$(ab)^n = a^{n}b^{n}$ in a unitary ring $A$ implies A commutative?












4















Here is my question :
Let $ninmathbb{N}$ and $A$ be a unitary ring such that for all $a, bin A$, we have $(ab)^{n} = a^{n}b^{n}$. For which value of n, we can affirm A commutative ?





For $n = 2$, I replaced $a$ by $a + 1$ in the identity, then $b$ by $b + 1$ in the new identity obtained for deducing $A$ commutative. I don't know if the same method works for general cases






abstract-algebra ring-theory







share|cite|improve this question




















  • 3




    We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
    – mathnoob
    Nov 27 '18 at 21:22
















4















Here is my question :
Let $ninmathbb{N}$ and $A$ be a unitary ring such that for all $a, bin A$, we have $(ab)^{n} = a^{n}b^{n}$. For which value of n, we can affirm A commutative ?





For $n = 2$, I replaced $a$ by $a + 1$ in the identity, then $b$ by $b + 1$ in the new identity obtained for deducing $A$ commutative. I don't know if the same method works for general cases






abstract-algebra ring-theory







share|cite|improve this question




















  • 3




    We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
    – mathnoob
    Nov 27 '18 at 21:22














4












4








4








Here is my question :
Let $ninmathbb{N}$ and $A$ be a unitary ring such that for all $a, bin A$, we have $(ab)^{n} = a^{n}b^{n}$. For which value of n, we can affirm A commutative ?





For $n = 2$, I replaced $a$ by $a + 1$ in the identity, then $b$ by $b + 1$ in the new identity obtained for deducing $A$ commutative. I don't know if the same method works for general cases






abstract-algebra ring-theory







share|cite|improve this question
















Here is my question :
Let $ninmathbb{N}$ and $A$ be a unitary ring such that for all $a, bin A$, we have $(ab)^{n} = a^{n}b^{n}$. For which value of n, we can affirm A commutative ?





For $n = 2$, I replaced $a$ by $a + 1$ in the identity, then $b$ by $b + 1$ in the new identity obtained for deducing $A$ commutative. I don't know if the same method works for general cases






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 20:58









RAM_3R

527214




527214










asked Nov 27 '18 at 20:54









曾靖國

3868




3868








  • 3




    We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
    – mathnoob
    Nov 27 '18 at 21:22














  • 3




    We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
    – mathnoob
    Nov 27 '18 at 21:22








3




3




We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
– mathnoob
Nov 27 '18 at 21:22




We know that if for any three consecutive number$ n,n+1,n+2$, it is true that $(ab)^n=a^nb^n$,$(ab)^{n+1}=a^{n+1}b^{n+1}$,$(ab)^{n+2}=a^{n+2}b^{n+2}$ then the associated group is commutative.
– mathnoob
Nov 27 '18 at 21:22










1 Answer
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We may assert that $A$ is commutative when $n=2$. Suppose $(ab)^2=a^2b^2$ for any $a,bin A$. Then
$$
a[a,b]b=a(ab-ba)b=a^2b^2-(ab)^2=0
$$
for any $a,bin A$. Hence
$$
a[a,b]=a[a,b](b+1)-a[a,b]b=a[a,b+1](b+1)-a[a,b]b=0
$$
for every $a,bin A$. In turn,
$$
[a,b]=(a+1)[a,b]-a[a,b]=(a+1)[a+1,b]-a[a,b]=0
$$
for any $a,bin A$, i.e. $A$ is commutative.



We cannot make the same assertion when $nge3$. Pick any fixed $nge3$. Then either $n-1$ or $n$ has a prime factor $p>2$. In other words, we must have $n=pk$ or $pk+1$ for some prime number $p>2$. Let $A$ be the set of all $p$-by-$p$ matrices over $GF(p)$ of the form $xI+M$ for some strictly upper triangular matrix $M$. Then $(xI+M)^p=x^pI$. It follows that
$$
[(xI+M)(yI+N)]^p=(xy)^pI=(xI+M)^p(yI+N)^p
$$
for every $xI+M,,yI+Nin A$. Consequently, when $rin{0,1}$,
begin{aligned}
&[(xI+M)(yI+N)]^{pk+r}\
&=[(xI+M)(yI+N)]^{pk}[(xI+M)(yI+N)]^r\
&=(xy)^{pk}[(xI+M)(yI+N)]^r\
&=(xI+M)^{pk}(xI+M)^r(yI+N)^r(yI+N)^{pk}\
&=(xI+M)^{pk+r}(yI+N)^{pk+r}.
end{aligned}
Hence we have exhibited a ring $A$ such that $(ab)^n=a^nb^n$ whenever $a,bin A$. However, as $p>2$, we have $MNne NM$ in general (such as when $M=E_{12}$ and $N=E_{23}$). Therefore our example $A$ is not commutative.






