If the square of every element of a ring is in the center, must the ring be commutative?

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Let $R$ be a ring with identity such that the square of any element belongs to the center of $R$. Is it necessary true that $R$ is commutative?



(I can show that for any $x,yin R$, $2(xy-yx) =0 $ but I cannot prove commutativity of $R$.)










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    up vote
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    down vote

    favorite
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    Let $R$ be a ring with identity such that the square of any element belongs to the center of $R$. Is it necessary true that $R$ is commutative?



    (I can show that for any $x,yin R$, $2(xy-yx) =0 $ but I cannot prove commutativity of $R$.)










    share|cite|improve this question

























      up vote
      8
      down vote

      favorite
      3









      up vote
      8
      down vote

      favorite
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      3





      Let $R$ be a ring with identity such that the square of any element belongs to the center of $R$. Is it necessary true that $R$ is commutative?



      (I can show that for any $x,yin R$, $2(xy-yx) =0 $ but I cannot prove commutativity of $R$.)










      share|cite|improve this question















      Let $R$ be a ring with identity such that the square of any element belongs to the center of $R$. Is it necessary true that $R$ is commutative?



      (I can show that for any $x,yin R$, $2(xy-yx) =0 $ but I cannot prove commutativity of $R$.)







      abstract-algebra ring-theory






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      edited Dec 8 at 7:36









      Eric Wofsey

      178k12202328




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      asked Dec 8 at 7:15









      Sara.T

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          Here's a counterexample. Consider the $mathbbF_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$. Explicitly, $R$ has $1,x,y,xy,yx$ as a basis and any product of basis elements that would give a word not in this set is $0$. Since $xyneq yx$, $R$ is not commutative.



          I now claim that the square of every element of $R$ is central. Indeed, for an element $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbbF_2$) we have
          $$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central. But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.






          share|cite|improve this answer






















          • Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
            – Ovi
            Dec 8 at 7:34










          • Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
            – Ovi
            Dec 8 at 7:36










          • Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
            – Sara.T
            Dec 8 at 7:45







          • 1




            @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
            – Eric Wofsey
            Dec 8 at 8:11











          • @EricWofsey thanks, I like that.
            – Sara.T
            Dec 8 at 8:24










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          1 Answer
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          1 Answer
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          active

          oldest

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          up vote
          7
          down vote



          accepted










          Here's a counterexample. Consider the $mathbbF_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$. Explicitly, $R$ has $1,x,y,xy,yx$ as a basis and any product of basis elements that would give a word not in this set is $0$. Since $xyneq yx$, $R$ is not commutative.



          I now claim that the square of every element of $R$ is central. Indeed, for an element $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbbF_2$) we have
          $$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central. But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.






          share|cite|improve this answer






















          • Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
            – Ovi
            Dec 8 at 7:34










          • Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
            – Ovi
            Dec 8 at 7:36










          • Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
            – Sara.T
            Dec 8 at 7:45







          • 1




            @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
            – Eric Wofsey
            Dec 8 at 8:11











          • @EricWofsey thanks, I like that.
            – Sara.T
            Dec 8 at 8:24














          up vote
          7
          down vote



          accepted










          Here's a counterexample. Consider the $mathbbF_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$. Explicitly, $R$ has $1,x,y,xy,yx$ as a basis and any product of basis elements that would give a word not in this set is $0$. Since $xyneq yx$, $R$ is not commutative.



          I now claim that the square of every element of $R$ is central. Indeed, for an element $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbbF_2$) we have
          $$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central. But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.






          share|cite|improve this answer






















          • Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
            – Ovi
            Dec 8 at 7:34










          • Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
            – Ovi
            Dec 8 at 7:36










          • Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
            – Sara.T
            Dec 8 at 7:45







          • 1




            @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
            – Eric Wofsey
            Dec 8 at 8:11











          • @EricWofsey thanks, I like that.
            – Sara.T
            Dec 8 at 8:24












          up vote
          7
          down vote



          accepted







          up vote
          7
          down vote



          accepted






          Here's a counterexample. Consider the $mathbbF_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$. Explicitly, $R$ has $1,x,y,xy,yx$ as a basis and any product of basis elements that would give a word not in this set is $0$. Since $xyneq yx$, $R$ is not commutative.



          I now claim that the square of every element of $R$ is central. Indeed, for an element $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbbF_2$) we have
          $$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central. But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.






          share|cite|improve this answer














          Here's a counterexample. Consider the $mathbbF_2$-algebra $R$ generated by two elements $x,y$ modulo relations that $x^2=y^2=0$ and every word of length $3$ formed by $x$ and $y$ is $0$. Explicitly, $R$ has $1,x,y,xy,yx$ as a basis and any product of basis elements that would give a word not in this set is $0$. Since $xyneq yx$, $R$ is not commutative.



          I now claim that the square of every element of $R$ is central. Indeed, for an element $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbbF_2$) we have
          $$r^2 = a^2 + bcxy + bcyx.$$ To show that such an element is central, it suffices to show that $xy+yx$ is central. But this is trivial, since $xy+yx$ annihilates both $x$ and $y$ on both sides.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 8 at 7:35

























          answered Dec 8 at 7:29









          Eric Wofsey

          178k12202328




          178k12202328











          • Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
            – Ovi
            Dec 8 at 7:34










          • Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
            – Ovi
            Dec 8 at 7:36










          • Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
            – Sara.T
            Dec 8 at 7:45







          • 1




            @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
            – Eric Wofsey
            Dec 8 at 8:11











          • @EricWofsey thanks, I like that.
            – Sara.T
            Dec 8 at 8:24
















          • Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
            – Ovi
            Dec 8 at 7:34










          • Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
            – Ovi
            Dec 8 at 7:36










          • Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
            – Sara.T
            Dec 8 at 7:45







          • 1




            @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
            – Eric Wofsey
            Dec 8 at 8:11











          • @EricWofsey thanks, I like that.
            – Sara.T
            Dec 8 at 8:24















          Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
          – Ovi
          Dec 8 at 7:34




          Since you are giving a specific example, I'm assuming that the answer is no, it doesn't have to be commutative?
          – Ovi
          Dec 8 at 7:34












          Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
          – Ovi
          Dec 8 at 7:36




          Ok just got that it's not commutative from your edit, was confused on that before lol since I thought it was commutative.
          – Ovi
          Dec 8 at 7:36












          Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
          – Sara.T
          Dec 8 at 7:45





          Thanks, Out of curiosity, is there a matrix representation for your algebra ? I mean an explicit matrix ring example.
          – Sara.T
          Dec 8 at 7:45





          1




          1




          @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
          – Eric Wofsey
          Dec 8 at 8:11





          @Sara.T: Well, you can just let it act on itself by left multiplication to get a representation by $5times 5$ matrices. If you further mod out $yx$ there is a nice representation by $3times 3$ matrices though, with $x$ being $beginpmatrix 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0endpmatrix$ and $y$ being $beginpmatrix 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0endpmatrix$. They generate the ring of upper-triangular $3times 3$ matrices with constant diagonal.
          – Eric Wofsey
          Dec 8 at 8:11













          @EricWofsey thanks, I like that.
          – Sara.T
          Dec 8 at 8:24




          @EricWofsey thanks, I like that.
          – Sara.T
          Dec 8 at 8:24

















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