lifting orthogonal idempotents (induction step)












3












$begingroup$


I'm trying to prove (by induction on $n$) that whenever $I$ is a nilpotent ideal of a ring $R$, and $r_1+I,ldots,r_n+I$ form a complete set of orthogonal idempotents in the quotient $R/I$, there exists a complete set of orthogonal idempotents $e_1,ldots, e_nin R$ such that for each $1leq ileq n$, $e_i+I=r_i+I$.



I'm fine with the base case, but am struggling with the inductive step.



This result appears as Corollary 7.5 on page 107 of Etingof's representation theory book, available here: http://www-math.mit.edu/~etingof/replect.pdf. In the (rather terse) proof, they suggest applying the inductive assumption to a particular subring, but I don't understand exactly what they mean. I'm wondering if anyone could explain Etingof's argument, or perhaps suggest a different approach.



Thanks very much.






ideals idempotents







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I'm trying to prove (by induction on $n$) that whenever $I$ is a nilpotent ideal of a ring $R$, and $r_1+I,ldots,r_n+I$ form a complete set of orthogonal idempotents in the quotient $R/I$, there exists a complete set of orthogonal idempotents $e_1,ldots, e_nin R$ such that for each $1leq ileq n$, $e_i+I=r_i+I$.



    I'm fine with the base case, but am struggling with the inductive step.



    This result appears as Corollary 7.5 on page 107 of Etingof's representation theory book, available here: http://www-math.mit.edu/~etingof/replect.pdf. In the (rather terse) proof, they suggest applying the inductive assumption to a particular subring, but I don't understand exactly what they mean. I'm wondering if anyone could explain Etingof's argument, or perhaps suggest a different approach.



    Thanks very much.






    ideals idempotents







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I'm trying to prove (by induction on $n$) that whenever $I$ is a nilpotent ideal of a ring $R$, and $r_1+I,ldots,r_n+I$ form a complete set of orthogonal idempotents in the quotient $R/I$, there exists a complete set of orthogonal idempotents $e_1,ldots, e_nin R$ such that for each $1leq ileq n$, $e_i+I=r_i+I$.



      I'm fine with the base case, but am struggling with the inductive step.



      This result appears as Corollary 7.5 on page 107 of Etingof's representation theory book, available here: http://www-math.mit.edu/~etingof/replect.pdf. In the (rather terse) proof, they suggest applying the inductive assumption to a particular subring, but I don't understand exactly what they mean. I'm wondering if anyone could explain Etingof's argument, or perhaps suggest a different approach.



      Thanks very much.






      ideals idempotents







      share|cite|improve this question











      $endgroup$




      I'm trying to prove (by induction on $n$) that whenever $I$ is a nilpotent ideal of a ring $R$, and $r_1+I,ldots,r_n+I$ form a complete set of orthogonal idempotents in the quotient $R/I$, there exists a complete set of orthogonal idempotents $e_1,ldots, e_nin R$ such that for each $1leq ileq n$, $e_i+I=r_i+I$.



      I'm fine with the base case, but am struggling with the inductive step.



      This result appears as Corollary 7.5 on page 107 of Etingof's representation theory book, available here: http://www-math.mit.edu/~etingof/replect.pdf. In the (rather terse) proof, they suggest applying the inductive assumption to a particular subring, but I don't understand exactly what they mean. I'm wondering if anyone could explain Etingof's argument, or perhaps suggest a different approach.



      Thanks very much.






      ideals idempotents




      ideals idempotents






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 14 '18 at 18:08







      mathdude1001

















      asked Dec 14 '18 at 17:29









      mathdude1001mathdude1001

      182




      182






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $pi_1,...,pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1oplus cdots oplus V_n$. Then $pi_1,...,pi_n$ are idempotents with $pi_1+cdots+pi_n=1$, $V_i=pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $pi_iApi_i$. Note that $pi_2+cdots+pi_n=1-pi_1$ is an idempotent too - it's the projection onto $V_2oplus cdots oplus V_n$ - and the algebra of matrices on $V_2oplus cdots oplus V_n$ is thus $(1-pi_1)A(1-pi_1)$.



          So now you have orthogonal idempotents $overline{e}_1,...,overline{e}_n$ in $A/I$ with $sum overline{e}_i=1$, and have a lift $e_1$ of $overline{e}_1$. Note that $overline{e}_2,...,overline{e}_n$ are ortohogonal idempotents in
          $$(1-overline{e}_1)A/I(1-overline{e}_1) = (1-e_1)A(1-e_1)/I$$ with $sum overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)subseteq A$. Note
          $$A = e_1Ae_1 oplus (1-e_1)A(1-e_1)$$
          and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
          $$e_1 + e_2 + cdots + e_n = 1.$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
            $endgroup$
            – Meow
            Dec 14 '18 at 18:20











          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%2f3039679%2flifting-orthogonal-idempotents-induction-step%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









          0












          $begingroup$

          The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $pi_1,...,pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1oplus cdots oplus V_n$. Then $pi_1,...,pi_n$ are idempotents with $pi_1+cdots+pi_n=1$, $V_i=pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $pi_iApi_i$. Note that $pi_2+cdots+pi_n=1-pi_1$ is an idempotent too - it's the projection onto $V_2oplus cdots oplus V_n$ - and the algebra of matrices on $V_2oplus cdots oplus V_n$ is thus $(1-pi_1)A(1-pi_1)$.



