Mixing
If $e^2=e$, then $(e+(1-e)re)$ is an idempotent.
$begingroup$
Suppose that in a unital ring $R$, that $e^2=e$, then I want to show that
$(e+(1-e)re)$ is an idempotent (that it's square is equal to itself) for any $ r in R$.
Attempt:
$$(e+(1-r)re)^2=(e+(1-e)re)(e+(1-e)re)
=(ee+e(1-e)re+(1-e)ree+(1-e)re(1-e)(re))$$
And now noting that $(e)(1-e)=(e-e^2)=0$,
$$=(e+(1-e)re)$$
Edit: The original problem had a typo, where I tried to show that $e+(1-r)(re)$ was an idempotent, which is false.
ring-theory
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
$endgroup$
|
show 5 more comments
$begingroup$
Suppose that in a unital ring $R$, that $e^2=e$, then I want to show that
$(e+(1-e)re)$ is an idempotent (that it's square is equal to itself) for any $ r in R$.
Attempt:
$$(e+(1-r)re)^2=(e+(1-e)re)(e+(1-e)re)
=(ee+e(1-e)re+(1-e)ree+(1-e)re(1-e)(re))$$
And now noting that $(e)(1-e)=(e-e^2)=0$,
$$=(e+(1-e)re)$$
Edit: The original problem had a typo, where I tried to show that $e+(1-r)(re)$ was an idempotent, which is false.
ring-theory
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
$endgroup$
$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
1
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
1
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34
|
show 5 more comments
$begingroup$
Suppose that in a unital ring $R$, that $e^2=e$, then I want to show that
$(e+(1-e)re)$ is an idempotent (that it's square is equal to itself) for any $ r in R$.
Attempt:
$$(e+(1-r)re)^2=(e+(1-e)re)(e+(1-e)re)
=(ee+e(1-e)re+(1-e)ree+(1-e)re(1-e)(re))$$
And now noting that $(e)(1-e)=(e-e^2)=0$,
$$=(e+(1-e)re)$$
Edit: The original problem had a typo, where I tried to show that $e+(1-r)(re)$ was an idempotent, which is false.
ring-theory
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
$endgroup$
Suppose that in a unital ring $R$, that $e^2=e$, then I want to show that
$(e+(1-e)re)$ is an idempotent (that it's square is equal to itself) for any $ r in R$.
Attempt:
$$(e+(1-r)re)^2=(e+(1-e)re)(e+(1-e)re)
=(ee+e(1-e)re+(1-e)ree+(1-e)re(1-e)(re))$$
And now noting that $(e)(1-e)=(e-e^2)=0$,
$$=(e+(1-e)re)$$
Edit: The original problem had a typo, where I tried to show that $e+(1-r)(re)$ was an idempotent, which is false.
ring-theory
ring-theory
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
edited Jan 20 at 2:18
IntegrateThis
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
asked Jan 19 at 23:01
IntegrateThisIntegrateThis
1,9111718
1,9111718
$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
1
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
1
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34
|
show 5 more comments
$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
1
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
1
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34
$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
1
1
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
1
1
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34
|
show 5 more comments
3 Answers
3
active
oldest
votes
$begingroup$
It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.
But suppose we look at $e + (1 - e)re$; we have
$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$
$= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! tag 1$
So $e + (1 - e)re$ is in fact idempotent; we have used
$e^2 = e tag 2$
in deriving (2), since it is equivalent to
$(1 - e)e = e(1 - e) = e - e^2 = 0. tag 3$
I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will
$e + (1 - e)re ne e? tag 4$
well, if and only if
$(1 - e)re ne 0, tag 5$
which is to say
$re ne ere; tag 6$
so if we want to generate new idempotents from old, we need find those $r in R$ such that (6) binds.
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
$endgroup$
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
add a comment |
$begingroup$
The claim is not true in $mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have
$$ e+(1-r)re = 1 + r -r^2 $$
and the polynomial on the RHS clearly doesn't take only idempotents as values.
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
add a comment |
$begingroup$
I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then
$$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.
But suppose we look at $e + (1 - e)re$; we have
$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$
$= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! tag 1$
So $e + (1 - e)re$ is in fact idempotent; we have used
$e^2 = e tag 2$
in deriving (2), since it is equivalent to
$(1 - e)e = e(1 - e) = e - e^2 = 0. tag 3$
I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will
$e + (1 - e)re ne e? tag 4$
well, if and only if
$(1 - e)re ne 0, tag 5$
which is to say
$re ne ere; tag 6$
so if we want to generate new idempotents from old, we need find those $r in R$ such that (6) binds.
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
$endgroup$
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
add a comment |
$begingroup$
It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.
But suppose we look at $e + (1 - e)re$; we have
$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$
$= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! tag 1$
So $e + (1 - e)re$ is in fact idempotent; we have used
$e^2 = e tag 2$
in deriving (2), since it is equivalent to
$(1 - e)e = e(1 - e) = e - e^2 = 0. tag 3$
I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will
$e + (1 - e)re ne e? tag 4$
well, if and only if
$(1 - e)re ne 0, tag 5$
which is to say
$re ne ere; tag 6$
so if we want to generate new idempotents from old, we need find those $r in R$ such that (6) binds.
