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Thread: Algebra, Problems For Fun (36)

  1. #1
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    Algebra, Problems For Fun (36)

    Definition: An element $\displaystyle r$ in a ring $\displaystyle R$ is called central if $\displaystyle r$ is in the center of $\displaystyle R,$ i.e. $\displaystyle rs=sr,$ for all $\displaystyle s \in R.$


    1) Suppose $\displaystyle R$ is a ring in which $\displaystyle x^2=0 \Longrightarrow x=0.$ Let $\displaystyle e \in R$ be an idempotent, i.e. $\displaystyle e^2=e.$ Prove that $\displaystyle e$ is central.

    Hint:
    Spoiler:
    Let $\displaystyle r \in R.$ What do you get if you expand $\displaystyle (er-ere)^2$ and $\displaystyle (re-ere)^2$ ?


    2) Use 1) to give a short proof of this very special case of Jacobson's theorem: if $\displaystyle x^3=x,$ for all $\displaystyle x$ in a ring with identity $\displaystyle R,$ then $\displaystyle R$ is commutative.

    Hint:
    Spoiler:
    clearly $\displaystyle r^2$ is idempotent, and thus central by 1), for any $\displaystyle r \in R.$ Show that $\displaystyle 2r$ and $\displaystyle 3r$ are also central.


    3) This time suppose $\displaystyle x^4=x,$ for all $\displaystyle x$ in a ring with identity $\displaystyle R.$ Use 1) to prove that $\displaystyle R$ is commutative.

    Hint:
    Spoiler:
    It's immediate that $\displaystyle r^2 + r$ is an idempotent, and thus central by 1), for all $\displaystyle r \in R.$ Put $\displaystyle r=a+b$ to show that $\displaystyle ab+ba$ is central for all $\displaystyle a,b \in R.$


    4) Challenge: Suppose $\displaystyle x^5=x,$ for all $\displaystyle x$ in a ring with identity $\displaystyle R.$ Use 1) to prove that $\displaystyle R$ is commutative.
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    Hint:
    Spoiler:
    It's immediate that $\displaystyle r^2 + r$ is an idempotent, and thus central by 1), for all $\displaystyle r \in R.$ Put $\displaystyle r=a+b$ to show that $\displaystyle ab+ba$ is central for all $\displaystyle a,b \in R.$
    Why is $\displaystyle r^2+r$ an idempotent? Isnít $\displaystyle (r^2+r)^2=r^2+r+2r^3?$
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    Quote Originally Posted by TheAbstractionist View Post
    Why is $\displaystyle r^2+r$ an idempotent? Isnít $\displaystyle (r^2+r)^2=r^2+r+2r^3?$
    because in $\displaystyle R$ we have $\displaystyle (-1)^4=-1,$ which gives us $\displaystyle 2=0.$ so $\displaystyle 2r^3=0.$
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    Senior Member TheAbstractionist's Avatar
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    Ah! I didnít realize that $\displaystyle R$ was a ring with multiplicative identity.
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    Senior Member TheAbstractionist's Avatar
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    Right. For the last one, $\displaystyle r^4$ is idempotent and therefore central. Now

    $\displaystyle (r+1)^5\,=\,r+1$

    $\displaystyle \implies\ 2r^3+2r^2+r\,=\,-r^4$ is central

    and

    $\displaystyle (r-1)^5\,=\,r-1$

    $\displaystyle \implies\ 2r^3-2r^2+r\,=\,r^4$ is central

    and so $\displaystyle (2r^3+2r^2+r)-(2r^3-2r^2+r)=4r^2$ and $\displaystyle (2r^3+2r^2+r)+(2r^3-2r^2+r)=4r^3+2r$ are central. Hence $\displaystyle 8(4r^2)=2^5r^2=2r^2$ is central. Moreover

    $\displaystyle 4r^3+6r^2+4r\,=\,(r+1)^4-r^4-1$

    is central. Hence $\displaystyle (4r^3+6r^2+4r)-(4r^3+2r)-3(2r^2)=2r$ is central. Hence $\displaystyle (4r^3+2r)-2r=4r^3$ is central and so $\displaystyle 8(4r^3)=2^5r^3=2r^3$ is central. Hence $\displaystyle (2r^3+2r^2+r)-2r^3-2r^2=r$ is central.
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    Quote Originally Posted by TheAbstractionist View Post

    $\displaystyle (r+1)^5\,=\,r+1$

    $\displaystyle \implies\ 2r^3+2r^2+r\,=\,-r^4$ is central
    we can't divide by 5. so we actually get: $\displaystyle 10r^3+10r^2+5r=-5r^4$ is central.
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    Senior Member TheAbstractionist's Avatar
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    Okay, letís try again.