          So now you have orthogonal idempotents $overline{e}_1,...,overline{e}_n$ in $A/I$ with $sum overline{e}_i=1$, and have a lift $e_1$ of $overline{e}_1$. Note that $overline{e}_2,...,overline{e}_n$ are ortohogonal idempotents in
          $$(1-overline{e}_1)A/I(1-overline{e}_1) = (1-e_1)A(1-e_1)/I$$ with $sum overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)subseteq A$. Note
          $$A = e_1Ae_1 oplus (1-e_1)A(1-e_1)$$
          and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
          $$e_1 + e_2 + cdots + e_n = 1.$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
            $endgroup$
            – Meow
            Dec 14 '18 at 18:20
















          0












          $begingroup$

          The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $pi_1,...,pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1oplus cdots oplus V_n$. Then $pi_1,...,pi_n$ are idempotents with $pi_1+cdots+pi_n=1$, $V_i=pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $pi_iApi_i$. Note that $pi_2+cdots+pi_n=1-pi_1$ is an idempotent too - it's the projection onto $V_2oplus cdots oplus V_n$ - and the algebra of matrices on $V_2oplus cdots oplus V_n$ is thus $(1-pi_1)A(1-pi_1)$.



          So now you have orthogonal idempotents $overline{e}_1,...,overline{e}_n$ in $A/I$ with $sum overline{e}_i=1$, and have a lift $e_1$ of $overline{e}_1$. Note that $overline{e}_2,...,overline{e}_n$ are ortohogonal idempotents in
          $$(1-overline{e}_1)A/I(1-overline{e}_1) = (1-e_1)A(1-e_1)/I$$ with $sum overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)subseteq A$. Note
          $$A = e_1Ae_1 oplus (1-e_1)A(1-e_1)$$
          and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
          $$e_1 + e_2 + cdots + e_n = 1.$$






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
            $endgroup$
            – Meow
            Dec 14 '18 at 18:20














          0












          0








          0





          $begingroup$

          The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $pi_1,...,pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1oplus cdots oplus V_n$. Then $pi_1,...,pi_n$ are idempotents with $pi_1+cdots+pi_n=1$, $V_i=pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $pi_iApi_i$. Note that $pi_2+cdots+pi_n=1-pi_1$ is an idempotent too - it's the projection onto $V_2oplus cdots oplus V_n$ - and the algebra of matrices on $V_2oplus cdots oplus V_n$ is thus $(1-pi_1)A(1-pi_1)$.



          So now you have orthogonal idempotents $overline{e}_1,...,overline{e}_n$ in $A/I$ with $sum overline{e}_i=1$, and have a lift $e_1$ of $overline{e}_1$. Note that $overline{e}_2,...,overline{e}_n$ are ortohogonal idempotents in
          $$(1-overline{e}_1)A/I(1-overline{e}_1) = (1-e_1)A(1-e_1)/I$$ with $sum overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)subseteq A$. Note
          $$A = e_1Ae_1 oplus (1-e_1)A(1-e_1)$$
          and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
          $$e_1 + e_2 + cdots + e_n = 1.$$






          share|cite|improve this answer









          $endgroup$



          The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $pi_1,...,pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1oplus cdots oplus V_n$. Then $pi_1,...,pi_n$ are idempotents with $pi_1+cdots+pi_n=1$, $V_i=pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $pi_iApi_i$. Note that $pi_2+cdots+pi_n=1-pi_1$ is an idempotent too - it's the projection onto $V_2oplus cdots oplus V_n$ - and the algebra of matrices on $V_2oplus cdots oplus V_n$ is thus $(1-pi_1)A(1-pi_1)$.



          So now you have orthogonal idempotents $overline{e}_1,...,overline{e}_n$ in $A/I$ with $sum overline{e}_i=1$, and have a lift $e_1$ of $overline{e}_1$. Note that $overline{e}_2,...,overline{e}_n$ are ortohogonal idempotents in
          $$(1-overline{e}_1)A/I(1-overline{e}_1) = (1-e_1)A(1-e_1)/I$$ with $sum overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)subseteq A$. Note
          $$A = e_1Ae_1 oplus (1-e_1)A(1-e_1)$$
          and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
          $$e_1 + e_2 + cdots + e_n = 1.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 18:15









          MeowMeow

          587314




          587314








          • 1




            $begingroup$
            P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
            $endgroup$
            – Meow
            Dec 14 '18 at 18:20














          • 1




            $begingroup$
            P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
            $endgroup$
            – Meow
            Dec 14 '18 at 18:20








          1




          1




          $begingroup$
          P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
          $endgroup$
          – Meow
          Dec 14 '18 at 18:20




          $begingroup$
          P.S. Idempotents give you a clearer way to look at basic things like Bezout's lemma and the eigenspace decomposition of a matrix.
          $endgroup$
          – Meow
          Dec 14 '18 at 18:20


















          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%2f3039679%2flifting-orthogonal-idempotents-induction-step%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