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Consider the set $S = \{0,1\} $ under XOR.
It is the case that $ x \oplus x = 0 \ \ \forall x \in S$.
I was wondering if there is a similar operation in some other domain such that $x*x*x = 0$ for all objects in the domain. Can we possibly generalize this?

user308485
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  • Surely there must be some other restrictions you want to put on the operation; otherwise, the constant operation (with result $0$) qualifies. – Brian Tung Jun 08 '16 at 23:28
  • What do you mean by "constant operation" ? And what restrictions do you suggest to make the problem interesting? – user308485 Jun 08 '16 at 23:39
  • I mean, if $x * y = 0$ for all $x, y$, that trivially satisfies your condition, right? But I don't think that's what you were looking for. So I assume you must want some restrictions on the operation that preclude that trivial answer. – Brian Tung Jun 08 '16 at 23:44
  • There is another function over $S$ that satisfy this. Let $00 = 1$ and $xy = 0$ otherwise. – Winther Jun 09 '16 at 00:01
  • In $\mathbb{Z}_6$, $2+2+2 \mod 6 =0$ – Mohammad W. Alomari Jun 09 '16 at 00:09
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    The collection of strictly lower triangular $3 \times 3$ matrices form an algebra which satisfy $xxx = 0$. In general, if $A$ is a strictly triangular $n\times n$ matrix, then $A^n = 0$. – achille hui Jun 09 '16 at 00:30
  • Slightly related example: Let $$W = \left{, \begin{bmatrix}1 & 0 & 0\x & 1 & 0\z & y & 1\end{bmatrix} \mid x,y,z \in \mathbb Z_3 ,\right}.$$ $W$ is a non-Abelian group of order $27$, and every element of $W$ has order $3$. For an Abelian example, take $(\mathbb Z_3)^n$, for any natural number $n$. – M. Vinay Jun 09 '16 at 01:36

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If $a$ is an element in the ring of polynomials $\mathbb{Z}_3[x]$, then $a+a+a=0$. This can be generalized to characteristic $p$, where $p$ is any prime.

Ashwin Ganesan
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  • This is the right answer. The example in the OP is an example of the general case. – Matt Samuel Jun 09 '16 at 03:56
  • In $\mathbb{Z}_n[x]$ ($n \ge 2$), $a+a+\cdots+a$ ($n$ times) is equal to 0. If the coefficient ring $\mathbb{Z}_n$ need not be a field, then $n$ need not be a prime. – Ashwin Ganesan Jun 09 '16 at 05:14