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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.

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  • $\begingroup$ What about the ring of sequences in $\mathbb Z_2$? Every element has additive order at most $2$ $\endgroup$ – Mathmo123 Jul 3 '14 at 22:17
  • $\begingroup$ yes, but this is a finite commutative ring and has characteristic $=2$ $\endgroup$ – MathMan Jul 3 '14 at 22:19
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    $\begingroup$ The ring of sequences.... i.e. (1,0,1,.....). Not finite. (I mean $\mathbb Z_2^{\mathbb N}$). The ring does have zero divisors. But it means that you need to think a bit more before you say "there must surely exist an element whose additive order is $\infty$" $\endgroup$ – Mathmo123 Jul 3 '14 at 22:19
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    $\begingroup$ In an infinite commutative ring without zero divisors, it is still possible that all elements have finite additive order. What can you say about the additive order, if there are no zero divisors? $\endgroup$ – Daniel Fischer Jul 3 '14 at 22:23
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    $\begingroup$ My counter example shows that your claim - that every infinite group has an element of infinite order - is incorrect. The problem is that you haven't used the condition that R has no zero divisors $\endgroup$ – Mathmo123 Jul 3 '14 at 22:25
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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?

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    $\begingroup$ Uhm .. $n \cdot (ab) = (n \cdot a)b = 0 = a(n \cdot b) \neq 0$ Hence, $O(a)$ must equal $O(b)$ . Did you mean this? $\endgroup$ – MathMan Jul 3 '14 at 22:42
  • $\begingroup$ Yep that's what I meant. $\endgroup$ – Mathmo123 Jul 3 '14 at 23:00
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Hint: either the characteristic of $R$ is $0$ or it is a prime number.


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.

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  • $\begingroup$ But, 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? $\endgroup$ – MathMan Jul 3 '14 at 22:38

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