Suppose that $R$ is a commutative ring with no zero divisors. Show that all non zero elements of $R$ have the same additive order.

Question

Suppose that $R$ is a commutative ring with no zero divisors. Show that all non zero elements of $R$ have the same additive order.

Attempt: CASE $1$ : When $R$ is finite commutative Ring

Every finite commutative ring with no zero divisors has a unity.

Hence, $\forall~~x \in R,~ n \cdot x = (n \cdot 1) ~x$ where $1$ is the multiplicative identity ( unity)

Hence, the additive order of every element in $R = $ Characteristic of $R$

CASE $2$ : When $R$ is an infinite commutative Ring

In this case, we can't argue that $R$ possesses a unity for sure. Hence, we can't talk about the characteristic as well. How do I proceed ahead ?

Thank you for your help.

Answer 1

Hint: suppose that $a, b \in R$, $a$ has additive order $n$ and $b$ has additive order $m > n$ or $\infty$. Consider the additive order of $ab$. Can you get a contradiction?

Answer 2

Hint: either the characteristic of $R$ is $0$ or it is a prime number.

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Let's not assume $R$ has a unit and is not the zero ring. Consider $$ I=\{n\in\mathbb{Z}:nr=0\text{ for some }r\in R, r\ne0\}. $$ If $n\in I$ and $m\in\mathbb{Z}$, then obviously $mn\in I$. If $m,n\in I$ and $mr=0$, $ns=0$, with $r\ne0$ and $s\ne0$, then $$ (m+n)(rs)=(mr)s+(ns)r=0 $$ and $rs\ne0$. Therefore $m+n\in I$ and we have proved that $I$ is an ideal of $\mathbb{Z}$, because obviously $0\in I$, as $R\ne\{0\}$. Also $1\notin I$.

Thus $I=k\mathbb{Z}$ for a unique $k\ge0$. If $k=ab$ with $1<a<k$ and $1<b<k$, then $ar\ne0$ and $br\ne0$ for any $r\in R$. Let $r\ne0$ with $kr=0$: then $ar\ne0$ and $br\ne0$ because $a,b\notin I$, but $$ (ar)(br)=(ab)r^2=(kr)r=0 $$ which is a contradiction. Thus either $k=0$ or $k$ is a prime.

In the first case $nr\ne0$ for every $n\in\mathbb{Z}$, $n\ne0$, and every $r\in R$, $r\ne0$, so that every nonzero element of $R$ has infinite order.

Suppose $k$ is prime; we want to show that $kr=0$, for every $r\in R$. Assume the contrary and let $kr_0=0$, with $r_0\ne0$, and $kr\ne0$: then $$ 0\ne r_0(kr)=(kr_0)r=0 $$ which is a contradiction.