Rings in which each element is a sum of $n$ commuting idempotents











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Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring such that every element of $R$ is a sum of $n$ pairwise commuting idempotents. (As usual, the class of nonunital rings includes the class of unital rings.)



In Theorem 1 of https://math.stackexchange.com/a/1258821 (scroll down to Section 2 for the theorem), I have shown that $left( n+1right) !x=0$ for all $xin R$. What else can be said about $R$ ?




Question 1. Is $R$ commutative?




This question has a positive answer for $n leq 1$ (this is Stone's famous theorem that every Boolean ring is commutative).



I would be quite happy with an answer that assumes $R$ to be unital. In fact, I think I have an argument showing that if the answer is positive for all unital rings $R$, then it is also positive for all nonunital rings $R$.



EDIT: Will Sawin seems to have discussed this in https://mathoverflow.net/a/142506 , though somewhat too telegraphically to fully understand (at least for me).



Remark. It is tempting to conjecture that $x^{n+1} = x$ for all $x in R$. And this conjecture indeed holds for $n leq 1$ (obviously) and for $n = 2$ (see Section 1 of https://math.stackexchange.com/a/1258821 for the proof). But it fails for $n = 3$. Indeed, the ring $mathbb{Z} / 4 mathbb{Z}$ has the property that every of its elements is a sum of $3$ pairwise commuting idempotents, but its element $x = 2$ does not satisfy $x^m = x$ for any $m > 1$.




Question 2. Are there integers $a$ and $b$ (depending on $n$ but not on $R$ and $x$) with $a > b > 0$ such that every $x in R$ is guaranteed to satisfy $x^a = x^b$ ?











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    Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring such that every element of $R$ is a sum of $n$ pairwise commuting idempotents. (As usual, the class of nonunital rings includes the class of unital rings.)



    In Theorem 1 of https://math.stackexchange.com/a/1258821 (scroll down to Section 2 for the theorem), I have shown that $left( n+1right) !x=0$ for all $xin R$. What else can be said about $R$ ?




    Question 1. Is $R$ commutative?




    This question has a positive answer for $n leq 1$ (this is Stone's famous theorem that every Boolean ring is commutative).



    I would be quite happy with an answer that assumes $R$ to be unital. In fact, I think I have an argument showing that if the answer is positive for all unital rings $R$, then it is also positive for all nonunital rings $R$.



    EDIT: Will Sawin seems to have discussed this in https://mathoverflow.net/a/142506 , though somewhat too telegraphically to fully understand (at least for me).



    Remark. It is tempting to conjecture that $x^{n+1} = x$ for all $x in R$. And this conjecture indeed holds for $n leq 1$ (obviously) and for $n = 2$ (see Section 1 of https://math.stackexchange.com/a/1258821 for the proof). But it fails for $n = 3$. Indeed, the ring $mathbb{Z} / 4 mathbb{Z}$ has the property that every of its elements is a sum of $3$ pairwise commuting idempotents, but its element $x = 2$ does not satisfy $x^m = x$ for any $m > 1$.




    Question 2. Are there integers $a$ and $b$ (depending on $n$ but not on $R$ and $x$) with $a > b > 0$ such that every $x in R$ is guaranteed to satisfy $x^a = x^b$ ?











    share|cite|improve this question


























      up vote
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      up vote
      1
      down vote

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      3





      Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring such that every element of $R$ is a sum of $n$ pairwise commuting idempotents. (As usual, the class of nonunital rings includes the class of unital rings.)



      In Theorem 1 of https://math.stackexchange.com/a/1258821 (scroll down to Section 2 for the theorem), I have shown that $left( n+1right) !x=0$ for all $xin R$. What else can be said about $R$ ?




      Question 1. Is $R$ commutative?




      This question has a positive answer for $n leq 1$ (this is Stone's famous theorem that every Boolean ring is commutative).



      I would be quite happy with an answer that assumes $R$ to be unital. In fact, I think I have an argument showing that if the answer is positive for all unital rings $R$, then it is also positive for all nonunital rings $R$.



      EDIT: Will Sawin seems to have discussed this in https://mathoverflow.net/a/142506 , though somewhat too telegraphically to fully understand (at least for me).



      Remark. It is tempting to conjecture that $x^{n+1} = x$ for all $x in R$. And this conjecture indeed holds for $n leq 1$ (obviously) and for $n = 2$ (see Section 1 of https://math.stackexchange.com/a/1258821 for the proof). But it fails for $n = 3$. Indeed, the ring $mathbb{Z} / 4 mathbb{Z}$ has the property that every of its elements is a sum of $3$ pairwise commuting idempotents, but its element $x = 2$ does not satisfy $x^m = x$ for any $m > 1$.




      Question 2. Are there integers $a$ and $b$ (depending on $n$ but not on $R$ and $x$) with $a > b > 0$ such that every $x in R$ is guaranteed to satisfy $x^a = x^b$ ?











      share|cite|improve this question















      Let $n$ be a nonnegative integer. Let $R$ be a nonunital ring such that every element of $R$ is a sum of $n$ pairwise commuting idempotents. (As usual, the class of nonunital rings includes the class of unital rings.)



      In Theorem 1 of https://math.stackexchange.com/a/1258821 (scroll down to Section 2 for the theorem), I have shown that $left( n+1right) !x=0$ for all $xin R$. What else can be said about $R$ ?




      Question 1. Is $R$ commutative?