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    1 Answer
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    active

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    2














    We may assert that $A$ is commutative when $n=2$. Suppose $(ab)^2=a^2b^2$ for any $a,bin A$. Then
    $$
    a[a,b]b=a(ab-ba)b=a^2b^2-(ab)^2=0
    $$
    for any $a,bin A$. Hence
    $$
    a[a,b]=a[a,b](b+1)-a[a,b]b=a[a,b+1](b+1)-a[a,b]b=0
    $$
    for every $a,bin A$. In turn,
    $$
    [a,b]=(a+1)[a,b]-a[a,b]=(a+1)[a+1,b]-a[a,b]=0
    $$
    for any $a,bin A$, i.e. $A$ is commutative.



    We cannot make the same assertion when $nge3$. Pick any fixed $nge3$. Then either $n-1$ or $n$ has a prime factor $p>2$. In other words, we must have $n=pk$ or $pk+1$ for some prime number $p>2$. Let $A$ be the set of all $p$-by-$p$ matrices over $GF(p)$ of the form $xI+M$ for some strictly upper triangular matrix $M$. Then $(xI+M)^p=x^pI$. It follows that
    $$
    [(xI+M)(yI+N)]^p=(xy)^pI=(xI+M)^p(yI+N)^p
    $$
    for every $xI+M,,yI+Nin A$. Consequently, when $rin{0,1}$,
    begin{aligned}
    &[(xI+M)(yI+N)]^{pk+r}\
    &=[(xI+M)(yI+N)]^{pk}[(xI+M)(yI+N)]^r\
    &=(xy)^{pk}[(xI+M)(yI+N)]^r\
    &=(xI+M)^{pk}(xI+M)^r(yI+N)^r(yI+N)^{pk}\
    &=(xI+M)^{pk+r}(yI+N)^{pk+r}.
    end{aligned}
    Hence we have exhibited a ring $A$ such that $(ab)^n=a^nb^n$ whenever $a,bin A$. However, as $p>2$, we have $MNne NM$ in general (such as when $M=E_{12}$ and $N=E_{23}$). Therefore our example $A$ is not commutative.






    share|cite|improve this answer




























      2














      We may assert that $A$ is commutative when $n=2$. Suppose $(ab)^2=a^2b^2$ for any $a,bin A$. Then
      $$
      a[a,b]b=a(ab-ba)b=a^2b^2-(ab)^2=0
      $$
      for any $a,bin A$. Hence
      $$
      a[a,b]=a[a,b](b+1)-a[a,b]b=a[a,b+1](b+1)-a[a,b]b=0
      $$
      for every $a,bin A$. In turn,
      $$
      [a,b]=(a+1)[a,b]-a[a,b]=(a+1)[a+1,b]-a[a,b]=0
      $$
      for any $a,bin A$, i.e. $A$ is commutative.



      We cannot make the same assertion when $nge3$. Pick any fixed $nge3$. Then either $n-1$ or $n$ has a prime factor $p>2$. In other words, we must have $n=pk$ or $pk+1$ for some prime number $p>2$. Let $A$ be the set of all $p$-by-$p$ matrices over $GF(p)$ of the form $xI+M$ for some strictly upper triangular matrix $M$. Then $(xI+M)^p=x^pI$. It follows that
      $$
      [(xI+M)(yI+N)]^p=(xy)^pI=(xI+M)^p(yI+N)^p
      $$
      for every $xI+M,,yI+Nin A$. Consequently, when $rin{0,1}$,
      begin{aligned}
      &[(xI+M)(yI+N)]^{pk+r}\
      &=[(xI+M)(yI+N)]^{pk}[(xI+M)(yI+N)]^r\
      &=(xy)^{pk}[(xI+M)(yI+N)]^r\
      &=(xI+M)^{pk}(xI+M)^r(yI+N)^r(yI+N)^{pk}\
      &=(xI+M)^{pk+r}(yI+N)^{pk+r}.
      end{aligned}
      Hence we have exhibited a ring $A$ such that $(ab)^n=a^nb^n$ whenever $a,bin A$. However, as $p>2$, we have $MNne NM$ in general (such as when $M=E_{12}$ and $N=E_{23}$). Therefore our example $A$ is not commutative.






