Prime Number iff Generates Principal Maximal Ideal

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Theorem

Let $\Z_{>0}$ be the set of strictly positive integers.

Let $p \in \Z_{>0}$.

Let $\ideal p$ be the principal ideal of $\Z$ generated by $p$.


Then $p$ is prime if and only if $\ideal p$ is a maximal ideal of $\Z$.


Proof

First, note that $\Z$ is a principal ideal domain, so all ideals are principal.


Sufficient Condition

Suppose $p$ is prime.

From Integer Divisor is Equivalent to Subset of Ideal, $m \divides n \iff \ideal n \subseteq \ideal m$.

But as $p$ is prime, the only divisors of $p$ are $1$ and $p$ itself.

By Natural Numbers Set Equivalent to Ideals of Integers, it follows that if $p$ is prime, then $\ideal p$ must be a maximal ideal.

$\Box$


Necessary Condition

Let $p \in \Z_{>0}$ such that $\ideal p$ is maximal.

Let $q \in \Z_{>0}$ be a divisor of $p$.

Then:

\(\ds q\) \(\divides\) \(\ds p\)
\(\ds \leadsto \ \ \) \(\ds \ideal p\) \(\subseteq\) \(\ds \ideal q\) Integer Divisor is Equivalent to Subset of Ideal
\(\ds \leadsto \ \ \) \(\ds \ideal q\) \(=\) \(\ds \ideal 1\) Definition of Maximal Ideal of Ring
\(\, \ds \lor \, \) \(\ds \ideal q\) \(=\) \(\ds \ideal p\)
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds 1\)
\(\, \ds \lor \, \) \(\ds q\) \(=\) \(\ds p\)

Thus $p$ is prime.

$\blacksquare$


Sources

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