lifting orthogonal idempotents (induction step)












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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







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    $begingroup$


    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






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      edited Dec 14 '18 at 18:08







      mathdude1001

















      asked Dec 14 '18 at 17:29









      mathdude1001mathdude1001

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          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





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          $begingroup$

          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

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          602314








          • 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


















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