If $a$ is an element of some ring and ${a^n; n=0,1,dots}$ is finite, then $a$ is invertible or a zero divisor...











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  • Every nonzero element in a finite ring is either a unit or a zero divisor

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I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.










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marked as duplicate by Bill Dubuque abstract-algebra
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Nov 21 at 14:19


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    What happens for $a=0$?
    – asdq
    Nov 21 at 9:17










  • Though not an exact dupe, many answers in the dupe apply here too.
    – Bill Dubuque
    Nov 21 at 15:15










  • e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
    – Bill Dubuque
    Nov 21 at 15:24

















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  • Every nonzero element in a finite ring is either a unit or a zero divisor

    6 answers




I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.










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marked as duplicate by Bill Dubuque abstract-algebra
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Nov 21 at 14:19


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.











  • 1




    What happens for $a=0$?
    – asdq
    Nov 21 at 9:17










  • Though not an exact dupe, many answers in the dupe apply here too.
    – Bill Dubuque
    Nov 21 at 15:15










  • e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
    – Bill Dubuque
    Nov 21 at 15:24















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This question already has an answer here:




  • Every nonzero element in a finite ring is either a unit or a zero divisor

    6 answers




I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.










share|cite|improve this question
















This question already has an answer here:




  • Every nonzero element in a finite ring is either a unit or a zero divisor

    6 answers




I'm supposed to prove that, in a ring, if $a$ is an element such that the set ${a^n; n=0,1,...}$ is finite, then $a$ is either invertible or a zero divisor. I don't understand how $a$ can not be invertible, since if the set is finite then there has to be an $i$ such that $a^i=1$.





This question already has an answer here:




  • Every nonzero element in a finite ring is either a unit or a zero divisor

    6 answers








abstract-algebra ring-theory






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edited Nov 21 at 14:13









Asaf Karagila

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asked Nov 21 at 9:08









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marked as duplicate by Bill Dubuque abstract-algebra
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Nov 21 at 14:19


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Nov 21 at 14:19


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.










  • 1




    What happens for $a=0$?
    – asdq
    Nov 21 at 9:17










  • Though not an exact dupe, many answers in the dupe apply here too.
    – Bill Dubuque
    Nov 21 at 15:15










  • e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
    – Bill Dubuque
    Nov 21 at 15:24
















  • 1




    What happens for $a=0$?
    – asdq
    Nov 21 at 9:17










  • Though not an exact dupe, many answers in the dupe apply here too.
    – Bill Dubuque
    Nov 21 at 15:15










  • e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
    – Bill Dubuque
    Nov 21 at 15:24










1




1




What happens for $a=0$?
– asdq
Nov 21 at 9:17




What happens for $a=0$?
– asdq
Nov 21 at 9:17












Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 at 15:15




Though not an exact dupe, many answers in the dupe apply here too.
– Bill Dubuque
Nov 21 at 15:15












e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 at 15:24






e.g. In $,Bbb Z/8,$ the positive powers of $2$ being $,2,4,0,0,ldots$ are not $equiv 1pmod{!8},$ being even $= 2^n !+ 8k$ $ $
– Bill Dubuque
Nov 21 at 15:24












2 Answers
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Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.



If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.






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    It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.



    Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.






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      2 Answers
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      2 Answers
      2






      active

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      active

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      active

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      up vote
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      Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.



      If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.






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        Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.



        If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.






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          up vote
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          up vote
          4
          down vote









          Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.



          If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.






          share|cite|improve this answer












          Since the set is finite there exist $m,n$ such that $a^m=a^n$. Without loss of generality assume that $m>n$. If $a$ is not a zero divisor then $0=a^m-a^n=a^n(a^{m-n}-1)$ implies that $a^{m-n}=1$ hence $a$ is invertible.



          If you consider the ring $k[x]/(x^2)$ for example, the powers of $x$ are finite and $x$ is a zero divisor.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 21 at 9:17









          Levent

          3,386825




          3,386825






















              up vote
              1
              down vote













              It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.



              Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.






              share|cite|improve this answer

























                up vote
                1
                down vote













                It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.



                Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.



                  Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.






                  share|cite|improve this answer












                  It is not true that if that set is finite, then $a^i=1$ for some $iinmathbb N$. Take, for instance, the ring of all $2times2$ rational matrices. Then each matrix $M$ of the form$$M=begin{bmatrix}1-d&frac{d-d^2}c\c&dend{bmatrix},$$($cinmathbb{Q}setminus{0}$, $dinmathbb Q$) is such that $M^2=M$. So, the set ${M^n,|,ninmathbb{Z}^+}$ is surely finite. However, there is no natural $i$ such that $M^i=operatorname{Id}$.



                  Of course, since $M^2=M$, you have $Mtimes(M-operatorname{Id})=0$ and therefore $M$ is a zero divisor.







                  share|cite|improve this answer












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                  share|cite|improve this answer










                  answered Nov 21 at 9:17









                  José Carlos Santos

                  141k19111207




                  141k19111207















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