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Polynomial reduction: polynomial in terms of other polynomials?

Consider each function below such as f, f2, f3 and f4 with the basis I. How can we express each f such that f_i=\sum a_i I_i and each a_i\geq 0?

Example

We demonstrate the polynomials below with M2 and Mathematica.

Macaulay2:

i1 : R=RR[x1,x2,x3,MonomialOrder=>Lex]; 
f=x3-x1*x2;
f2=x3*x2-x1;
f3=x1-0.2;
f4=x1-x3+0.8;

i5 : I=ideal(x1-0.2,-x1+0.5,x2,-x2+1,x3-1,-x3+1); G=gb(I);

We can express f3 with elements of I, namely with zeroth term

i11 : I_0==f3

o11 = true

We can express f4 with I_5 and I_0

i17 : I_5+I_0==f4

o17 = true

Can we express f and f2 with I?

Mathematica: f and f-2 cannot be expressed in terms of the I but f-1 can be expressed in I but negative terms so cannot use Handelman's theorem on it.

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but

  • f-2 is not non-negative (choose x3=1,x1=2 so 1-0-2=-1<0)

  • f is non-negative (x3=1 so 1-x1x2>0) and

  • f-1 is not non-negative (x3=1,x2>0 so -x1x2<0).

and by Handelman's theorem, all computations are inconclusive because the the third term -x1 is negative. More about Mathematica aspects here.

How can we express a polynomial in terms of other polynomials and each quotient term is positive like PolynomialReduce in Mathematica but each quotient term positive?

Solution

  • Note that in this answer, I am using your terminology, in which R is the polynomial ring and RR is the ring of real numbers. I should also say that almost never use the ring RR, since computations in macaulay2 over the real numbers are not always reliable, always use the ring of rationals QQ or a positive characteristic field like QQ/(101).

    Your f and f2 polynomials are not linear, so you can not even write them as a linear combination of I_0,...,I_5 (i.e. the generators of I). Furthermore the ideal I as you defined it contains a scalar so it is what mathematicians call the unit ideal. It means I=R, that is the whole polynomial ring. So you can write f and f2 as a combination of I_0,...,I_5 but not a linear one. It means that f = \sum g_i I_i with g_i polynomials where at least one of them is not a number.

    Remark. For an arbitrary ring R, the elements are usually called scalars, but when R is a polynomial ring, let's say R=RR[x_1,...x_n] then usually the constant polynomials (which are exactly the real numbers, i.e. elements of RR) are called scalars. This is just a common and of course confusing terminology.

    Here is an example,

    i2 : R=QQ[x_1,x_2]

o2 = R

o2 : PolynomialRing

i3 : I=ideal(x_1-1,x_2,x_1+1)

o3 = ideal (x  - 1, x , x  + 1)
             1       2   1

o3 : Ideal of R

i4 : I == R

o4 = true

i5 : J = ideal(x_1,x_2)

o5 = ideal (x , x )
             1   2

o5 : Ideal of R

i6 : J == R

o6 = false

    You see that the ideal I has x_1-1,x_2,x_1+1 so the element (x_1+1)-(x_1-1) = 2 also belongs to I, so I has a constant polynomial which is a unit element (a unit element in a ring is an element that has an inverse) which implies that I=R. For a proof of this fact visit, https://math.stackexchange.com/questions/552173/if-an-ideal-contains-the-unit-then-it-is-the-whole-ring

    On the other hand J does not have any constant polynomial, so J is not the whole ring R.