Set of Integer Multiples is Integral Ideal

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Theorem

Let $m \in \Z$ be an integer.

Let $m \Z$ denote the set of integer multiples of $m$.


Then $m \Z$ is an integral ideal.


Proof

First note that $m \times 0 \in m \Z$ whatever $m$ may be.

Thus $m \Z \ne \O$.


Let $a, b \in m \Z$.

Then:

\(\ds a + b\) \(=\) \(\ds m j + m k\) for some $j, k \in \Z$ by definition of $m \Z$
\(\ds \) \(=\) \(\ds m \paren {j + k}\)
\(\ds \) \(\in\) \(\ds m \Z\)

and:

\(\ds a - b\) \(=\) \(\ds m j - m k\) for some $j, k \in \Z$ by definition of $m \Z$
\(\ds \) \(=\) \(\ds m \paren {j - k}\)
\(\ds \) \(\in\) \(\ds m \Z\)


Let $r \in \Z$.

Then:

\(\ds r a\) \(=\) \(\ds r \paren {m j}\) for some $j \in \Z$ by definition of $m \Z$
\(\ds \) \(=\) \(\ds m \paren {r j}\)
\(\ds \) \(\in\) \(\ds m \Z\)

Thus the conditions for $m \Z$ to be an integral ideal are fulfilled.

Hence the result.

$\blacksquare$


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