# orders of elements in integral domain

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\newtheorem*{thmplain}{Theorem}
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\begin{thmplain}
Let\, $(D,\,+,\,\cdot)$\, be an integral domain, i.e. a commutative ring with non-zero unity 1 and no zero divisors.\, All non-zero elements of $D$ have the same \PMlinkname{order}{OrderGroup} in the additive group\, $(D,\,+)$.
\end{thmplain}

{\em Proof.}\, Let $a$ be arbitrary non-zero element.\, Any \PMlinkname{multiple}{GeneralAssociativity} $na$ may be  written as
$$na = n(1a) = \underbrace{1a+1a+\cdots+1a}_{n} = (\underbrace{1+1+\cdots+1}_{n})a = (n1)a.$$
Thus, because\, $a \ne 0$\, and there are no zero divisors, an equation\, $na = 0$\, is \PMlinkname{equivalent}{Equivalent3} with the equation\, $n1 = 0$.\, So $a$ must have the same \PMlinkescapetext{order} as the unity of $D$.

\textbf{Note.}\, The \PMlinkescapetext{order} of the unity element is the \PMlinkname{characteristic}{Characteristic} of the integral domain, which is 0 or a positive prime number.
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