If 2 is a unit in a ring R, then there exists a bijection between the idempotents of the ring and the...
$begingroup$
Suppose $R$ is a ring with unity $1_R$, and suppose $2=1_R+1_R$ is a unit. I am asked to show that the map $sigma: {e:e^{2}=e} rightarrow {u:u^{2}=1},$ $sigma(e)=1-2e$ is a bijection. Here is my attempt: First, the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse. Injectivity is clear. To show surjectivity, suppose $uin R$ with $u^{2}=1$. Consider the element $2^{-1}(1-u)$. $1-2(2^{-1}(1-u))=1-(1-u)=u$, so if we can show $2^{-1}(1-u)$ is an idempotent, we're done. To establish this, note first that since $u$ commutes with $2$ it must also commute with $2^{-1}$. Hence we may write $(2^{-1}(1-u))^{2}=(2^{-1})^{2}(1-u)^{2}=(2^{-1})^{2}(1-2u+u^{2})=(2^{-1})^{2}(2-2u)=2^{-1}(1-u)$, as desired. Is my proof correct? I welcome any comments or criticism.
abstract-algebra proof-verification ring-theory
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
$endgroup$
|
show 3 more comments
$begingroup$
Suppose $R$ is a ring with unity $1_R$, and suppose $2=1_R+1_R$ is a unit. I am asked to show that the map $sigma: {e:e^{2}=e} rightarrow {u:u^{2}=1},$ $sigma(e)=1-2e$ is a bijection. Here is my attempt: First, the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse. Injectivity is clear. To show surjectivity, suppose $uin R$ with $u^{2}=1$. Consider the element $2^{-1}(1-u)$. $1-2(2^{-1}(1-u))=1-(1-u)=u$, so if we can show $2^{-1}(1-u)$ is an idempotent, we're done. To establish this, note first that since $u$ commutes with $2$ it must also commute with $2^{-1}$. Hence we may write $(2^{-1}(1-u))^{2}=(2^{-1})^{2}(1-u)^{2}=(2^{-1})^{2}(1-2u+u^{2})=(2^{-1})^{2}(2-2u)=2^{-1}(1-u)$, as desired. Is my proof correct? I welcome any comments or criticism.
abstract-algebra proof-verification ring-theory
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
$endgroup$
4
$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
4
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
1
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
1
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
4
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44
|
show 3 more comments
$begingroup$
Suppose $R$ is a ring with unity $1_R$, and suppose $2=1_R+1_R$ is a unit. I am asked to show that the map $sigma: {e:e^{2}=e} rightarrow {u:u^{2}=1},$ $sigma(e)=1-2e$ is a bijection. Here is my attempt: First, the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse. Injectivity is clear. To show surjectivity, suppose $uin R$ with $u^{2}=1$. Consider the element $2^{-1}(1-u)$. $1-2(2^{-1}(1-u))=1-(1-u)=u$, so if we can show $2^{-1}(1-u)$ is an idempotent, we're done. To establish this, note first that since $u$ commutes with $2$ it must also commute with $2^{-1}$. Hence we may write $(2^{-1}(1-u))^{2}=(2^{-1})^{2}(1-u)^{2}=(2^{-1})^{2}(1-2u+u^{2})=(2^{-1})^{2}(2-2u)=2^{-1}(1-u)$, as desired. Is my proof correct? I welcome any comments or criticism.
abstract-algebra proof-verification ring-theory
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
$endgroup$
Suppose $R$ is a ring with unity $1_R$, and suppose $2=1_R+1_R$ is a unit. I am asked to show that the map $sigma: {e:e^{2}=e} rightarrow {u:u^{2}=1},$ $sigma(e)=1-2e$ is a bijection. Here is my attempt: First, the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse. Injectivity is clear. To show surjectivity, suppose $uin R$ with $u^{2}=1$. Consider the element $2^{-1}(1-u)$. $1-2(2^{-1}(1-u))=1-(1-u)=u$, so if we can show $2^{-1}(1-u)$ is an idempotent, we're done. To establish this, note first that since $u$ commutes with $2$ it must also commute with $2^{-1}$. Hence we may write $(2^{-1}(1-u))^{2}=(2^{-1})^{2}(1-u)^{2}=(2^{-1})^{2}(1-2u+u^{2})=(2^{-1})^{2}(2-2u)=2^{-1}(1-u)$, as desired. Is my proof correct? I welcome any comments or criticism.
