If the square of every element of a ring is in the center, must the ring be commutative?
up vote
8
down vote
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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
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
add a comment |
up vote
8
down vote
favorite
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
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
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
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
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
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
edited Dec 8 at 7:36
Eric Wofsey
177k12202328
177k12202328
asked Dec 8 at 7:15
Sara.T
1449
asked Dec 8 at 7:15
Sara.T
1449
asked Dec 8 at 7:15
Sara.T
1449
1449
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Dec 8 at 7:29
Eric Wofsey
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
|
show 2 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Dec 8 at 7:29
Eric Wofsey
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
|
show 2 more comments
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.
answered Dec 8 at 7:29
Eric Wofsey
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
|
show 2 more comments
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.
answered Dec 8 at 7:29
Eric Wofsey
177k12202328
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.
answered Dec 8 at 7:29
Eric Wofsey
177k12202328
edited Dec 8 at 7:35
answered Dec 8 at 7:29
Eric Wofsey
177k12202328
answered Dec 8 at 7:29
Eric Wofsey
177k12202328
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
|
show 2 more comments
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
|
show 2 more comments
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