Set of Integer Multiples is Integral Ideal
Jump to navigation
Jump to search
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.
$\blacksquare$
Sources
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {2-1}$ Euclid's Division Lemma: Exercise $4$