abstract-algebra proof-verification ring-theory
abstract-algebra proof-verification ring-theory
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
asked Dec 19 '18 at 2:10
Alex SangerAlex Sanger
10329
10329
4
$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
4
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
1
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
1
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
4
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44
|
show 3 more comments
4
$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
4
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
1
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
1
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
4
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44
4
4
$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
4
4
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
1
1
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
1
1
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
4
4
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44
|
show 3 more comments
1 Answer
1
active
oldest
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$begingroup$
the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse.
Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."
To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $emapsto 1-2e$ defines a function since it is the composition of functions $emapsto 2e$ and $emapsto 1-e$."
That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.
Otherwise, I think you have covered all the bases well.
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
$endgroup$
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
add a comment |
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$begingroup$
the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse.
Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."
To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $emapsto 1-2e$ defines a function since it is the composition of functions $emapsto 2e$ and $emapsto 1-e$."
That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.
Otherwise, I think you have covered all the bases well.
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
$endgroup$
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
add a comment |
$begingroup$
the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse.
Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."
To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $emapsto 1-2e$ defines a function since it is the composition of functions $emapsto 2e$ and $emapsto 1-e$."
That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.
Otherwise, I think you have covered all the bases well.
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
$endgroup$
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
add a comment |
$begingroup$
the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse.
Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."
To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $emapsto 1-2e$ defines a function since it is the composition of functions $emapsto 2e$ and $emapsto 1-e$."
That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.
Otherwise, I think you have covered all the bases well.
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
$endgroup$
the map is well-defined since if $e$ is an idempotent, $(1-2e)^{2}=1-4e+4e^{2}=1-4e+4e=1$, so the image of any idempotent under $sigma$ is indeed self-inverse.
Hm? That doesn't say anything about the rule being well-defined. It just simply confirms that the relation's codomain lies in the set of self-inverse elements. Useful, but it is not "why the map is well-defined."
To say that it is well-defined, you'd need to start with $e=f$ and conclude that $1-2e=1-2f$. Fortunately, this is easy. You can just say "the rule $emapsto 1-2e$ defines a function since it is the composition of functions $emapsto 2e$ and $emapsto 1-e$."
That the latter two things are functions are inherent in the definition of binary operations of a ring, so there is no question about that.
Otherwise, I think you have covered all the bases well.
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
edited Dec 19 '18 at 15:08
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
answered Dec 19 '18 at 14:50
rschwiebrschwieb
107k12103252
107k12103252
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
add a comment |
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
1
1
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
$begingroup$
Sorry, initially I misinterpreted the adjectives on something and had a different problem a few minutes ago. After re-reading, I changed my mind about the problem I was seeing.
$endgroup$
– rschwieb
Dec 19 '18 at 15:09
1
1
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
$begingroup$
I see, I think I was confused about what "well-defined" meant here. What I wanted to do was emphasize that it may not be immediately apparent that $1-2e$ is its own inverse.
$endgroup$
– Alex Sanger
Dec 19 '18 at 21:28
add a comment |
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$begingroup$
It looks fine. I would avoid saying that something "is clear". If it's easy, just give the proof. In this case, injectivity requires using the invertibility of $2$ (the only condition of the problem) so it's worth spelling it out.
$endgroup$
– Slade
Dec 19 '18 at 2:17
4
$begingroup$
Also, use paragraphs!
$endgroup$
– Slade
Dec 19 '18 at 2:18
1
$begingroup$
I think it looks fine. I think the clear stuff really is clear and would leave it. This is a matter of audience.
$endgroup$
– Randall
Dec 19 '18 at 2:18
1
$begingroup$
Why does $u$ commute with $2$?
$endgroup$
– Chris Custer
Dec 19 '18 at 2:21
4
$begingroup$
@Randall I think that "injectivity follows from the invertibility of $2$" is just as easy to write as "injectivity is clear" and is much more informative regarding the structure of the proof. There's no purpose in obscuring where the conditions are used.
$endgroup$
– Slade
Dec 19 '18 at 2:44