    $\displaystyle 10r^3+10r^2+5r=-5r^4$ and $\displaystyle 10r^3-10r^2+5r=-5r^4
    $ are central, so $\displaystyle (10r^3+10r^2+5r)-(10r^3-10r^2+5r)=20r^2$ and $\displaystyle (10r^3+10r^2+5r)+(10r^3-10r^2+5r)=20r^3+10r$ are central. So $\displaystyle 2(20r^2)=(2^5+8)r^2=10r^2$ is central. Moreover

    $\displaystyle 4r^3+6r^2+4r\,=\,(r+1)^4-r^4-1$

    is central. Hence $\displaystyle 5(4r^3+6r^2+4r)-(20r^3+10r)-3(10r^2)=10r$ is central. Hence $\displaystyle (20r^3+10r)-10r=20r^3$ is central, and so $\displaystyle 10r^3$ is central. Hence $\displaystyle (10r^3+10r^2+5r)-10r^3-10r^2=5r$ is central. Now

    $\displaystyle 4r^3-6r^2+4r\,=\,-(r+1)^4+r^4+1$

    is also central, and so $\displaystyle (4r^3+6r^2+4r)-(4r^3-6r^2+4r)=12r^2$ is central, and so $\displaystyle 12r^2-10r^2=2r^2$ is central. Replacing $\displaystyle r$ by $\displaystyle r+1$ gives that $\displaystyle 2(r+1)^2$ is central. Hence $\displaystyle 2(r+1)^2-2r^2-2=4r$ is central.

    At last! Hence $\displaystyle r=5r-4r$ is central.
    Last edited by TheAbstractionist; Jul 31st 2009 at 04:04 AM.
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    that's a nice work! there are two points, which won't have any effect on your solution:

    Quote Originally Posted by TheAbstractionist View Post

    Hence $\displaystyle 5(4r^3+6r^2+4r)-(20r^3+10r)-3(10r^2)=10r$ is central.
    or since $\displaystyle 30 = 0$ in R, we have $\displaystyle 5(4r^3 + 6r^2 + 4r) - (20r^3 + 10r)=10r.$


    $\displaystyle 4r^3-6r^2+4r\,=\,-(r+1)^4+r^4+1$
    the right hand side should be $\displaystyle -(r-1)^4+r^4+1.$
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    My question is related to:

    2) Use 1) to give a short proof of this very special case of Jacobson's theorem: if for all in a ring with identity then is commutative.
    Hint: Spoiler:
    clearly is idempotent, and thus central by 1), for any Show that and are also central.


    Let center of the ring be C. $\displaystyle C=\{r \in R | rs=sr \forall s \in R\}$
    Clearly C is a sub-ring of R

    Also $\displaystyle 1 , r^2 \in C$ $\displaystyle \forall r \in R$
    To prove $\displaystyle 2r \in C$ $\displaystyle \forall r \in R$
    Consider $\displaystyle (1+r)^2 - 1 - r^2 = 2r$
    From the closure property of sub-ring we get $\displaystyle 2r \in C$
    Is my approach correct?

    I am having trouble in proving $\displaystyle 3r \in C$. Any push plz?
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    Quote Originally Posted by aman_cc View Post

    Let center of the ring be C. $\displaystyle C=\{r \in R | rs=sr \forall s \in R\}$
    Clearly C is a sub-ring of R

    Also $\displaystyle 1 , r^2 \in C$ $\displaystyle \forall r \in R$
    To prove $\displaystyle 2r \in C$ $\displaystyle \forall r \in R$
    Consider $\displaystyle (1+r)^2 - 1 - r^2 = 2r$
    From the closure property of sub-ring we get $\displaystyle 2r \in C$
    Is my approach correct?
    yes.


