Additive group of local rings Announcing the arrival of Valued Associate #679: Cesar Manara ...



Additive group of local rings



Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Krull's intersection theorem for commutative local not necessarily noetherian ringsIs the class of additive groups of rings axiomatizable?Characterization of non-commutative local rings of orders 64 and 128Local rings with simple radicalA question on local ringsIs every commutative group structure underlying at least one (unitary, commutative) ring structureProjecting solutions of Hermitian forms over local ringsautomorphisms of local rings vs local change of coordinatesQuotients of rings with finite free additive groupWhen is a zero dimensional local ring a chain ring?












4












$begingroup$


Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






    ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings







      share|cite|improve this question









      $endgroup$




      Is there a theory or characterization for those finite $p$-groups that can be considered as the additive group of a finite local commutative ring with identity?






      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings




      ac.commutative-algebra ra.rings-and-algebras abelian-groups local-rings






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 6 hours ago









      Lisa_KLisa_K

      604




      604






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            5 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            5 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            4 hours ago






          • 1




            $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            4 hours ago












          Your Answer








          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "504"
          };
          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%2fmathoverflow.net%2fquestions%2f329877%2fadditive-group-of-local-rings%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









          5












          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            5 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            5 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            4 hours ago






          • 1




            $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            4 hours ago
















          5












          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            5 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            5 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            4 hours ago






          • 1




            $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            4 hours ago














          5












          5








          5





          $begingroup$

          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.






          share|cite|improve this answer











          $endgroup$



          Any nonzero finite abelian $ p $ - group is the additive group of a commutative local ring.



          Proof : Let $ G $ be such a group, by the structure theorem you can write it as $ mathbb{Z}/p^k mathbb{Z} times M $ where $ p^k = exp (G) $. Then $ M $ naturally has the structure of a $ mathbb{Z}/p^k mathbb{Z} $ - module.



          Now define a multiplication on $ G $ by $ (x, m)(y, n) := (xy, ym + xn) $. Clearly $ (1,0) $ is a unity.



          Now let's prove it's local with maximal ideal $ { (x,m) mid p $ divides $x } $.



          Indeed clearly this is a proper ideal, and now if $ p $ doesn't divide $ x $ then $ x $ is invertible modulo $ p^k $, let $ y $ be its inverse. Then $(x,m) (y, -y^2m) = (xy, -xy^2m + ym) = (1,0) $ so $ (x,m) $ is invertible : therefore the complement of our ideal is the set of nonunits, which implies that our ring is local.



          This example can be generalized of course : whenever $ R $ is a ring, $ M $ an $ R $ - module, you can "adjoin" $ M $ as an ideal to $ R $, this is where my construction comes from.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 5 hours ago

























          answered 5 hours ago









          MaxMax

          6191619




          6191619








          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            5 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            5 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            4 hours ago






          • 1




            $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            4 hours ago














          • 1




            $begingroup$
            Precisely every nonzero such group.
            $endgroup$
            – YCor
            5 hours ago










          • $begingroup$
            @YCor : indeed, let me correct that
            $endgroup$
            – Max
            5 hours ago










          • $begingroup$
            What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
            $endgroup$
            – LSpice
            4 hours ago






          • 1




            $begingroup$
            @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
            $endgroup$
            – Max
            4 hours ago








          1




          1




          $begingroup$
          Precisely every nonzero such group.
          $endgroup$
          – YCor
          5 hours ago




          $begingroup$
          Precisely every nonzero such group.
          $endgroup$
          – YCor
          5 hours ago












          $begingroup$
          @YCor : indeed, let me correct that
          $endgroup$
          – Max
          5 hours ago




          $begingroup$
          @YCor : indeed, let me correct that
          $endgroup$
          – Max
          5 hours ago












          $begingroup$
          What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
          $endgroup$
          – LSpice
          4 hours ago




          $begingroup$
          What is the $M$ in $G = mathbb Z/p^kmathbb Z times M$?
          $endgroup$
          – LSpice
          4 hours ago




          1




          1




          $begingroup$
          @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
          $endgroup$
          – Max
          4 hours ago




          $begingroup$
          @LSpice : you can see it as $G/(mathbb{Z}/p^kmathbb{Z})$ for instance; it comes from the structure theorem for finite abelian groups
          $endgroup$
          – Max
          4 hours ago


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to MathOverflow!


          • 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%2fmathoverflow.net%2fquestions%2f329877%2fadditive-group-of-local-rings%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

          “%fieldName is a required field.”, in Magento2 REST API Call for GET Method Type The Next...

          Image
          How seriously should I take size and weight limits of hand luggage? Can a virus destroy the BIOS of a modern computer? Can I hook these wires up to find the connection to a dead outlet? How to keep a dark protagonist who wants to keep his humanity dark? How does a dynamic QR code work? My singleton can be called multiple times How to detect integer overflow in C? Can a PhD from a non-TU9 German university become a professor in a TU9 university? How to travel Japan while expressing milk? Is it a bad idea to plug the other end of ESD strap to wall ground? How to show a landlord what we have in s
          Read more

          How to change City field to a dropdown in Checkout step Magento 2Magento 2 : How to change UI field(s)...

          Image
          Does the US political system, in principle, allow for a no-party system? Is it a Cyclops number? "Nobody" knows! Understanding a proof about a set being closed Rationale to prefer local variables over instance variables? Movie: boy escapes the real world and goes to a fantasy world with big furry trolls Why aren't there more Gauls like Obelix? I reported the illegal activity of my boss to his boss. My boss found out. Now I am being punished. What should I do? Solving Linear Matrix Recurrences What is the generally accepted pronunciation of “topoi”? Is it safe to abruplty remove arduino power
          Read more

          夢乃愛華...

          Image
          1995年出生在世人物日本AV女優S1女優埼玉縣出身人物 日本AV女優S1TeStar千葉縣松戶市立旭町中學校松戶市立松戶高等學校淑德大學S1 body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} ゆめの あいか 夢乃 あいか AV女优 别名 安藤琴乃（本名） [1] 民族 大和族 出生 ( 1995-09-18 ) 1995年9月18日 （ 23歲） 出道地点   日本宮城縣 → 千葉縣 → 埼玉縣 活跃年代 2013年 - 血型 B型 体形（2013年 [2] 数值） 身高／体重 149 厘米 / ― 公斤 三围 79 - 52 - 78 cm 罩杯 G 体形备注 巨乳 计量单位换算 身高／体重 4英尺11英寸 / ― 磅 三围 31 - 20 - 31 英寸 活动 拍摄类型 AV 模特儿类型 水着 其他职业 鯛魚燒店員（求學時期） [1] 業餘攝影會模特兒（求學時期） [3] 教学：Template:AV女优/doc 夢乃愛華 （日语： 夢乃 あいか ／ ゆめの あいか   Yumeno Aika ，1995年9月18日 － [1] ）是日本AV女優，现为S1专属女优，所屬經紀公司為TeStar。 目录 1 簡歷 2 演出作品 2.1 成人影片 2.2 形象影片 2.3 綜藝節目 2.4 網上節目 3 其他產品 3.1 雜誌 3.2 寫真集 3.3 網上寫真 4 相關條目 5 參考資料 6 外部連結 簡歷 夢乃愛華出身於宮城縣，有一位姊姊 [3] ，後與家人移居千葉縣，曾就讀當地的松戶市立旭町中學校及松戶市立松戶高等學校 [1] 。她中學一年級時身高已停止增長，而高中初期胸部罩杯亦只有 B，但隨著「初體驗」和均衡飲食帶來的生理影響，三年級畢業那刻罩杯則升至 F [3] 。
          Read more