Calculating $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ over $mathbb{Z}$












4














Calculating $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ over $mathbb{Z}$:



I just need someone to confirm that I have calculated $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ correctly. I know this is an easy, common calculation, however I cannot find any actual answer to confirm my results against anywhere. Which is very frustrating.



I have calculated: $ operatorname{Ext}^0(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, $ operatorname{Ext}^1(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, and $ operatorname{Ext}^n(mathbb{Z}/2, mathbb{Z}/2) = 0$ for $n geq 2$. Is this correct?



I used the projective resolution:



$...rightarrow 0 rightarrow 0 rightarrow mathbb{Z} = P_1 xrightarrow {cdot 2} mathbb{Z} = P_0 xrightarrow {cdot 1} mathbb{Z}/2$



which yielded Hom groups:



$...leftarrow 0 leftarrow 0 leftarrow operatorname{Hom}(mathbb{Z},mathbb{Z}/2) xleftarrow {cdot 2} operatorname{Hom}(mathbb{Z},mathbb{Z}/2) leftarrow 0$



which is equivalent to:



$...leftarrow 0 leftarrow 0 leftarrow mathbb{Z}/2 xleftarrow {cdot 0} mathbb{Z}/2 leftarrow 0$



The homology at the 0th and 1st points is then $mathbb{Z}/2$, and $0$ elsewhere.






homological-algebra







share|cite|improve this question





























    4














    Calculating $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ over $mathbb{Z}$:



    I just need someone to confirm that I have calculated $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ correctly. I know this is an easy, common calculation, however I cannot find any actual answer to confirm my results against anywhere. Which is very frustrating.



    I have calculated: $ operatorname{Ext}^0(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, $ operatorname{Ext}^1(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, and $ operatorname{Ext}^n(mathbb{Z}/2, mathbb{Z}/2) = 0$ for $n geq 2$. Is this correct?



    I used the projective resolution:



    $...rightarrow 0 rightarrow 0 rightarrow mathbb{Z} = P_1 xrightarrow {cdot 2} mathbb{Z} = P_0 xrightarrow {cdot 1} mathbb{Z}/2$



    which yielded Hom groups:



    $...leftarrow 0 leftarrow 0 leftarrow operatorname{Hom}(mathbb{Z},mathbb{Z}/2) xleftarrow {cdot 2} operatorname{Hom}(mathbb{Z},mathbb{Z}/2) leftarrow 0$



    which is equivalent to:



    $...leftarrow 0 leftarrow 0 leftarrow mathbb{Z}/2 xleftarrow {cdot 0} mathbb{Z}/2 leftarrow 0$



    The homology at the 0th and 1st points is then $mathbb{Z}/2$, and $0$ elsewhere.






    homological-algebra







    share|cite|improve this question



























      4












      4








      4







      Calculating $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ over $mathbb{Z}$:



      I just need someone to confirm that I have calculated $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ correctly. I know this is an easy, common calculation, however I cannot find any actual answer to confirm my results against anywhere. Which is very frustrating.



      I have calculated: $ operatorname{Ext}^0(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, $ operatorname{Ext}^1(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, and $ operatorname{Ext}^n(mathbb{Z}/2, mathbb{Z}/2) = 0$ for $n geq 2$. Is this correct?



      I used the projective resolution:



      $...rightarrow 0 rightarrow 0 rightarrow mathbb{Z} = P_1 xrightarrow {cdot 2} mathbb{Z} = P_0 xrightarrow {cdot 1} mathbb{Z}/2$



      which yielded Hom groups:



      $...leftarrow 0 leftarrow 0 leftarrow operatorname{Hom}(mathbb{Z},mathbb{Z}/2) xleftarrow {cdot 2} operatorname{Hom}(mathbb{Z},mathbb{Z}/2) leftarrow 0$



      which is equivalent to:



      $...leftarrow 0 leftarrow 0 leftarrow mathbb{Z}/2 xleftarrow {cdot 0} mathbb{Z}/2 leftarrow 0$



      The homology at the 0th and 1st points is then $mathbb{Z}/2$, and $0$ elsewhere.






      homological-algebra







      share|cite|improve this question















      Calculating $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ over $mathbb{Z}$:



      I just need someone to confirm that I have calculated $ operatorname{Ext}(mathbb{Z}/2, mathbb{Z}/2)$ correctly. I know this is an easy, common calculation, however I cannot find any actual answer to confirm my results against anywhere. Which is very frustrating.



