Projective Resolution of $C[x,y,t]/(x^2+y^2-t)$

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I need to find a free projective resolution of $mathbbC[x,y,t]/(x^2+y^2-t)$ as a $mathbbC[t]$-module. A projective resolution of an $R$-module $M$ is a long exact sequence
$$...rightarrow P_nrightarrow P_n-1rightarrow...rightarrow P_0rightarrow Mrightarrow 0.$$ Here each $P_i$ is required to be a projective $R$-module (hence the name). I require a free resolution, so the $P_i$ must be free as $C[t]$-modules. I can do simple examples, such as projective resolutions of $mathbbZ/2$ as a $mathbbZ$ module, i.e
$$0rightarrow mathbbZ xrightarrow2 mathbbZ rightarrow mathbbZ/2rightarrow 0.$$



I can do slightly harder example as well, but am at a loss whenever it comes to more complex examples. It would be a great help if some one could describe a projective resolution of this $R$-module, and even better if they could give some general techniques for tackling these sorts of questions.



Note: This is not a homework question, but rather me trying to understand some algebraic geometry machinery, and this came up in the process.




algebraic-geometry homological-algebra exact-sequence projective-module free-modules




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    up vote
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    down vote

    favorite












    I need to find a free projective resolution of $mathbbC[x,y,t]/(x^2+y^2-t)$ as a $mathbbC[t]$-module. A projective resolution of an $R$-module $M$ is a long exact sequence
    $$...rightarrow P_nrightarrow P_n-1rightarrow...rightarrow P_0rightarrow Mrightarrow 0.$$ Here each $P_i$ is required to be a projective $R$-module (hence the name). I require a free resolution, so the $P_i$ must be free as $C[t]$-modules. I can do simple examples, such as projective resolutions of $mathbbZ/2$ as a $mathbbZ$ module, i.e
    $$0rightarrow mathbbZ xrightarrow2 mathbbZ rightarrow mathbbZ/2rightarrow 0.$$



    I can do slightly harder example as well, but am at a loss whenever it comes to more complex examples. It would be a great help if some one could describe a projective resolution of this $R$-module, and even better if they could give some general techniques for tackling these sorts of questions.



    Note: This is not a homework question, but rather me trying to understand some algebraic geometry machinery, and this came up in the process.




    algebraic-geometry homological-algebra exact-sequence projective-module free-modules




    share|cite|improve this question





















      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I need to find a free projective resolution of $mathbbC[x,y,t]/(x^2+y^2-t)$ as a $mathbbC[t]$-module. A projective resolution of an $R$-module $M$ is a long exact sequence
      $$...rightarrow P_nrightarrow P_n-1rightarrow...rightarrow P_0rightarrow Mrightarrow 0.$$ Here each $P_i$ is required to be a projective $R$-module (hence the name). I require a free resolution, so the $P_i$ must be free as $C[t]$-modules. I can do simple examples, such as projective resolutions of $mathbbZ/2$ as a $mathbbZ$ module, i.e
      $$0rightarrow mathbbZ xrightarrow2 mathbbZ rightarrow mathbbZ/2rightarrow 0.$$



      I can do slightly harder example as well, but am at a loss whenever it comes to more complex examples. It would be a great help if some one could describe a projective resolution of this $R$-module, and even better if they could give some general techniques for tackling these sorts of questions.



      Note: This is not a homework question, but rather me trying to understand some algebraic geometry machinery, and this came up in the process.




      algebraic-geometry homological-algebra exact-sequence projective-module free-modules




      share|cite|improve this question











      I need to find a free projective resolution of $mathbbC[x,y,t]/(x^2+y^2-t)$ as a $mathbbC[t]$-module. A projective resolution of an $R$-module $M$ is a long exact sequence
      $$...rightarrow P_nrightarrow P_n-1rightarrow...rightarrow P_0rightarrow Mrightarrow 0.$$ Here each $P_i$ is required to be a projective $R$-module (hence the name). I require a free resolution, so the $P_i$ must be free as $C[t]$-modules. I can do simple examples, such as projective resolutions of $mathbbZ/2$ as a $mathbbZ$ module, i.e
      $$0rightarrow mathbbZ xrightarrow2 mathbbZ rightarrow mathbbZ/2rightarrow 0.$$



      I can do slightly harder example as well, but am at a loss whenever it comes to more complex examples. It would be a great help if some one could describe a projective resolution of this $R$-module, and even better if they could give some general techniques for tackling these sorts of questions.



