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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.

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    $\begingroup$ Where did you read the claim referenced by "It's said"? What is the definition of characteristic you use? $\endgroup$ – coffeemath Aug 4 at 12:01
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    $\begingroup$ Are you thinking of fields? $\endgroup$ – badjohn Aug 4 at 12:06
  • $\begingroup$ 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)) $\endgroup$ – Parthiv Basu Aug 4 at 12:08
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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.

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  • $\begingroup$ Alright. I added a short proof as well. $\endgroup$ – Magma Aug 4 at 14:36
  • $\begingroup$ Good point. Mentioned. $\endgroup$ – Magma Aug 4 at 19:47

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