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
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
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add a comment |
$begingroup$
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
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
$endgroup$
add a comment |
$begingroup$
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
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
$endgroup$
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
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
edited Dec 14 '18 at 18:08
mathdude1001
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
asked Dec 14 '18 at 17:29
mathdude1001mathdude1001
182
182
add a comment |
add a comment |
1 Answer
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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
$$e_1 + e_2 + cdots + e_n = 1.$$
answered Dec 14 '18 at 18:15
MeowMeow
602314
$endgroup$
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
add a comment |
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1 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
$$e_1 + e_2 + cdots + e_n = 1.$$
answered Dec 14 '18 at 18:15
MeowMeow
602314
$endgroup$
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
add a comment |
$begingroup$
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
$$e_1 + e_2 + cdots + e_n = 1.$$
answered Dec 14 '18 at 18:15
MeowMeow
602314
$endgroup$
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
add a comment |
$begingroup$
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
$$e_1 + e_2 + cdots + e_n = 1.$$
answered Dec 14 '18 at 18:15
MeowMeow
602314
$endgroup$
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
$$e_1 + e_2 + cdots + e_n = 1.$$
answered Dec 14 '18 at 18:15
MeowMeow
602314
answered Dec 14 '18 at 18:15
MeowMeow
602314
answered Dec 14 '18 at 18:15
MeowMeow
602314
answered Dec 14 '18 at 18:15
MeowMeow
602314
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
add a comment |
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
add a comment |
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