      Note: This is not a homework question, but rather me trying to understand some algebraic geometry machinery, and this came up in the process.





      algebraic-geometry homological-algebra exact-sequence projective-module free-modules





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      asked Aug 1 at 11:42









      noetherianpickles

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          $Bbb C [x,y,t] $ is a free $Bbb C [t] $-module, hence we have a free resolution over $Bbb C [t] $:
          $$0toBbb C [x,y,t]toBbb C [x,y,t]toBbb C [x,y,t]/langle x^2+y^2-trangleto0 $$
          where the first map is the multiplication by $x^2+y^2-t $, while the second is the canonical projection.






          share|cite|improve this answer





















          • Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
            – noetherianpickles
            Aug 1 at 14:40










          • This situation arise with quotients by principal ideals.
            – Fabio Lucchini
            Aug 1 at 15:57










          Your Answer




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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

          oldest

          votes








          up vote
          3
          down vote



          accepted










          $Bbb C [x,y,t] $ is a free $Bbb C [t] $-module, hence we have a free resolution over $Bbb C [t] $:
          $$0toBbb C [x,y,t]toBbb C [x,y,t]toBbb C [x,y,t]/langle x^2+y^2-trangleto0 $$
          where the first map is the multiplication by $x^2+y^2-t $, while the second is the canonical projection.






          share|cite|improve this answer





















          • Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
            – noetherianpickles
            Aug 1 at 14:40










          • This situation arise with quotients by principal ideals.
            – Fabio Lucchini
            Aug 1 at 15:57














          up vote
          3
          down vote



          accepted










          $Bbb C [x,y,t] $ is a free $Bbb C [t] $-module, hence we have a free resolution over $Bbb C [t] $:
          $$0toBbb C [x,y,t]toBbb C [x,y,t]toBbb C [x,y,t]/langle x^2+y^2-trangleto0 $$
          where the first map is the multiplication by $x^2+y^2-t $, while the second is the canonical projection.






          share|cite|improve this answer





















          • Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
            – noetherianpickles
            Aug 1 at 14:40










          • This situation arise with quotients by principal ideals.
            – Fabio Lucchini
            Aug 1 at 15:57












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          $Bbb C [x,y,t] $ is a free $Bbb C [t] $-module, hence we have a free resolution over $Bbb C [t] $:
          $$0toBbb C [x,y,t]toBbb C [x,y,t]toBbb C [x,y,t]/langle x^2+y^2-trangleto0 $$
          where the first map is the multiplication by $x^2+y^2-t $, while the second is the canonical projection.






          share|cite|improve this answer













          $Bbb C [x,y,t] $ is a free $Bbb C [t] $-module, hence we have a free resolution over $Bbb C [t] $:
          $$0toBbb C [x,y,t]toBbb C [x,y,t]toBbb C [x,y,t]/langle x^2+y^2-trangleto0 $$
          where the first map is the multiplication by $x^2+y^2-t $, while the second is the canonical projection.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 1 at 14:13









          Fabio Lucchini

          5,56911025




          5,56911025











          • Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
            – noetherianpickles
            Aug 1 at 14:40










          • This situation arise with quotients by principal ideals.
            – Fabio Lucchini
            Aug 1 at 15:57
















          • Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
            – noetherianpickles
            Aug 1 at 14:40










          • This situation arise with quotients by principal ideals.
            – Fabio Lucchini
            Aug 1 at 15:57















          Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
          – noetherianpickles
          Aug 1 at 14:40




          Oh, of course! Multiplication by the prime ideal (x^2+y^2-t), so it is very similar to forming a projective resolution of Z/2 over Z.
          – noetherianpickles
          Aug 1 at 14:40












          This situation arise with quotients by principal ideals.
          – Fabio Lucchini
          Aug 1 at 15:57




          This situation arise with quotients by principal ideals.
          – Fabio Lucchini
          Aug 1 at 15:57












           

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