# If $x^3 =x$ then $6x=0$ in a ring

Let $R$ be a ring with unity where

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

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

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It holds even when $R$ has no unit. In fact the ring must be commutative. – PAD Dec 12 '12 at 21:30
So if $x^n=x$, then multiplication by $(m^n-m)$, with $m$ natural, annihilates the ring element. In this case you you present $6=2^3-2=0$. – NikolajK Jul 31 '13 at 8:50

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 $0$.

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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? – Temitope.A Dec 12 '12 at 16:44
Of course, it is still true in a ring without identity, by computing $x+x=(x+x)^3$. – Thomas Andrews Dec 12 '12 at 16:47
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. – Tobias Kildetoft Dec 12 '12 at 16:49
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. – Tobias Kildetoft Dec 12 '12 at 16:50
@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. – Hurkyl Dec 12 '12 at 16:57

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 $f(n\cdot 1_R)=0$? – Thomas Andrews Dec 12 '12 at 20:14
@Thomas Yes, I've clarified that, thanks. – Bill Dubuque Dec 12 '12 at 21:30
Fascinating approach! – Manny Reyes Dec 12 '12 at 21:39
(+1) I feel you are nearby. Like the Sun behind the Dark clouds. :-) – Babak S. Jun 4 '13 at 5:34

$(x + x)^3 = x^3 + 3x^3 + 3x^3 + x^3$ 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.

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