Decomposing an ideal using Macaulay2












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I give Macaulay2 the ideal $I=(y^2, x) in Q[x , y]$ and then I put decompose I. The result is $(x , y)$ but I do not understand why. Does it mean that $I = (x , y)$? but that is not true, because we can not create $y$ in $I$.










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    $begingroup$


    I give Macaulay2 the ideal $I=(y^2, x) in Q[x , y]$ and then I put decompose I. The result is $(x , y)$ but I do not understand why. Does it mean that $I = (x , y)$? but that is not true, because we can not create $y$ in $I$.










    share|cite|improve this question











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      $begingroup$


      I give Macaulay2 the ideal $I=(y^2, x) in Q[x , y]$ and then I put decompose I. The result is $(x , y)$ but I do not understand why. Does it mean that $I = (x , y)$? but that is not true, because we can not create $y$ in $I$.










      share|cite|improve this question











      $endgroup$




      I give Macaulay2 the ideal $I=(y^2, x) in Q[x , y]$ and then I put decompose I. The result is $(x , y)$ but I do not understand why. Does it mean that $I = (x , y)$? but that is not true, because we can not create $y$ in $I$.







      ideals macaulay2






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      edited Dec 9 '18 at 9:36









      Rodrigo de Azevedo

      13.1k41960




      13.1k41960










      asked Sep 26 '16 at 18:04









      M. SaM. Sa

      495




      495






















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          $begingroup$

          From the documentation:




          decompose is a synonym for minimalPrimes.



          This function computes the minimal associated primes of the ideal I using characteristic sets. Geometrically, it decomposes the algebraic set defined by I.




          So $(x,y)$ is the minimal associated prime of $I = (y^2,x)$. (It's just the radical of $I$, since $I$ is a primary ideal.)






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            From the documentation:




            decompose is a synonym for minimalPrimes.



            This function computes the minimal associated primes of the ideal I using characteristic sets. Geometrically, it decomposes the algebraic set defined by I.




            So $(x,y)$ is the minimal associated prime of $I = (y^2,x)$. (It's just the radical of $I$, since $I$ is a primary ideal.)






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              From the documentation:




              decompose is a synonym for minimalPrimes.



              This function computes the minimal associated primes of the ideal I using characteristic sets. Geometrically, it decomposes the algebraic set defined by I.




              So $(x,y)$ is the minimal associated prime of $I = (y^2,x)$. (It's just the radical of $I$, since $I$ is a primary ideal.)






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                From the documentation:




                decompose is a synonym for minimalPrimes.



                This function computes the minimal associated primes of the ideal I using characteristic sets. Geometrically, it decomposes the algebraic set defined by I.




                So $(x,y)$ is the minimal associated prime of $I = (y^2,x)$. (It's just the radical of $I$, since $I$ is a primary ideal.)






                share|cite|improve this answer









                $endgroup$



                From the documentation:




                decompose is a synonym for minimalPrimes.



                This function computes the minimal associated primes of the ideal I using characteristic sets. Geometrically, it decomposes the algebraic set defined by I.




                So $(x,y)$ is the minimal associated prime of $I = (y^2,x)$. (It's just the radical of $I$, since $I$ is a primary ideal.)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 26 '16 at 18:25









                arkeetarkeet

                5,185923




                5,185923






























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