    I am having trouble in proving $\displaystyle 3r \in C$. Any push plz?
    $\displaystyle 1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r.$ thus $\displaystyle 3r=-3r^2 \in C.$
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    Quote Originally Posted by NonCommAlg View Post
    yes.



    $\displaystyle 1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r.$ thus $\displaystyle 3r=-3r^2 \in C.$
    Thanks. It look so easy when you see it

    I'm now stuck with
    3) This time suppose for all in a ring with identity Use 1) to prove that is commutative.

    Hint: Spoiler:
    It's immediate that is an idempotent, and thus central by 1), for all Put to show that is central for all


    As suggested in the hint - I could prove the following:

    $\displaystyle \forall a,b \in R$
    1. $\displaystyle a+a = 0$
    2. $\displaystyle a^2 + a \in C$
    3. $\displaystyle ab+ba \in C$

    How do I show R is commutative? A push again plz?

    Infact I was trying to show that ab+ba=0. Then using (1) above I will show
    ab = ba and hence will be done. But no success.
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    Quote Originally Posted by aman_cc View Post
    Thanks. It look so easy when you see it

    I'm now stuck with
    3) This time suppose for all in a ring with identity Use 1) to prove that is commutative.

    Hint: Spoiler:
    It's immediate that is an idempotent, and thus central by 1), for all Put to show that is central for all


    As suggested in the hint - I could prove the following:

    $\displaystyle \forall a,b \in R$
    1. $\displaystyle a+a = 0$
    2. $\displaystyle a^2 + a \in C$
    3. $\displaystyle ab+ba \in C$

    How do I show R is commutative? A push again plz?

    Infact I was trying to show that ab+ba=0. Then using (1) above I will show
    ab = ba and hence will be done. But no success.
    so $\displaystyle a^2b + ba^2 \in C,$ for all $\displaystyle a,b \in R.$ thus $\displaystyle a^2(a^2b + ba^2)=(a^2b + ba^2)a^2,$ which gives us: $\displaystyle ab=ba.$
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    Quote Originally Posted by NonCommAlg View Post
    so $\displaystyle a^2b + ba^2 \in C,$ for all $\displaystyle a,b \in R.$ thus $\displaystyle a^2(a^2b + ba^2)=(a^2b + ba^2)a^2,$ which gives us: $\displaystyle ab=ba.$
    Thanks. And now for the last part.

    4) Challenge: Suppose for all in a ring with identity Use 1) to prove that is commutative.

    I completely follow TheAbstractionist proof. I have just two question.

    1. I would never have been able to do this proof. Is there a structured line of attack which you were following here? Or this just a matter of lot of practice and off-course lot of gray-matter?

    2. Do we end at 5? Or this is true in general, where $\displaystyle x^n=x$?
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    Quote Originally Posted by aman_cc View Post
    Thanks. And now for the last part.

    4) Challenge: Suppose for all in a ring with identity Use 1) to prove that is commutative.

    I completely follow TheAbstractionist proof. I have just two question.

    1. I would never have been able to do this proof. Is there a structured line of attack which you were following here? Or this just a matter of lot of practice and off-course lot of gray-matter?
    the idea is to look for idempotents.

    2. Do we end at 5? Or this is true in general, where $\displaystyle x^n=x$?
    the result is true for all $\displaystyle n.$ the interesting thing is that $\displaystyle n$ can even change as the element $\displaystyle x$ changes. so, the Jacobson's theorem says:

    if $\displaystyle R$ is a ring with identity such that for every $\displaystyle x \in R,$ there exists an integer $\displaystyle n(x) \geq 2$ such that $\displaystyle x^{n(x)}=x,$ then $\displaystyle R$ is commutative.

    of course, understanding the proof of this theorem requires a fairly deep knowledge of ring theory. there are, however, special cases of

    this theorem which can be proved quite easily. for example if $\displaystyle R$ is finite or, more genrally, Artinian, then the theorem is just a quick result

    of Artin-Wedderburn theorem.
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