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Let $R$ be a ring with unity where

$$x^3=x,\;\;\; \forall x \in R$$

How do I prove that $$x+x+x+x+x+x=0$$

  • 2
    It holds even when $R$ has no unit. In fact the ring must be commutative.2012-12-12
  • 2
    So if $x^n=x$, then multiplication by $(m^n-m)$, with $m$ natural, annihilates the ring element. In this case you you present $6=2^3-2=0$.2013-07-31

3 Answers 3

26

Simply calculate $2 = 1+1 = (1+1)^3 = 8$ so $6 = 0$ (here I use an integer $n$ to mean the unit added to itself $n$ times). Now note that any element added to itself $6$ times is the same as $6$ times that element, which is then $0$.

  • 0
    So easy, I've spent more than 30 minutes working on $x$ and transformation of x. How did it occour you to work on the unity? Is it some sort of reminiscence of similar exercices or have you taught something n particular?2012-12-12
  • 8
    Of course, it is still true in a ring without identity, by computing $x+x=(x+x)^3$.2012-12-12
  • 0
    Indeed, this is a very general idea when working with rings with a unit. The characteristic of such a ring is the smallest number of times you need to add $1$ to itself in order to get $0$ (if no such number exists, we call the characteristic $0$). A unital ring satisfies an equation $nx = 0$ for all $x$ iff the characteristic divides $n$ by a similar argument to the one above.2012-12-12
  • 0
    This also ties in with the concept of the prime subring of such a ring, which is the subring consisting of all elements of the form $1+1+\dots=1$ and which is isomorphic to $\mathbb{Z}/n\mathbb{Z}$ where $n$ is the characteristic of the ring.2012-12-12
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    @Temitope: You have only one tool to use: "$x^3 = x$". So you simply plug in the simplest things you know to get more information out of the tool. $2$ is about the fourth simplest thing you know. $2x$ is about the fifth simplest thing you know if you need something that isn't constant.2012-12-12
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Hint $\rm\,\ \forall x\!: f(x) = 0\:\Rightarrow\:\forall n\in \Bbb Z\!: f(n) = 0\ (in\ R)\:\Rightarrow\: char\, R\mid\, gcd(f(\Bbb Z))$

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    Do you mean $f(n)=0$ in $R$ - that is, really, that $f(n\cdot 1_R)=0$?2012-12-12
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    @Thomas Yes, I've clarified that, thanks.2012-12-12
7

$(x + x)^3 = x^3 + 3x^3 + 3x^3 + x^3$ by the binomial theorem. Now use the condition that $x^3 = x$ for all elements in the ring to conclude that $2x = 8x$, from which the desired conclusion follows.