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 $R$.)






abstract-algebra ring-theory







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    up vote
    8
    down vote

    favorite
    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 $R$.)






    abstract-algebra ring-theory







    share|cite|improve this question


























      up vote
      8
      down vote

      favorite
      3









      up vote
      8
      down vote

      favorite
      3






      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 $R$.)






      abstract-algebra ring-theory







      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




      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 Dec 8 at 7:36









      Eric Wofsey

      177k12202328




      177k12202328










      asked Dec 8 at 7:15









      Sara.T

      1449




      1449






















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          up vote
          7
          down vote



          accepted










          Here's a counterexample. Consider the $mathbb{F}_2$-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 $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbb{F}_2$) 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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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          up vote
          7
          down vote



          accepted










          Here's a counterexample. Consider the $mathbb{F}_2$-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 $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbb{F}_2$) 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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 $mathbb{F}_2$-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 $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbb{F}_2$) 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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 $mathbb{F}_2$-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 $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbb{F}_2$) 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 $mathbb{F}_2$-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 $r=a+bx+cy+dxy+eyxin R$ ($a,b,c,d,einmathbb{F}_2$) 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

          177k12202328




          177k12202328












          • 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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 $begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0end{pmatrix}$ and $y$ being $begin{pmatrix} 0 & 0 & 0 \ 0 & 0 & 1 \ 0 & 0 & 0end{pmatrix}$. 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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