Algebra, Problems For Fun (38)

**NonCommAlg** · August 2nd 2009, 06:44 PM

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.

**Swlabr** · August 4th 2009, 02:52 AM

Originally Posted by NonCommAlg

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.

**NonCommAlg** · August 4th 2009, 05:21 PM

Originally Posted by Swlabr

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 $x \in R.$

**Swlabr** · August 4th 2009, 10:13 PM

Originally Posted by NonCommAlg

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.

**NonCommAlg** · August 4th 2009, 10:44 PM

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:

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

$=x^{2^{n+1} + 2} + x^{2^n + 2} - x^{2^{n+1}+1} - x^{2^n + 1}=x^2 +x^2 - x - x = 2x^2-2x = 0.$

**Swlabr** · August 4th 2009, 10:53 PM

Originally Posted by NonCommAlg

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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