# ideal of elements with finite order

Theorem.  The set of all elements of a ring, which have a finite order in the additive group  of the ring, is a (two-sided) ideal of the ring.

Proof.  Let $S$ be the set of the elements with finite order in the ring $R$.  Denote by $o(x)$ the order of $x$.  Take arbitrary elements $a,\,b$ of the set $S$.

If  $\operatorname{lcm}(o(a),\,o(b))=n=ko(a)=lo(b)$,  then

 $n(a-b)=na-nb=ko(a)a-lo(b)b=k\cdot 0-l\cdot 0=0-0=0.$

Thus  $o(a-b)\leqq n<\infty$  and so  $a-b\in S$.

For any element $r$ of $R$ we have

 $o(a)(ra)=\underbrace{ra+ra+\ldots+ra}_{o(a)}=r(\underbrace{a+a+\ldots+a}_{o(a)% })=r(o(a)a)=r\cdot 0=0.$

Therefore, $o(ra)\leqq o(a)<\infty$  and $ra\in S$.  Similarly,  $ar\in S$.

Since $S$ satisfies the conditions for an ideal, the theorem has been proven.

 Title ideal of elements with finite order Canonical name IdealOfElementsWithFiniteOrder Date of creation 2013-03-22 17:52:30 Last modified on 2013-03-22 17:52:30 Owner pahio (2872) Last modified by pahio (2872) Numerical id 8 Author pahio (2872) Entry type Theorem Classification msc 20A05 Classification msc 16D25 Related topic OrderGroup Related topic Lcm Related topic Multiple Related topic OrdersOfElementsInIntegralDomain Related topic CharacteristicOfFiniteRing