# Integral Ideal is Set of Integer Multiples

## Theorem

Let $J$ be an integral ideal.

Then $J$ is in the form of a set of integer multiples $m \Z$ for some $m \in \Z$.

## Proof

By definition, $J$ satisfies the following conditions:

- $(1): \quad n, m \in J \implies m + n \in J, m - n \in J$

- $(2): \quad n \in J, r \in \Z \implies r n \in J$

First note that the null ideal $\set 0$ is an integral ideal.

This is of the form $0 \Z$.

Let $J \ne \set 0$.

Then $\exists a \in J: a \ne 0$.

As $0 \in \Z$ it follows from $(2)$ that:

- $0 \times a = 0 \in J$

Thus from $(1)$ we have that:

- $0 - a \in J$

so both $a \in J$ and $-a \in J$

Thus there exists $a \in J: a > 0$.

Consider the set $S = \set {x \in J: x > 0}$.

$S$ is bounded below by $0$, for example.

$S$ is not empty, because it has been established that $a \in S$.

By Set of Integers Bounded Below has Smallest Element, $S$ has a smallest element, $m$, say.

Consider $y \in m \Z$.

By definition of $m \Z$, $y = r m$ for some $r \in \Z$.

Thus, by $(2)$, $y \in J$.

So by definition of subset, $m \Z \subseteq J$.

It remains to be demonstrated that $J \subseteq m \Z$.

Aiming for a contradiction, suppose $k \in J: k \notin m \Z$.

As $k \in J$, it follows from $(1)$ that $0 - k = -k \in J$ also.

Either $k > 0$ or $-k > 0$.

Without loss of generality, let $k > 0$.

If $k < m$, then that contradicts the statement that $m \in J$ is the smallest element of $J$ that is (strictly) positive.

So $k > m$.

Consider the set $T = \set {x \in m \Z: x > k}$.

$T$ is not empty, as $m \paren {k + 1}$, for example, is bound to be in $T$ (even if $m = 1$, the smallest it can be).

By Set of Integers Bounded Below has Smallest Element, $T$ has a smallest element, $m s$, say.

Then:

- $m \paren {s - 1} < k < m s$

As $k > m$ it follows also that $m \paren {s - 1} > 0$.

Equality cannot happen because $k \notin m \Z$.

But we have that $k \in J$ and so:

- $m s - k \in J$

and:

- $k - m \paren {s - 1} \in J$

But $m s - k < m$, otherwise $m \paren {s - 1} > k$ which contradicts the statement that $m s \in J$ is the smallest element of $T$

Thus $m s - k \in J$ and $0 < m s - k < m$.

This contradicts the statement that $m$ is the smallest element of $J$ that is (strictly) positive.

Hence by Proof by Contradiction there is no such $k$.

Hence $J \subseteq m \Z$ and the result follows.

$\blacksquare$

## Sources

- 1971: George E. Andrews:
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