      I have calculated: $ operatorname{Ext}^0(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, $ operatorname{Ext}^1(mathbb{Z}/2, mathbb{Z}/2) = mathbb{Z}/2$, and $ operatorname{Ext}^n(mathbb{Z}/2, mathbb{Z}/2) = 0$ for $n geq 2$. Is this correct?



      I used the projective resolution:



      $...rightarrow 0 rightarrow 0 rightarrow mathbb{Z} = P_1 xrightarrow {cdot 2} mathbb{Z} = P_0 xrightarrow {cdot 1} mathbb{Z}/2$



      which yielded Hom groups:



      $...leftarrow 0 leftarrow 0 leftarrow operatorname{Hom}(mathbb{Z},mathbb{Z}/2) xleftarrow {cdot 2} operatorname{Hom}(mathbb{Z},mathbb{Z}/2) leftarrow 0$



      which is equivalent to:



      $...leftarrow 0 leftarrow 0 leftarrow mathbb{Z}/2 xleftarrow {cdot 0} mathbb{Z}/2 leftarrow 0$



      The homology at the 0th and 1st points is then $mathbb{Z}/2$, and $0$ elsewhere.






      homological-algebra




      homological-algebra






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Bernard

      118k639112




      118k639112










      asked 2 days ago









      DavenDaven

      36619




      36619






















          1 Answer
          1






          active

          oldest

          votes


















          2














          Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $mathbb Z/2$ by itself are the trivial one and $mathbb Z/4$, up to isomorphism, and this is isomorphic to $mathbb Z/2$, of course. Note, however, that if you try to do the same with $mathbb Z/p$ (the very same computation works), then you get Ext is $mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $mathbb Z/p^2$ which are not-isomorphic. Can you find them?






          share|cite|improve this answer























          • Thank you! Your extra comment will be very helpful for furthering my understanding
            – Daven
            2 days ago











          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%2f3062900%2fcalculating-operatornameext-mathbbz-2-mathbbz-2-over-mathbbz%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









          2














          Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $mathbb Z/2$ by itself are the trivial one and $mathbb Z/4$, up to isomorphism, and this is isomorphic to $mathbb Z/2$, of course. Note, however, that if you try to do the same with $mathbb Z/p$ (the very same computation works), then you get Ext is $mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $mathbb Z/p^2$ which are not-isomorphic. Can you find them?






          share|cite|improve this answer























          • Thank you! Your extra comment will be very helpful for furthering my understanding
            – Daven
            2 days ago
















          2














          Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $mathbb Z/2$ by itself are the trivial one and $mathbb Z/4$, up to isomorphism, and this is isomorphic to $mathbb Z/2$, of course. Note, however, that if you try to do the same with $mathbb Z/p$ (the very same computation works), then you get Ext is $mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $mathbb Z/p^2$ which are not-isomorphic. Can you find them?






          share|cite|improve this answer























          • Thank you! Your extra comment will be very helpful for furthering my understanding
            – Daven
            2 days ago














          2












          2








          2






          Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $mathbb Z/2$ by itself are the trivial one and $mathbb Z/4$, up to isomorphism, and this is isomorphic to $mathbb Z/2$, of course. Note, however, that if you try to do the same with $mathbb Z/p$ (the very same computation works), then you get Ext is $mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $mathbb Z/p^2$ which are not-isomorphic. Can you find them?






          share|cite|improve this answer














          Yes, that's fine. Another way to compute Ext is to note the only two possible extensions of $mathbb Z/2$ by itself are the trivial one and $mathbb Z/4$, up to isomorphism, and this is isomorphic to $mathbb Z/2$, of course. Note, however, that if you try to do the same with $mathbb Z/p$ (the very same computation works), then you get Ext is $mathbb Z/p$: there are $p-1$ nontrivial extensions with the same underlying middle term $mathbb Z/p^2$ which are not-isomorphic. Can you find them?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Pedro Tamaroff♦Pedro Tamaroff

          96.3k10151296




          96.3k10151296












          • Thank you! Your extra comment will be very helpful for furthering my understanding
            – Daven
            2 days ago


















          • Thank you! Your extra comment will be very helpful for furthering my understanding
            – Daven
            2 days ago
















          Thank you! Your extra comment will be very helpful for furthering my understanding
          – Daven
          2 days ago




          Thank you! Your extra comment will be very helpful for furthering my understanding
          – Daven
          2 days ago


















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3062900%2fcalculating-operatornameext-mathbbz-2-mathbbz-2-over-mathbbz%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

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
          Read more

          The Binding of Isaac: Rebirth/Afterbirth

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me
          Read more