Natural Numbers Set Equivalent to Ideals of Integers

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Theorem

Let the mapping $\psi: \N \to$ the set of all ideals of $\Z$ be defined as:

$\forall b \in \N: \psi \left({b}\right) = \left({b}\right)$

where $\left({b}\right)$ 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 $b \in \left({b}\right)$, but $b \notin \left({c}\right)$, because from Principal Ideals of Integers‎, $c$ is the smallest positive integer in $\left({c}\right)$.

Thus $\left({b}\right) \ne \left({c}\right)$.

It is also apparent that $b > 0 \implies \left({b}\right) \ne \left({0}\right)$ as $\left({0}\right) = \left\{{0}\right\}$.

Thus $\psi$ is injective.


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

$\blacksquare$


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