ideal of elements with finite orderIdealOfElementsWithFiniteOrder2008-03-01 16:29:252008-03-05 17:17:28Theoremideal% this is the default PlanetMath preamble. as your knowledge
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\DeclareMathOperator{\lcm}{lcm}\textbf{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.\\
{\em 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\; $\lcm(o(a),\,o(b)) = n = ko(a) = lo(b)$,\, then
$$n(a-b) = na-nb = ko(a)a-lo(b)b = k\cdot0-l\cdot0 = 0-0 = 0.$$
Thus\, $o(a-b) \leqq n < \infty$\, and so\, $a-b \in S$.
For any element $r$ of $R$ we have
$$o(a)(ra) = \underbrace{ra+ra+\ldots+ra}_{o(a)} = r(\underbrace{a+a+\ldots+a}_{o(a)}) =
r(o(a)a) = r\cdot0 = 0.$$
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.