      share|cite|improve this answer


























        2












        2








        2






        We may assert that $A$ is commutative when $n=2$. Suppose $(ab)^2=a^2b^2$ for any $a,bin A$. Then
        $$
        a[a,b]b=a(ab-ba)b=a^2b^2-(ab)^2=0
        $$
        for any $a,bin A$. Hence
        $$
        a[a,b]=a[a,b](b+1)-a[a,b]b=a[a,b+1](b+1)-a[a,b]b=0
        $$
        for every $a,bin A$. In turn,
        $$
        [a,b]=(a+1)[a,b]-a[a,b]=(a+1)[a+1,b]-a[a,b]=0
        $$
        for any $a,bin A$, i.e. $A$ is commutative.



        We cannot make the same assertion when $nge3$. Pick any fixed $nge3$. Then either $n-1$ or $n$ has a prime factor $p>2$. In other words, we must have $n=pk$ or $pk+1$ for some prime number $p>2$. Let $A$ be the set of all $p$-by-$p$ matrices over $GF(p)$ of the form $xI+M$ for some strictly upper triangular matrix $M$. Then $(xI+M)^p=x^pI$. It follows that
        $$
        [(xI+M)(yI+N)]^p=(xy)^pI=(xI+M)^p(yI+N)^p
        $$
        for every $xI+M,,yI+Nin A$. Consequently, when $rin{0,1}$,
        begin{aligned}
        &[(xI+M)(yI+N)]^{pk+r}\
        &=[(xI+M)(yI+N)]^{pk}[(xI+M)(yI+N)]^r\
        &=(xy)^{pk}[(xI+M)(yI+N)]^r\
        &=(xI+M)^{pk}(xI+M)^r(yI+N)^r(yI+N)^{pk}\
        &=(xI+M)^{pk+r}(yI+N)^{pk+r}.
        end{aligned}
        Hence we have exhibited a ring $A$ such that $(ab)^n=a^nb^n$ whenever $a,bin A$. However, as $p>2$, we have $MNne NM$ in general (such as when $M=E_{12}$ and $N=E_{23}$). Therefore our example $A$ is not commutative.






        share|cite|improve this answer














        We may assert that $A$ is commutative when $n=2$. Suppose $(ab)^2=a^2b^2$ for any $a,bin A$. Then
        $$
        a[a,b]b=a(ab-ba)b=a^2b^2-(ab)^2=0
        $$
        for any $a,bin A$. Hence
        $$
        a[a,b]=a[a,b](b+1)-a[a,b]b=a[a,b+1](b+1)-a[a,b]b=0
        $$
        for every $a,bin A$. In turn,
        $$
        [a,b]=(a+1)[a,b]-a[a,b]=(a+1)[a+1,b]-a[a,b]=0
        $$
        for any $a,bin A$, i.e. $A$ is commutative.



        We cannot make the same assertion when $nge3$. Pick any fixed $nge3$. Then either $n-1$ or $n$ has a prime factor $p>2$. In other words, we must have $n=pk$ or $pk+1$ for some prime number $p>2$. Let $A$ be the set of all $p$-by-$p$ matrices over $GF(p)$ of the form $xI+M$ for some strictly upper triangular matrix $M$. Then $(xI+M)^p=x^pI$. It follows that
        $$
        [(xI+M)(yI+N)]^p=(xy)^pI=(xI+M)^p(yI+N)^p
        $$
        for every $xI+M,,yI+Nin A$. Consequently, when $rin{0,1}$,
        begin{aligned}
        &[(xI+M)(yI+N)]^{pk+r}\
        &=[(xI+M)(yI+N)]^{pk}[(xI+M)(yI+N)]^r\
        &=(xy)^{pk}[(xI+M)(yI+N)]^r\
        &=(xI+M)^{pk}(xI+M)^r(yI+N)^r(yI+N)^{pk}\
        &=(xI+M)^{pk+r}(yI+N)^{pk+r}.
        end{aligned}
        Hence we have exhibited a ring $A$ such that $(ab)^n=a^nb^n$ whenever $a,bin A$. However, as $p>2$, we have $MNne NM$ in general (such as when $M=E_{12}$ and $N=E_{23}$). Therefore our example $A$ is not commutative.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 29 '18 at 3:45

























        answered Nov 29 '18 at 2:51









        William McGonagall

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