# Set of Integer Multiples is Integral Ideal

## 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.

$\blacksquare$