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
$endgroup$
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
add a comment |
$begingroup$
It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.
But suppose we look at $e + (1 - e)re$; we have
$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$
$= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! tag 1$
So $e + (1 - e)re$ is in fact idempotent; we have used
$e^2 = e tag 2$
in deriving (2), since it is equivalent to
$(1 - e)e = e(1 - e) = e - e^2 = 0. tag 3$
I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will
$e + (1 - e)re ne e? tag 4$
well, if and only if
$(1 - e)re ne 0, tag 5$
which is to say
$re ne ere; tag 6$
so if we want to generate new idempotents from old, we need find those $r in R$ such that (6) binds.
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
$endgroup$
It is clear from Henning Makholm's answer, and also the comments, that the assertion that $e + (1 - r)re$ is idempotent, given that $e^2 = e$, is false.
But suppose we look at $e + (1 - e)re$; we have
$(e + (1 - e)re)^2 = (e + (1 - e)re)(e + (1 - e)re)$
$= e^2 + e(1 - e)re + (1 - e)re^2 + (1 - e)re(1 - e)re = e + (1 -e)re! tag 1$
So $e + (1 - e)re$ is in fact idempotent; we have used
$e^2 = e tag 2$
in deriving (2), since it is equivalent to
$(1 - e)e = e(1 - e) = e - e^2 = 0. tag 3$
I think this is a useful result since it allows us to construct potentially many idempotents from $e$. When will
$e + (1 - e)re ne e? tag 4$
well, if and only if
$(1 - e)re ne 0, tag 5$
which is to say
$re ne ere; tag 6$
so if we want to generate new idempotents from old, we need find those $r in R$ such that (6) binds.
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
answered Jan 19 at 23:28
Robert LewisRobert Lewis
47.5k23067
47.5k23067
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
add a comment |
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
1
1
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
In particular we won't be getting any new idempotents this way if the ring is commutative -- or if only $e$ is central.
$endgroup$
– Henning Makholm
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
$begingroup$
@HenningMakholm; indeed!
$endgroup$
– Robert Lewis
Jan 19 at 23:31
1
1
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
$begingroup$
Thanks for the help, sorry for the typo
$endgroup$
– IntegrateThis
Jan 19 at 23:33
add a comment |
$begingroup$
The claim is not true in $mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have
$$ e+(1-r)re = 1 + r -r^2 $$
and the polynomial on the RHS clearly doesn't take only idempotents as values.
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
add a comment |
$begingroup$
The claim is not true in $mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have
$$ e+(1-r)re = 1 + r -r^2 $$
and the polynomial on the RHS clearly doesn't take only idempotents as values.
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
$endgroup$
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
add a comment |
$begingroup$
The claim is not true in $mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have
$$ e+(1-r)re = 1 + r -r^2 $$
and the polynomial on the RHS clearly doesn't take only idempotents as values.
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
$endgroup$
The claim is not true in $mathbb R$ with $e=1$ (which certainly satisfies $e^2=e$). In that case we have
$$ e+(1-r)re = 1 + r -r^2 $$
and the polynomial on the RHS clearly doesn't take only idempotents as values.
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
answered Jan 19 at 23:12
Henning MakholmHenning Makholm
241k17308546
241k17308546
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
add a comment |
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
$begingroup$
I think there's an error in my textbook then, thanks for the help.
$endgroup$
– IntegrateThis
Jan 19 at 23:14
add a comment |
$begingroup$
I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then
$$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then
$$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
$endgroup$
add a comment |
$begingroup$
I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then
$$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
$endgroup$
I concur with Robert Lewis' answer but propose a slightly easier calculation: if $a = e + (1-e)re = (1+r-er)e$, then
$$a^2=(1+r-er)e(1+r-er)e=(1+r-er)(e+er-e^2r)e=(1+r-er)e^2=(1+r-er)e=a$$
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
answered Jan 19 at 23:44
Maestro13Maestro13
1,081724
1,081724
add a comment |
add a comment |
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$begingroup$
What is your "$r$" here?
$endgroup$
– Sorin Tirc
Jan 19 at 23:06
$begingroup$
@SorinTirc: The goal is "for any $rin R$".
$endgroup$
– Henning Makholm
Jan 19 at 23:09
$begingroup$
@HenningMakholm: I strongly doubt that such a relation holds in any unitaly ring. Just take a ring of matrices, does it seem likely?
$endgroup$
– Sorin Tirc
Jan 19 at 23:11
1
$begingroup$
You might consider editing your post in that case!
$endgroup$
– Robert Lewis
Jan 19 at 23:30
1
$begingroup$
@RobertLewis sorry was eating dinner. My bad everybody, my mistake.
$endgroup$
– IntegrateThis
Jan 19 at 23:34