Can Macaulay2 do computations with symbolic parameters?
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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 $J_t=langle tx^2+yz,ty^2 +xz,tz^2+xyranglesubset mathbb{Q}[x,y,z]$ with $tinBbb C$. So long as $t^3neq 1$ 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 $x^2 yin J_t$. 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?
polynomials ring-theory commutative-algebra ideals macaulay2
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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 $J_t=langle tx^2+yz,ty^2 +xz,tz^2+xyranglesubset mathbb{Q}[x,y,z]$ with $tinBbb C$. So long as $t^3neq 1$ 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 $x^2 yin J_t$. 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?
polynomials ring-theory commutative-algebra ideals macaulay2
$endgroup$
add a comment |
$begingroup$
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 $J_t=langle tx^2+yz,ty^2 +xz,tz^2+xyranglesubset mathbb{Q}[x,y,z]$ with $tinBbb C$. So long as $t^3neq 1$ 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 $x^2 yin J_t$. 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?
polynomials ring-theory commutative-algebra ideals macaulay2
$endgroup$
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 $J_t=langle tx^2+yz,ty^2 +xz,tz^2+xyranglesubset mathbb{Q}[x,y,z]$ with $tinBbb C$. So long as $t^3neq 1$ 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 $x^2 yin J_t$. 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?
polynomials ring-theory commutative-algebra ideals macaulay2
polynomials ring-theory commutative-algebra ideals macaulay2
edited Dec 8 '18 at 15:05
Rodrigo de Azevedo
12.9k41857
12.9k41857
asked Nov 22 '16 at 17:57
SemiclassicalSemiclassical
11.1k32465
11.1k32465
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1 Answer
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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
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
answered Nov 23 '16 at 0:58
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