If 2 is a unit in a ring R, then there exists a bijection between the idempotents of the ring and the...












3












$begingroup$


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







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$endgroup$








  • 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
















3












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














3












3








3





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






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oldest

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1












$begingroup$


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.






share|cite|improve this answer











$endgroup$









  • 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












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1 Answer
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1












$begingroup$


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.






share|cite|improve this answer











$endgroup$









  • 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
















1












$begingroup$


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.






share|cite|improve this answer











$endgroup$









  • 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














1












1








1





$begingroup$


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.






share|cite|improve this answer











$endgroup$




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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 15:08

























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














  • 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


















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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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