Results 1 to 14 of 14

Math Help - Algebra, Problems For Fun (36)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (36)

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


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

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


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

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


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

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


    4) Challenge: Suppose x^5=x, for all x in a ring with identity R. Use 1) to prove that R is commutative.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Quote Originally Posted by NonCommAlg View Post
    Hint:
    Spoiler:
    It's immediate that r^2 + r is an idempotent, and thus central by 1), for all r \in R. Put r=a+b to show that ab+ba is central for all a,b \in R.
    Why is r^2+r an idempotent? Isnít (r^2+r)^2=r^2+r+2r^3?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post
    Why is r^2+r an idempotent? Isnít (r^2+r)^2=r^2+r+2r^3?
    because in R we have (-1)^4=-1, which gives us 2=0. so 2r^3=0.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Ah! I didnít realize that R was a ring with multiplicative identity.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Right. For the last one, r^4 is idempotent and therefore central. Now

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

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

    and

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

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

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

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

    is central. Hence (4r^3+6r^2+4r)-(4r^3+2r)-3(2r^2)=2r is central. Hence (4r^3+2r)-2r=4r^3 is central and so 8(4r^3)=2^5r^3=2r^3 is central. Hence (2r^3+2r^2+r)-2r^3-2r^2=r is central.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by TheAbstractionist View Post

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

    \implies\ 2r^3+2r^2+r\,=\,-r^4 is central
    we can't divide by 5. so we actually get: 10r^3+10r^2+5r=-5r^4 is central.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member TheAbstractionist's Avatar
    Joined
    Apr 2009
    Posts
    328
    Thanks
    1
    Okay, letís try again.

    10r^3+10r^2+5r=-5r^4 and 10r^3-10r^2+5r=-5r^4<br />
are central, so (10r^3+10r^2+5r)-(10r^3-10r^2+5r)=20r^2 and (10r^3+10r^2+5r)+(10r^3-10r^2+5r)=20r^3+10r are central. So 2(20r^2)=(2^5+8)r^2=10r^2 is central. Moreover

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

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

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

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

    At last! Hence r=5r-4r is central.
    Last edited by TheAbstractionist; July 31st 2009 at 04:04 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 5(4r^3+6r^2+4r)-(20r^3+10r)-3(10r^2)=10r is central.
    or since 30 = 0 in R, we have 5(4r^3 + 6r^2 + 4r) - (20r^3 + 10r)=10r.


    4r^3-6r^2+4r\,=\,-(r+1)^4+r^4+1
    the right hand side should be -(r-1)^4+r^4+1.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    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. C=\{r \in R | rs=sr \forall s \in R\}
    Clearly C is a sub-ring of R

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

    I am having trouble in proving 3r \in C. Any push plz?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by aman_cc View Post

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

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


    I am having trouble in proving 3r \in C. Any push plz?
    1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r. thus 3r=-3r^2 \in C.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by NonCommAlg View Post
    yes.



    1+r=(1+r)^3 = 1+3r+3r^2 + r^3 = 1+3r+3r^2+r. thus 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:

    \forall a,b \in R
    1. a+a = 0
    2. a^2 + a \in C
    3. 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.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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:

    \forall a,b \in R
    1. a+a = 0
    2. a^2 + a \in C
    3. 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 a^2b + ba^2 \in C, for all a,b \in R. thus a^2(a^2b + ba^2)=(a^2b + ba^2)a^2, which gives us: ab=ba.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Super Member
    Joined
    Apr 2009
    Posts
    678
    Thanks
    1
    Quote Originally Posted by NonCommAlg View Post
    so a^2b + ba^2 \in C, for all a,b \in R. thus a^2(a^2b + ba^2)=(a^2b + ba^2)a^2, which gives us: 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 x^n=x?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 x^n=x?
    the result is true for all n. the interesting thing is that n can even change as the element x changes. so, the Jacobson's theorem says:

    if R is a ring with identity such that for every x \in R, there exists an integer n(x) \geq 2 such that x^{n(x)}=x, then 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 R is finite or, more genrally, Artinian, then the theorem is just a quick result

    of Artin-Wedderburn theorem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Algebra, Problems For Fun (35)
    Posted in the Advanced Algebra Forum
    Replies: 7
    Last Post: July 21st 2009, 02:47 AM
  2. Algebra, Problems For Fun (7)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 18th 2009, 06:24 AM
  3. 2 more algebra problems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 5th 2009, 03:18 AM
  4. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: November 26th 2006, 06:42 PM
  5. Few more Algebra problems..
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 20th 2006, 06:54 AM

Search Tags


/mathhelpforum @mathhelpforum