If a ring has no zero-divisors and it contains a nonzero element $b$ such that $b^2 = b$, then show that $b$...











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  • Idempotents in a ring without unity (rng) and no zero divisors.

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Suppose that $mathbb{R}$ is a ring with no zero-divisors and that $mathbb{R}$ contains a nonzero element $b$ such that $b^2 = b$. Show that $b$ is the unity for $mathbb{R}$.




I attempted to show that $ab=a$ as $b$ is the unity for that ring. Now, with that assumption, $$(ab)^2=a^2$$
$$implies abab=a^2$$



Multiplying both sides by $b$ and using $b^2=b$ we get:



$$abab=a^2b=a(ab)$$



Not only can we not take inverse on both sides (as it is not mentioned), but this kinda seems like a circular proof. I proved what I assumed so it is definitely incorrect.



I am unable to come up with any other method. What is the correct proof here?










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Nov 15 at 12:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.



















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    This question already has an answer here:




    • Idempotents in a ring without unity (rng) and no zero divisors.

      1 answer





    Suppose that $mathbb{R}$ is a ring with no zero-divisors and that $mathbb{R}$ contains a nonzero element $b$ such that $b^2 = b$. Show that $b$ is the unity for $mathbb{R}$.




    I attempted to show that $ab=a$ as $b$ is the unity for that ring. Now, with that assumption, $$(ab)^2=a^2$$
    $$implies abab=a^2$$



    Multiplying both sides by $b$ and using $b^2=b$ we get:



    $$abab=a^2b=a(ab)$$



    Not only can we not take inverse on both sides (as it is not mentioned), but this kinda seems like a circular proof. I proved what I assumed so it is definitely incorrect.



    I am unable to come up with any other method. What is the correct proof here?










    share|cite|improve this question













    marked as duplicate by rschwieb abstract-algebra
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    Nov 15 at 12:00


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

















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      up vote
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      down vote

      favorite












      This question already has an answer here:




      • Idempotents in a ring without unity (rng) and no zero divisors.

        1 answer





      Suppose that $mathbb{R}$ is a ring with no zero-divisors and that $mathbb{R}$ contains a nonzero element $b$ such that $b^2 = b$. Show that $b$ is the unity for $mathbb{R}$.




      I attempted to show that $ab=a$ as $b$ is the unity for that ring. Now, with that assumption, $$(ab)^2=a^2$$
      $$implies abab=a^2$$



      Multiplying both sides by $b$ and using $b^2=b$ we get:



      $$abab=a^2b=a(ab)$$



      Not only can we not take inverse on both sides (as it is not mentioned), but this kinda seems like a circular proof. I proved what I assumed so it is definitely incorrect.



      I am unable to come up with any other method. What is the correct proof here?










      share|cite|improve this question














      This question already has an answer here:




      • Idempotents in a ring without unity (rng) and no zero divisors.

        1 answer





      Suppose that $mathbb{R}$ is a ring with no zero-divisors and that $mathbb{R}$ contains a nonzero element $b$ such that $b^2 = b$. Show that $b$ is the unity for $mathbb{R}$.




      I attempted to show that $ab=a$ as $b$ is the unity for that ring. Now, with that assumption, $$(ab)^2=a^2$$
      $$implies abab=a^2$$



      Multiplying both sides by $b$ and using $b^2=b$ we get:



      $$abab=a^2b=a(ab)$$



      Not only can we not take inverse on both sides (as it is not mentioned), but this kinda seems like a circular proof. I proved what I assumed so it is definitely incorrect.



      I am unable to come up with any other method. What is the correct proof here?





      This question already has an answer here:




      • Idempotents in a ring without unity (rng) and no zero divisors.

        1 answer








      abstract-algebra ring-theory






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      asked Nov 15 at 4:24









      Gaurang Tandon

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      Nov 15 at 12:00


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by rschwieb abstract-algebra
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      Nov 15 at 12:00


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          3 Answers
          3






          active

          oldest

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          up vote
          6
          down vote



          accepted










          Let $ain R$ an arbitrary, nonzero element. Then $ba = b^2a implies (b^2-b)a =0 implies b^2-b=0implies b(b-1)=0 implies b=0 text{ or } b=1$, but since we have assumed $bneq 0$ we have $b=1$ (unity in the ring)



          Edit:



          There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.






          share|cite|improve this answer























          • I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
            – Gaurang Tandon
            Nov 15 at 4:30






          • 3




            Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
            – Theo C.
            Nov 15 at 4:31










          • @GaurangTandon when someone teaches something you should make sure they teach you why
            – Prince M
            Nov 15 at 6:43


















          up vote
          2
          down vote













          One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $bnot=0$ we get $ab-a=0$ as desired.



          Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.






          share|cite|improve this answer




























            up vote
            0
            down vote













            Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $ain R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $bne 0$ by hypothesis. Hence, $b=e$.






            share|cite|improve this answer




























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              6
              down vote



              accepted










              Let $ain R$ an arbitrary, nonzero element. Then $ba = b^2a implies (b^2-b)a =0 implies b^2-b=0implies b(b-1)=0 implies b=0 text{ or } b=1$, but since we have assumed $bneq 0$ we have $b=1$ (unity in the ring)



              Edit:



              There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.






              share|cite|improve this answer























              • I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
                – Gaurang Tandon
                Nov 15 at 4:30






              • 3




                Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
                – Theo C.
                Nov 15 at 4:31










              • @GaurangTandon when someone teaches something you should make sure they teach you why
                – Prince M
                Nov 15 at 6:43















              up vote
              6
              down vote



              accepted










              Let $ain R$ an arbitrary, nonzero element. Then $ba = b^2a implies (b^2-b)a =0 implies b^2-b=0implies b(b-1)=0 implies b=0 text{ or } b=1$, but since we have assumed $bneq 0$ we have $b=1$ (unity in the ring)



              Edit:



              There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.






              share|cite|improve this answer























              • I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
                – Gaurang Tandon
                Nov 15 at 4:30






              • 3




                Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
                – Theo C.
                Nov 15 at 4:31










              • @GaurangTandon when someone teaches something you should make sure they teach you why
                – Prince M
                Nov 15 at 6:43













              up vote
              6
              down vote



              accepted







              up vote
              6
              down vote



              accepted






              Let $ain R$ an arbitrary, nonzero element. Then $ba = b^2a implies (b^2-b)a =0 implies b^2-b=0implies b(b-1)=0 implies b=0 text{ or } b=1$, but since we have assumed $bneq 0$ we have $b=1$ (unity in the ring)



              Edit:



              There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.






              share|cite|improve this answer














              Let $ain R$ an arbitrary, nonzero element. Then $ba = b^2a implies (b^2-b)a =0 implies b^2-b=0implies b(b-1)=0 implies b=0 text{ or } b=1$, but since we have assumed $bneq 0$ we have $b=1$ (unity in the ring)



              Edit:



              There actually isn't really any need for the arbitrary element $a$ -- we could just have easily have said $b^2=b implies b^2-b=0$ and worked from there. Sorry for the extraneous detail.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Nov 15 at 4:35

























              answered Nov 15 at 4:29









              Theo C.

              20928




              20928












              • I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
                – Gaurang Tandon
                Nov 15 at 4:30






              • 3




                Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
                – Theo C.
                Nov 15 at 4:31










              • @GaurangTandon when someone teaches something you should make sure they teach you why
                – Prince M
                Nov 15 at 6:43


















              • I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
                – Gaurang Tandon
                Nov 15 at 4:30






              • 3




                Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
                – Theo C.
                Nov 15 at 4:31










              • @GaurangTandon when someone teaches something you should make sure they teach you why
                – Prince M
                Nov 15 at 6:43
















              I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
              – Gaurang Tandon
              Nov 15 at 4:30




              I was taught in a ring if $ab=0$ we cannot assume $a=0$ or $b=0$
              – Gaurang Tandon
              Nov 15 at 4:30




              3




              3




              Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
              – Theo C.
              Nov 15 at 4:31




              Correct! But, in your problem statement you assume that the ring has no zero divisors. So since we have $(b^2-b)a =0$ and we assume $aneq0$, we must have that $b^2-b=0$
              – Theo C.
              Nov 15 at 4:31












              @GaurangTandon when someone teaches something you should make sure they teach you why
              – Prince M
              Nov 15 at 6:43




              @GaurangTandon when someone teaches something you should make sure they teach you why
              – Prince M
              Nov 15 at 6:43










              up vote
              2
              down vote













              One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $bnot=0$ we get $ab-a=0$ as desired.



              Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.






              share|cite|improve this answer

























                up vote
                2
                down vote













                One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $bnot=0$ we get $ab-a=0$ as desired.



                Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.






                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $bnot=0$ we get $ab-a=0$ as desired.



                  Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.






                  share|cite|improve this answer












                  One has to show that $ab=a$ for all $a$. Consider the product $x:=(ab-a)b$. By distributivity $x=ab^2-ab=ab-ab=0$. Since $bnot=0$ we get $ab-a=0$ as desired.



                  Note that it is not necessary to assume that the ring has a multiplicative unit. This follows from the assumptions.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 15 at 4:57









                  Jens Schwaiger

                  1,404127




                  1,404127






















                      up vote
                      0
                      down vote













                      Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $ain R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $bne 0$ by hypothesis. Hence, $b=e$.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $ain R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $bne 0$ by hypothesis. Hence, $b=e$.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $ain R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $bne 0$ by hypothesis. Hence, $b=e$.






                          share|cite|improve this answer












                          Let $e$ be the unit element of $R$. Then $e^2=e$ and $ae=a=ea$ for all $ain R$. Thus $b(e-b) = be-b^2= b-b=0$ and so $e-b = 0$ since $R$ has no zero divisors and $bne 0$ by hypothesis. Hence, $b=e$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 15 at 6:37









                          Wuestenfux

                          2,4991410




                          2,4991410















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