 22:11   #2
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Quote:
Originally Posted by fivemack View Post
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.
What about using a Groebner basis f1', f2', expressing G with respect to that base (giving q1', q2'), and then transforming back?
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Old 2012-01-20, 22:29   #3
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What about using a Groebner basis f1', f2', expressing G with respect to that base (giving q1', q2'), and then transforming back?
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 2012-01-20 at 22:31
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Old 2012-01-20, 23:33   #4
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Quote:
Originally Posted by fivemack View Post
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.
From googling for "degree bounds on Bezout polynomials", I have found this article.
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Old 2012-01-21, 00:09   #5
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fivemack,

That should follow directly from the fact that f_1 and f_2 are homogeneous.

Given G, write it as G=\sum_{i} G_i, where G_i is the homogeneous part of G in degree i.

If G=q_1*f_1 + q_2*f2 and q_{i,j} is the homogeneous part of q_i in degree j then G_i=q_{1,i-2}f_1 + q_{2,i-2}f_2. (To see this, just look at all of the possible terms of a given degree.)

In particular, you can choose the degrees on the q's to be bounded by \deg(G)-2.

Last fiddled with by Zeta-Flux on 2012-01-21 at 00:09
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Old 2012-01-21, 08:06   #6
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From googling for "degree bounds on Bezout polynomials", I have found this article.
This mainly considers zero-dimensional zero loci. Your problem has too many variables for that, sorry. Yet another interesting article is this one which seems to cover your problem.
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Old 2012-01-21, 08:51   #7
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This mainly considers zero-dimensional zero loci. Your problem has too many variables for that, sorry. Yet another interesting article is this one which seems to cover your problem.
In particular, theorem 1.2 (p. 104) of the above article by Andersson mostly confirms Zeta-Flux's claim.

Last fiddled with by ccorn on 2012-01-21 at 08:52
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Old 2012-01-21, 10:21   #8
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fivemack,

That should follow directly from the fact that f_1 and f_2 are homogeneous.

Given G, write it as G=\sum_{i} G_i, where G_i is the homogeneous part of G in degree i.

If G=q_1*f_1 + q_2*f2 and q_{i,j} is the homogeneous part of q_i in degree j then G_i=q_{1,i-2}f_1 + q_{2,i-2}f_2. (To see this, just look at all of the possible terms of a given degree.)

In particular, you can choose the degrees on the q's to be bounded by \deg(G)-2.
I see. Even if higher-degree parts of q1, q2 would lead to a complete cancellation in the linear combination, those parts do not influence lower-degree parts of the linear combination, so they can as well be left out.

Last fiddled with by ccorn on 2012-01-21 at 10:23 Reason: Replaced "terms" by "parts" where appropriate
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Old 2012-01-21, 11:37   #9
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Thanks very much. I feel silly now, I should have been able to construct zeta-flux's argument on my own ...
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Old 2012-01-21, 15:30   #10
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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 counter-examples. So I'm just very familiar with this kind of argument.

Last fiddled with by Zeta-Flux on 2012-01-21 at 15:31
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Old 2012-01-22, 11:01   #11
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Thanks to Zeta-Flux, the original question has been answered.

However---Fivemack: 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 degree-1 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 2012-01-22 at 11:08
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