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

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

Thus    and so  $a-b\in S$.

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

$$o(a)(ra)=\underset{o(a)}{\underbrace{ra+ra+\mathrm{\dots }+ra}}=r(\underset{o(a)}{\underbrace{a+a+\mathrm{\dots }+a}})=r(o(a)a)=r\cdot 0=0.$$

Therefore,   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