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
    $\begingroup$ It holds even when $R$ has no unit. In fact the ring must be commutative. $\endgroup$
    – PAD
    Dec 12, 2012 at 21:30
  • 2
    $\begingroup$ So if $x^n=x$, then multiplication by $(m^n-m)$, with $m$ natural, annihilates the ring element. In this case you you present $6=2^3-2=0$. $\endgroup$
    – Nikolaj-K
    Jul 31, 2013 at 8:50

3 Answers 3


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 $0$.

  • $\begingroup$ 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? $\endgroup$
    – Temitope.A
    Dec 12, 2012 at 16:44
  • 8
    $\begingroup$ Of course, it is still true in a ring without identity, by computing $x+x=(x+x)^3$. $\endgroup$ Dec 12, 2012 at 16:47
  • $\begingroup$ 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. $\endgroup$ Dec 12, 2012 at 16:49
  • $\begingroup$ 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. $\endgroup$ Dec 12, 2012 at 16:50
  • 2
    $\begingroup$ @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. $\endgroup$
    – user14972
    Dec 12, 2012 at 16:57

Hint $\rm\,\ \forall x\!: f(x) = 0\:\Rightarrow\:\forall n\in \Bbb Z\!: f(n) = 0\ (in\ R)\:\Rightarrow\: char\, R\mid\, gcd(f(\Bbb Z))$

  • $\begingroup$ Do you mean $f(n)=0$ in $R$ - that is, really, that $f(n\cdot 1_R)=0$? $\endgroup$ Dec 12, 2012 at 20:14
  • $\begingroup$ @Thomas Yes, I've clarified that, thanks. $\endgroup$ Dec 12, 2012 at 21:30

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.