Can Macaulay2 do computations with symbolic parameters?












4












$begingroup$


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







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Dec 8 '18 at 15:05









      Rodrigo de Azevedo

      12.9k41857




      12.9k41857










      asked Nov 22 '16 at 17:57









      SemiclassicalSemiclassical

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      11.1k32465






















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          $begingroup$

          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





          share|cite|improve this answer











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            $begingroup$

            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





            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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





              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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





                share|cite|improve this answer











                $endgroup$



                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






                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                answered Nov 23 '16 at 0:58


























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