Computing $operatorname{Tor}_*^{R}(mathbb{Z},mathbb{Z})$ for $R=mathbb Z[C_n]$.












4












$begingroup$



I would like to compute $operatorname{Tor}_k^{R}(mathbb{Z},mathbb{Z})$, where $R = mathbb{Z}[x]/(x^n -1)=mathbb Z[X]$.




I think I am near to do so, but I can not figure out the last step.
I started looking for a free resolution of $mathbb{Z}$ as an $R$-module, and found that



$$cdotslongrightarrow R longrightarrow R longrightarrow mathbb{Z} longrightarrow 0$$



where $d_0: Rlongrightarrow mathbb{Z}: f(X) mapsto f(1)$, and for $igeqslant 1$, $d_{2i-1}: Rlongrightarrow R$ is multiplication by $X-1$ and $d_{2i}: Rlongrightarrow R$ by $1+X+cdots+X^{n-1}$



In this way I have a free resolution of $mathbb{Z}$ as an $R$-module. Now I can apply the functor $ * otimes mathbb{Z}$ to the free resolution (projective too):



$$C^*:cdots longrightarrow R otimes mathbb{Z} longrightarrow R otimes mathbb{Z} longrightarrow 0$$



where the functions defined above do the same in the first component and apply identity in the second component. Now I know that there is an isomorphism between $R otimes N$ and $N$ if $N$ is an $R$-module. Then I would obtain something like



$$cdots longrightarrow mathbb{Z} longrightarrow mathbb{Z} longrightarrow 0$$



which it may looks easier to find the $n$-th cohomology object of the cochain complex. But I do not know how to find $ker(d_notimes 1)$ neither $operatorname{im}(d_notimes 1)$.






abstract-algebra commutative-algebra homological-algebra







share|cite|improve this question











$endgroup$

















    4












    $begingroup$



    I would like to compute $operatorname{Tor}_k^{R}(mathbb{Z},mathbb{Z})$, where $R = mathbb{Z}[x]/(x^n -1)=mathbb Z[X]$.




    I think I am near to do so, but I can not figure out the last step.
    I started looking for a free resolution of $mathbb{Z}$ as an $R$-module, and found that



    $$cdotslongrightarrow R longrightarrow R longrightarrow mathbb{Z} longrightarrow 0$$



    where $d_0: Rlongrightarrow mathbb{Z}: f(X) mapsto f(1)$, and for $igeqslant 1$, $d_{2i-1}: Rlongrightarrow R$ is multiplication by $X-1$ and $d_{2i}: Rlongrightarrow R$ by $1+X+cdots+X^{n-1}$



    In this way I have a free resolution of $mathbb{Z}$ as an $R$-module. Now I can apply the functor $ * otimes mathbb{Z}$ to the free resolution (projective too):



    $$C^*:cdots longrightarrow R otimes mathbb{Z} longrightarrow R otimes mathbb{Z} longrightarrow 0$$



    where the functions defined above do the same in the first component and apply identity in the second component. Now I know that there is an isomorphism between $R otimes N$ and $N$ if $N$ is an $R$-module. Then I would obtain something like



    $$cdots longrightarrow mathbb{Z} longrightarrow mathbb{Z} longrightarrow 0$$



    which it may looks easier to find the $n$-th cohomology object of the cochain complex. But I do not know how to find $ker(d_notimes 1)$ neither $operatorname{im}(d_notimes 1)$.






    abstract-algebra commutative-algebra homological-algebra







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$



      I would like to compute $operatorname{Tor}_k^{R}(mathbb{Z},mathbb{Z})$, where $R = mathbb{Z}[x]/(x^n -1)=mathbb Z[X]$.




      I think I am near to do so, but I can not figure out the last step.
      I started looking for a free resolution of $mathbb{Z}$ as an $R$-module, and found that



      $$cdotslongrightarrow R longrightarrow R longrightarrow mathbb{Z} longrightarrow 0$$



      where $d_0: Rlongrightarrow mathbb{Z}: f(X) mapsto f(1)$, and for $igeqslant 1$, $d_{2i-1}: Rlongrightarrow R$ is multiplication by $X-1$ and $d_{2i}: Rlongrightarrow R$ by $1+X+cdots+X^{n-1}$



      In this way I have a free resolution of $mathbb{Z}$ as an $R$-module. Now I can apply the functor $ * otimes mathbb{Z}$ to the free resolution (projective too):



      $$C^*:cdots longrightarrow R otimes mathbb{Z} longrightarrow R otimes mathbb{Z} longrightarrow 0$$



      where the functions defined above do the same in the first component and apply identity in the second component. Now I know that there is an isomorphism between $R otimes N$ and $N$ if $N$ is an $R$-module. Then I would obtain something like



      $$cdots longrightarrow mathbb{Z} longrightarrow mathbb{Z} longrightarrow 0$$



      which it may looks easier to find the $n$-th cohomology object of the cochain complex. But I do not know how to find $ker(d_notimes 1)$ neither $operatorname{im}(d_notimes 1)$.






      abstract-algebra commutative-algebra homological-algebra







      share|cite|improve this question











      $endgroup$





      I would like to compute $operatorname{Tor}_k^{R}(mathbb{Z},mathbb{Z})$, where $R = mathbb{Z}[x]/(x^n -1)=mathbb Z[X]$.




