Natural Numbers Set Equivalent to Ideals of Integers


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Theorem

Let $S$ be the set of all ideals of $\Z$.

Let the mapping $\psi: \N \to S$ be defined as:

$\forall b \in \N: \map \psi b = \ideal b$

where $\ideal b$ is the principal ideal of $\Z$ generated by $b$.


Then $\psi$ is a bijection.


Proof

First we show that $\psi$ is injective.

Suppose $0 < b < c$.

Then $b \in \ideal b$, but $b \notin \ideal c$, because from Principal Ideals of Integers‎, $c$ is the smallest positive integer in $\ideal c$.

Thus $\ideal b \ne \ideal c$.

It is also apparent that $b > 0 \implies \ideal b \ne \ideal 0$ as $\ideal 0 = \set 0$.

Thus $\psi$ is injective.


Surjectivity follows from Principal Ideals of Integers: every integer is the smallest strictly positive element of a principal ideal of $\Z$.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.3$
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