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Math Help - Algebra, Problems For Fun (38)

  1. #1
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    Algebra, Problems For Fun (38)

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

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

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

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

    \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 a \leq 2^n then either \binom{2^n}{a} is even with 0 < a < 2^n or a \in \{0, 2^n\} and \binom{2^n}{a}=1.

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

    Therefore, (x^2-x)^{2^n+1} = (x^{2^n+2}+2r_1x^{2^n+1}+2r_2x^{2^n} + \ldots + 2r_{n-1}x^{3} -x^2) - (x^{2^n+1}+2r_1x^{2^n}+2r_2x^{2^n-1} + \ldots + 2r_{n-1}x^{2} -x). Inserting this into our ring we see that this is actually just equal to (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.
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  3. #3
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    Quote Originally Posted by Swlabr View Post

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

    remember, we're also given that 2x=0, for all x \in R.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    that's not correct! we have (x^{2}-x)^{2^n+1} = (x^2- x)(x^2-x)^{2^n}. besides R doesn't necessarily have 1, as i mentioned in the problem. your idea is correct though and it needs a little fixing.

    remember, we're also given that 2x=0, for all x \in R.
    I was unsure if this was valid or not - surely (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 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 x^{2^n+1} at the start, not just an x. Although because everything becomes zero it doesn't really matter.
    Last edited by Swlabr; August 4th 2009 at 10:28 PM.
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    i probably wasn't very clear in my previous post. in your proof you said (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

    (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 n > 0. as you mentioned 2 \mid \binom{2^n}{k}, for all 0 < k < 2^n. therefore, since 2r=0 for all

    r \in R, we have:

    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})

    =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.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    i probably wasn't very clear in my previous post. in your proof you said (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

    (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 n > 0. as you mentioned 2 \mid \binom{2^n}{k}, for all 0 < k < 2^n. therefore, since 2r=0 for all

    r \in R, we have:

    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})

    =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...
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