# characteristic of ring

I’m a bit lost on the characteristic of a ring. It’s said the additive order of every non-zero element is the same.

But, consider ring $$\mathbb{Z}_2\times \mathbb{Z}_{3}$$, element $$(0,1)$$ has order $$3$$, but $$(1,0)$$ has order $$2$$, they are different.

Where is my misunderstanding?

— update:

I’m sorry, I missed the line in the original notes that the ring shall be entire.

• Where did you read the claim referenced by "It's said"? What is the definition of characteristic you use? – coffeemath Aug 4 at 12:01
• Are you thinking of fields? – badjohn Aug 4 at 12:06
• It has to be the least common multiple. I think you are misunderstanding the definition of exponent of a periodic group. Read wiki (en.m.wikipedia.org/wiki/Characteristic_(algebra)) – Parthiv Basu Aug 4 at 12:08

Fields work that way, more generally domains. The additive order of every nonzero element of a domain $$R$$ is the same.
Proof: Let $$n$$ be additive order of $$1_R$$, i.e. the smallest positive integer such that $$n \cdot 1_R = 0_R$$ (note that every ring is a $$\mathbb Z$$-algebra so we can multiply with integers). Then every element $$x\in R$$ with $$x \neq 0_R$$ satisfies $$n \cdot x = n \cdot 1_R \cdot x = 0_R \cdot x = 0_R$$, so the additive order $$k$$ of $$x$$ divides $$n$$. But if $$k < n$$, then $$0_R = k \cdot x = (k \cdot 1_R) \cdot x$$, so $$x$$ is a right zero divisor. But domains do not have zero divisors, so $$k = n$$.
The previous argument assumed that $$1_R$$ has finite order. But if the order of $$1_R$$ is infinite, then a similar argument shows that the order of all nonzero $$x$$ must be infinite.
But if $$R$$ is not a domain, this is not true in general. You've found an example yourself.