      This question has a positive answer for $n leq 1$ (this is Stone's famous theorem that every Boolean ring is commutative).



      I would be quite happy with an answer that assumes $R$ to be unital. In fact, I think I have an argument showing that if the answer is positive for all unital rings $R$, then it is also positive for all nonunital rings $R$.



      EDIT: Will Sawin seems to have discussed this in https://mathoverflow.net/a/142506 , though somewhat too telegraphically to fully understand (at least for me).



      Remark. It is tempting to conjecture that $x^{n+1} = x$ for all $x in R$. And this conjecture indeed holds for $n leq 1$ (obviously) and for $n = 2$ (see Section 1 of https://math.stackexchange.com/a/1258821 for the proof). But it fails for $n = 3$. Indeed, the ring $mathbb{Z} / 4 mathbb{Z}$ has the property that every of its elements is a sum of $3$ pairwise commuting idempotents, but its element $x = 2$ does not satisfy $x^m = x$ for any $m > 1$.




      Question 2. Are there integers $a$ and $b$ (depending on $n$ but not on $R$ and $x$) with $a > b > 0$ such that every $x in R$ is guaranteed to satisfy $x^a = x^b$ ?








      abstract-algebra ring-theory noncommutative-algebra positive-characteristic boolean-ring






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      edited Nov 25 at 21:32

























      asked Nov 25 at 20:18









      darij grinberg

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          The answer to Question 2 is yes (but for uninteresting reasons).



          Fix $n$. Let $N=(n+1)!$ and consider the ring $T=mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $bar t^a=bar t^b$, where $bar t=t+Kin T/K$.



          Now suppose $R$ satisfies the given condition, and consider any $xin R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=mathbb Z_Noplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/Kto R^1$ with $bar tmapsto x$. Thus $x^a=x^b$.






          share|cite|improve this answer





















          • Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
            – darij grinberg
            Nov 25 at 23:05












          • btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
            – stewbasic
            Nov 25 at 23:07











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          up vote
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          The answer to Question 2 is yes (but for uninteresting reasons).



          Fix $n$. Let $N=(n+1)!$ and consider the ring $T=mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $bar t^a=bar t^b$, where $bar t=t+Kin T/K$.



          Now suppose $R$ satisfies the given condition, and consider any $xin R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=mathbb Z_Noplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/Kto R^1$ with $bar tmapsto x$. Thus $x^a=x^b$.






          share|cite|improve this answer





















          • Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
            – darij grinberg
            Nov 25 at 23:05












          • btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
            – stewbasic
            Nov 25 at 23:07















          up vote
          2
          down vote













          The answer to Question 2 is yes (but for uninteresting reasons).



          Fix $n$. Let $N=(n+1)!$ and consider the ring $T=mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $bar t^a=bar t^b$, where $bar t=t+Kin T/K$.



          Now suppose $R$ satisfies the given condition, and consider any $xin R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=mathbb Z_Noplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/Kto R^1$ with $bar tmapsto x$. Thus $x^a=x^b$.






          share|cite|improve this answer





















          • Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
            – darij grinberg
            Nov 25 at 23:05












          • btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
            – stewbasic
            Nov 25 at 23:07













          up vote
          2
          down vote










          up vote
          2
          down vote









          The answer to Question 2 is yes (but for uninteresting reasons).



          Fix $n$. Let $N=(n+1)!$ and consider the ring $T=mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $bar t^a=bar t^b$, where $bar t=t+Kin T/K$.



          Now suppose $R$ satisfies the given condition, and consider any $xin R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=mathbb Z_Noplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/Kto R^1$ with $bar tmapsto x$. Thus $x^a=x^b$.






          share|cite|improve this answer












          The answer to Question 2 is yes (but for uninteresting reasons).



          Fix $n$. Let $N=(n+1)!$ and consider the ring $T=mathbb Z_N[t]$. Let $K$ be the ideal of $T$ generated by $p(t)=t(t-1)ldots(t-n)$. Since $p(t)$ is monic, $T/K$ is finite. Therefore there exist integers $a>b>0$ with $bar t^a=bar t^b$, where $bar t=t+Kin T/K$.



          Now suppose $R$ satisfies the given condition, and consider any $xin R$. By your linked answer we have $NR=0$ and $p(x)=0$. Let $R^1=mathbb Z_Noplus R$ with multiplication $(n,r)(m,s)=(nm,ns+mr+rs)$. This is a unital ring and $NR^1=0$, so we have obtain a homomorphism $T/Kto R^1$ with $bar tmapsto x$. Thus $x^a=x^b$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 23:03









          stewbasic

          5,7231926




          5,7231926












          • Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
            – darij grinberg
            Nov 25 at 23:05












          • btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
            – stewbasic
            Nov 25 at 23:07


















          • Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
            – darij grinberg
            Nov 25 at 23:05












          • btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
            – stewbasic
            Nov 25 at 23:07
















          Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
          – darij grinberg
          Nov 25 at 23:05






          Nice! Just as a remark for future readers: $mathbb{Z}_N$ means the commutative ring $mathbb{Z} / N mathbb{Z}$ here.
          – darij grinberg
          Nov 25 at 23:05














          btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
          – stewbasic
          Nov 25 at 23:07




          btw we can actually get a homomorphism from $T/langle p(tq(t))mid q(t)in Trangle$. The latter ring may be quite small...
          – stewbasic
          Nov 25 at 23:07


















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