      I think I am near to do so, but I can not figure out the last step.
      I started looking for a free resolution of $mathbb{Z}$ as an $R$-module, and found that



      $$cdotslongrightarrow R longrightarrow R longrightarrow mathbb{Z} longrightarrow 0$$



      where $d_0: Rlongrightarrow mathbb{Z}: f(X) mapsto f(1)$, and for $igeqslant 1$, $d_{2i-1}: Rlongrightarrow R$ is multiplication by $X-1$ and $d_{2i}: Rlongrightarrow R$ by $1+X+cdots+X^{n-1}$



      In this way I have a free resolution of $mathbb{Z}$ as an $R$-module. Now I can apply the functor $ * otimes mathbb{Z}$ to the free resolution (projective too):



      $$C^*:cdots longrightarrow R otimes mathbb{Z} longrightarrow R otimes mathbb{Z} longrightarrow 0$$



      where the functions defined above do the same in the first component and apply identity in the second component. Now I know that there is an isomorphism between $R otimes N$ and $N$ if $N$ is an $R$-module. Then I would obtain something like



      $$cdots longrightarrow mathbb{Z} longrightarrow mathbb{Z} longrightarrow 0$$



      which it may looks easier to find the $n$-th cohomology object of the cochain complex. But I do not know how to find $ker(d_notimes 1)$ neither $operatorname{im}(d_notimes 1)$.






      abstract-algebra commutative-algebra homological-algebra




      abstract-algebra commutative-algebra homological-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 27 '18 at 16:39









      user26857

      39.3k124183




      39.3k124183










      asked Dec 27 '18 at 15:23









      idriskameniidriskameni

      642319




      642319






















          1 Answer
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          3












          $begingroup$

          Note that the isomorphism $Rotimes_R mathbb Zto mathbb Z $ is induced from the left action, so that it sends $rotimes nto rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:mathbb Ztomathbb Z$, while multiplication by $1+cdots+x^{n-1}$ goes to multiplication by $n:mathbb Ztomathbb Z$. Can you continue?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
            $endgroup$
            – idriskameni
            Dec 27 '18 at 15:43






          • 1




            $begingroup$
            @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 27 '18 at 15:47













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          1 Answer
          1






          active

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          active

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          active

          oldest

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          3












          $begingroup$

          Note that the isomorphism $Rotimes_R mathbb Zto mathbb Z $ is induced from the left action, so that it sends $rotimes nto rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:mathbb Ztomathbb Z$, while multiplication by $1+cdots+x^{n-1}$ goes to multiplication by $n:mathbb Ztomathbb Z$. Can you continue?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
            $endgroup$
            – idriskameni
            Dec 27 '18 at 15:43






          • 1




            $begingroup$
            @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 27 '18 at 15:47


















          3












          $begingroup$

          Note that the isomorphism $Rotimes_R mathbb Zto mathbb Z $ is induced from the left action, so that it sends $rotimes nto rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:mathbb Ztomathbb Z$, while multiplication by $1+cdots+x^{n-1}$ goes to multiplication by $n:mathbb Ztomathbb Z$. Can you continue?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
            $endgroup$
            – idriskameni
            Dec 27 '18 at 15:43






          • 1




            $begingroup$
            @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 27 '18 at 15:47
















          3












          3








          3





          $begingroup$

          Note that the isomorphism $Rotimes_R mathbb Zto mathbb Z $ is induced from the left action, so that it sends $rotimes nto rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:mathbb Ztomathbb Z$, while multiplication by $1+cdots+x^{n-1}$ goes to multiplication by $n:mathbb Ztomathbb Z$. Can you continue?






          share|cite|improve this answer









          $endgroup$



          Note that the isomorphism $Rotimes_R mathbb Zto mathbb Z $ is induced from the left action, so that it sends $rotimes nto rn$. If $r$ is a polynomial, then the action is given by evaluation at $1$, that is, $rn = r(1)n$. Under this map, multiplication by $x-1$ goes to the zero map $0:mathbb Ztomathbb Z$, while multiplication by $1+cdots+x^{n-1}$ goes to multiplication by $n:mathbb Ztomathbb Z$. Can you continue?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 27 '18 at 15:29









          Pedro Tamaroff♦Pedro Tamaroff

          96.9k10153297




          96.9k10153297












          • $begingroup$
            Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
            $endgroup$
            – idriskameni
            Dec 27 '18 at 15:43






          • 1




            $begingroup$
            @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 27 '18 at 15:47




















          • $begingroup$
            Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
            $endgroup$
            – idriskameni
            Dec 27 '18 at 15:43






          • 1




            $begingroup$
            @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
            $endgroup$
            – Pedro Tamaroff♦
            Dec 27 '18 at 15:47


















          $begingroup$
          Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
          $endgroup$
          – idriskameni
          Dec 27 '18 at 15:43




          $begingroup$
          Nice! Then, from what I have understood, for $d_k otimes 1$ where $k$ is odd, $Im(d_kotimes 1)=mathbb{Z}$ and $ker(d_k otimes 1)={0}$. And for the even should be the other way. Am I right?
          $endgroup$
          – idriskameni
          Dec 27 '18 at 15:43




          1




          1




          $begingroup$
          @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 27 '18 at 15:47






          $begingroup$
          @idriskameni The image of multiplication by $n$ is $nmathbb Z$, not $mathbb Z$. It kernel is zero. The image of the zero map is zero, of course, and the kernel is $mathbb Z$. So you get $0$ and $mathbb Z/nmathbb Z$ repeated with period 2.
          $endgroup$
          – Pedro Tamaroff♦
          Dec 27 '18 at 15:47




















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