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
          1






          active

          oldest

          votes


















          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













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054043%2fcomputing-operatornametor-r-mathbbz-mathbbz-for-r-mathbb-zc-n%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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




















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3054043%2fcomputing-operatornametor-r-mathbbz-mathbbz-for-r-mathbb-zc-n%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more