I don't understand how $\{am + pn : m, n \in \mathbb{Z}\}$ is equal $\langle 1 \rangle$, doesn't $\langle 1 \rangle$ contains all integer of $\mathbb{Z}$? the passage I got it from -
Prime ideal property. If $p$ is prime and the ideal $\langle p \rangle$ contains $\langle ab \rangle$, then $\langle p \rangle$ contains $\langle a \rangle$ or $\langle p \rangle$ contains $\langle b \rangle$ Suppose $\langle a \rangle \nsubseteq \langle p \rangle$, so we have to prove $\langle b \rangle \subseteq \langle p \rangle$.
Proof: Since the ideal $\{am + pn : m, n \in \mathbb{Z}\}$ contains both $\langle p \rangle$ and $\langle a \rangle$, and $\langle a \rangle \nsubseteq \langle p \rangle, {am +pn : m,n \in \mathbb{Z}}$ can only equal $\langle 1 \rangle$. This means $1 = am+pn$ for some $m, n \in \mathbb{Z}...$
Fitst $\langle1\rangle=\Bbb{Z}$.
Next $\langle a\rangle\not\subseteq \langle p\rangle$ implies $p\not\mid a$. Thus $a$ and $p$ are relatively prime. So there exists integers $x$ and $y$ such that $ax+py=1$. This gives any integer $n$ can be written as integral linear combination of $a$ and $p$, since $n=anx+pny$. Thus $\Bbb{Z}\subseteq \{am+pn:m,n\in\Bbb{Z}\}$. And $\{am+pn:m,n\in\Bbb{Z}\}\subseteq \Bbb{Z}$ very easily. So $\Bbb{Z}=\langle1\rangle= \{am+pn:m,n\in\Bbb{Z}\}$.