# Ring with each element finite order but not of finite characteristic

What is example of ring without identity with each element of finite additive order but not of finite characteristic.

Motivation: A ring with identity and having identity of finite additive order is always of finite characteristic

One More Similar Questions: ring without zero divisor with each element of finite additive order but not of finite characteristic.

One More Question: Ring with multiplicative identity with one element of finite additive order(obviously it can't be multiplicative identity of ring) but not of finite characteristic

• How do you define order in a ring? – user207868 Feb 1 '15 at 10:06
• @LeonAragones presumably additive order, since $R$ is an abelian group under addition. – Adam Hughes Feb 1 '15 at 10:09
• @I have made edit also. Sorry have forgotten to mention – Sushil Feb 1 '15 at 10:12

Let

$$R=\bigoplus_{n=1}^\infty \Bbb Z/n\Bbb Z$$

Then $R$ has elements of arbitrarily large order, hence cannot have a finite characteristic. Note that elements of the direct sum only have finitely many coordinates non-zero, so if $N$ is the largest such for $x\in R$, we have $N!x=0$, i.e. the order of $x$ divides $N!$ and is--in particular--finite.

• @Sushil That is not true, the direct sum is not the same as the direct product. – Adam Hughes Feb 1 '15 at 10:12
• @Sushil check out en.wikipedia.org/wiki/Direct_sum for details on how direct sums work and how they contrast with direct products. If you look, you can also see directly that $R$ is non-empty, it contains sequences, which you can explicitly construct, such as $(1,0,0,\ldots)$. – Adam Hughes Feb 1 '15 at 10:15
• @Sushil: The ring written down contains sequences $(a_n)_{n\geq1}$ where $a_n\in\Bbb Z/n\Bbb Z$ and for sufficiently large $n$ the class $a_n$ is the class of $0$. You can compute with these sequences componentwise. It is clearly a non-trivial commutative ring (without unit). – Marc van Leeuwen Feb 1 '15 at 10:21
• @AdamHughes: I was just pointing out that this example does not contradict the true statement in the question that for rings with identity this does not happen. – Marc van Leeuwen Feb 1 '15 at 10:23
• @Sushil not possible: since every element has finite order, but the ring does not have finite characteristic, there exist infinitely many elements of different order (since the orders must go to infinity to make sure there is no characteristic). Select two elements, $x_1,x_2$ of different orders, $n_1, n_2$. There are two cases, either one divides the other, eg. $n_1|n_2$ or $\gcd(n_1,n_2)\not\in \{n_1,n_2\}$ for either $i$. In the latter case $n_2x_1$ is a zero divisor with co-divisor $n_1x_2$. In the former $n_1x_2$ is a zero divisor with co-divisor $x_1$. – Adam Hughes Feb 1 '15 at 21:00

(This answers what the question was before the requirement that the ring does not have a unit was added)

If every element in a unital ring $R$ has finite order, in particular the unit element $1_R$ has some finite order $n$ and then $nx=0$ for all $x\in R$. It follows that the orders of the elements of $R$ cannot be unbounded.

• Right but question is for ring without unity – Sushil Feb 1 '15 at 10:24
• Mariano, The question was always for no unit, the op just didn't express it very well. (I don't think your answer needs to go away, but it should be said that the content was confusing, not different) – Adam Hughes Feb 1 '15 at 10:36
• Well, it was not well expressed in that the requirement for non-unitality was omited! As evinced by what happened, that omission completely changed the question. – Mariano Suárez-Álvarez Feb 1 '15 at 10:38
• @MarianoSuárez-Alvarez I suppose one could view it that way, the ambiguity was very strong in the actual question. The clue was that the motivation given was exactly the case you prove, though, so it was probably meant to be reasoned that it wouldn't make sense for him to be asking for a proof of the case which he's already taken as motivation! :-) – Adam Hughes Feb 1 '15 at 10:47
• In any case, it's not really important, this is just my thorough side making note since I know some people like such things pointed out. – Adam Hughes Feb 1 '15 at 10:50