Results 1 to 6 of 6

Thread: Algebra, Problems For Fun (38)

  1. #1
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7

    Algebra, Problems For Fun (38)

    Let $\displaystyle R$ be a ring which is not necessarily commutative or with identity element. Suppose that there exists an integer $\displaystyle n \geq 0$ such that $\displaystyle 2x = 0$ and $\displaystyle x^{2^n +1} = x,$ for all $\displaystyle x \in R.$

    Prove that $\displaystyle x^2=x,$ for all $\displaystyle x \in R,$ and hence $\displaystyle R$ is commutative.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    Let $\displaystyle R$ be a ring which is not necessarily commutative or with identity element. Suppose that there exists an integer $\displaystyle n \geq 0$ such that $\displaystyle 2x = 0$ and $\displaystyle x^{2^n +1} = x,$ for all $\displaystyle x \in R.$

    Prove that $\displaystyle x^2=x,$ for all $\displaystyle x \in R,$ and hence $\displaystyle R$ is commutative.
    Firstly note that $\displaystyle (x^2-x)^{2^n+1} = 0 \Rightarrow x^2-x=0 \Rightarrow x^2=x$ holds. Thus, I shall prove that $\displaystyle (x^2-x)^{2^n+1} = 0$, which will prove the result.

    Let us forget about rings for a moment and look at $\displaystyle (x^2-x)^{2^n+1}$. This is clearly equal to $\displaystyle x(x-1)(x-1)^{2^n}$. So,

    $\displaystyle \binom{2^n}{a} = (1/{a!})2^n(2^n-1)\ldots (2^n-a+1)$. As this is always a whole number and as $\displaystyle a \leq 2^n$ then either $\displaystyle \binom{2^n}{a}$ is even with $\displaystyle 0 < a < 2^n$ or $\displaystyle a \in \{0, 2^n\}$ and $\displaystyle \binom{2^n}{a}=1$.

    Thus, $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x-1)^{2^n} = x(x-1)(x^{2^n}+2r_1x^{2^n-1}+2r_2x^{2^n-2} + \ldots $ $\displaystyle + 2r_{n-1}x^{1} -1)$, $\displaystyle r_i \in \mathbb{Z}$. This expands to $\displaystyle (x^{2^n+2}+2r_1x^{2^n+1}+2r_2x^{2^n} + \ldots $$\displaystyle + 2r_{n-1}x^{3} -x^2) - (x^{2^n+1}+2r_1x^{2^n}+2r_2x^{2^n-1} + \ldots $ $\displaystyle + 2r_{n-1}x^{2} -x)$.

    Therefore, $\displaystyle (x^2-x)^{2^n+1} = (x^{2^n+2}+2r_1x^{2^n+1}+2r_2x^{2^n} + \ldots $ $\displaystyle + 2r_{n-1}x^{3} -x^2) - (x^{2^n+1}+2r_1x^{2^n}+2r_2x^{2^n-1} + \ldots $ $\displaystyle + 2r_{n-1}x^{2} -x)$. Inserting this into our ring we see that this is actually just equal to $\displaystyle (x^{2^n+2} -x^2) - (x^{2^n+1} -x) = x.x^{2^n+1} -x^2 - x^{2^n+1} + x = x^2-x^2-x+x=0$, as required.

    Thus, by Problem 36, every element is central and so the ring is commutative.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

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

    Thus, $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x-1)^{2^n} = .... $
    that's not correct! we have $\displaystyle (x^{2}-x)^{2^n+1} = (x^2- x)(x^2-x)^{2^n}.$ besides $\displaystyle R$ doesn't necessarily have $\displaystyle 1,$ as i mentioned in the problem. your idea is correct though and it needs a little fixing.

    remember, we're also given that $\displaystyle 2x=0,$ for all $\displaystyle x \in R.$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    that's not correct! we have $\displaystyle (x^{2}-x)^{2^n+1} = (x^2- x)(x^2-x)^{2^n}.$ besides $\displaystyle R$ doesn't necessarily have $\displaystyle 1,$ as i mentioned in the problem. your idea is correct though and it needs a little fixing.

    remember, we're also given that $\displaystyle 2x=0,$ for all $\displaystyle x \in R.$
    I was unsure if this was valid or not - surely $\displaystyle (x^2-x)^k$ is just the polynomial? Although I used 1s when I expanded it the expanded version does not use a 1 (I took the time to say when we were exiting and re-entering the realm of our ring). I used 1s in my expansion of it, but if someone had taken the time to expand it properly without removing the $\displaystyle x$-term they would get the same answer (as it is just a polynomial) but their method would still be valid. Thus, although my method is invalid w.r.t. the ring the answer is valid, and thus my solution is still valid?

    EDIT: I have just picked up another problem with my solution. It should be $\displaystyle x^{2^n+1}$ at the start, not just an $\displaystyle x$. Although because everything becomes zero it doesn't really matter.
    Last edited by Swlabr; Aug 4th 2009 at 10:28 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    i probably wasn't very clear in my previous post. in your proof you said $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x-1)^{2^n},$ which is clearly not correct even if we assume the ring has 1. the correct one is

    $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x^2-x)^{2^n}.$ anyway, since there is nothing to prove if n = 0, we may assume that $\displaystyle n > 0.$ as you mentioned $\displaystyle 2 \mid \binom{2^n}{k},$ for all $\displaystyle 0 < k < 2^n.$ therefore, since $\displaystyle 2r=0$ for all

    $\displaystyle r \in R,$ we have:

    $\displaystyle x^2 - x = (x^2 - x)^{2^n + 1} = (x^2 - x)(x^2 - x)^{2^n}=(x^2 - x)(x^{2^{n+1}} + x^{2^n})$

    $\displaystyle =x^{2^{n+1} + 2} + x^{2^n + 2} - x^{2^{n+1}+1} - x^{2^n + 1}=x^2 +x^2 - x - x = 2x^2-2x = 0.$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by NonCommAlg View Post
    i probably wasn't very clear in my previous post. in your proof you said $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x-1)^{2^n},$ which is clearly not correct even if we assume the ring has 1. the correct one is

    $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x^2-x)^{2^n}.$ anyway, since there is nothing to prove if n = 0, we may assume that $\displaystyle n > 0.$ as you mentioned $\displaystyle 2 \mid \binom{2^n}{k},$ for all $\displaystyle 0 < k < 2^n.$ therefore, since $\displaystyle 2r=0$ for all

    $\displaystyle r \in R,$ we have:

    $\displaystyle x^2 - x = (x^2 - x)^{2^n + 1} = (x^2 - x)(x^2 - x)^{2^n}=(x^2 - x)(x^{2^{n+1}} + x^{2^n})$

    $\displaystyle =x^{2^{n+1} + 2} + x^{2^n + 2} - x^{2^{n+1}+1} - x^{2^n + 1}=x^2 +x^2 - x - x = 2x^2-2x = 0.$
    Ah yes - I did notice that problem, but only noticed it (and edited my post to that effect) about 10 mins before you posted...
    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: Jul 21st 2009, 01:47 AM
  2. Algebra, Problems For Fun (7)
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 18th 2009, 05:24 AM
  3. 2 more algebra problems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 5th 2009, 02:18 AM
  4. Need Help With All these Algebra Problems.
    Posted in the Algebra Forum
    Replies: 10
    Last Post: Nov 26th 2006, 05:42 PM
  5. Few more Algebra problems..
    Posted in the Algebra Forum
    Replies: 4
    Last Post: Jun 20th 2006, 05:54 AM

Search Tags


/mathhelpforum @mathhelpforum