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I'm trying to figure out how to use Macaulay2 to do some ideal membership computations, and I'm running into a problem with symbolic parameters. Here is a practical example.

Consider the family of ideals $J_t=\langle tx^2+yz,ty^2 +xz,tz^2+xy\rangle\subset \mathbb{Q}[x,y,z]$ with $t\in\Bbb C$. So long as $t^3\neq 1$ we have $$x^2 y = \frac{t^2 y}{t^3+1}(tx^2+yz)-\frac{t z}{t^3+1}(ty^2+xz)+\frac{x}{t^3+1}(tz^2+xy)$$ and therefore $x^2 y\in J_t$. I would like to reproduce this in Macaulay2 (web browser version here).

If I pick a specific value of the parameter $t$, this isn't terribly hard: I construct the ideal $J$, and then compute the coordinates of the polynomial $x^2 y$ in $J$ by x^2*y // J. For $t=2$ we have

i1 : R=QQ[x,y,z];
i2 : t=2;
i3 : J=ideal(x^2*t+y*z,y^2*t+x*z,z^2*t+x*y);
o3 : Ideal of R
i4 : x^2*y // gens J
o4 = {2} | 4/9y  |
     {2} | -2/9z |
     {2} | 1/9x  |
             3       1
o4 : Matrix R  <--- R

which reproduces the decomposition $$x^2 y =\frac{4y}{9}(2x^2+yz)-\frac{2z}{9}(2y^2+xz)+\frac{x}{9}(2z^2+zy).$$

However, it would be preferable to have an algorithm that works for arbitrary symbolic $t$. Is there any way to do this within Macaulay2?

Rodrigo de Azevedo
  • 23,223
Semiclassical
  • 18,592

1 Answers1

4

There is in fact an easy solution to this, which I'll credit to Thomas Khale's answer to a Google groups question. First, we do K=frac(QQ[t]) to initialize $K$ as a rational fraction field in $t$. We then take the underlying polynomial ring not to be $\mathbb{Q}[x,y,z]$ but rather $K[x,y,z]$ via R=K[x,y,z]. This permits Macaulay2 to decompose $x^2 y$ in the ideal $J_t$ with coefficients being rational functions of $t$.

Here's the modified code:

i1 : K=frac(QQ[t])
o1 = K
o1 : FractionField
i2 : R=K[x,y,z];
i3 : J=ideal(x^2*t+y*z,y^2*t+x*z,z^2*t+x*y);
o3 : Ideal of R
i4 : x^2*y // gens J
o4 = {2} | t2/(t3+1)y |
     {2} | -t/(t3+1)z |
     {2} | 1/(t3+1)x  |
             3       1
o4 : Matrix R  <--- R
Semiclassical
  • 18,592