The prime ideals of a field and of the integers

A proper ideal of a field cannot contain a nonzero element, because any such element is a unit and would force $1$ into the ideal. Hence the only proper ideal of $F$ is $(0)$, and $(0)$ is prime because $F$ has no zero divisors.

Let $I⊴\mathbb{Z}$. If $I\mathrm{\ne }(0)$, choose the least positive integer $n\in I$ using [L1]. For any $m\in I$, [L4] gives $m=qn+r$ with $0\le r<n$; since $r=m-qn\in I$ and $n$ was the least positive element of $I$, one must have $r=0$, so every element of $I$ is a multiple of $n$ and therefore $I=(n)$.

If $n=ab$ with $1<a,b<n$, then $(a+(n))(b+(n))=0+(n)$ in $\mathbb{Z}\mathrm{/}(n)$ while neither factor is zero, so [L2] shows $(n)$ is not prime. If $n=p$ is prime, then [L3] makes $\mathbb{Z}\mathrm{/}(p)$ a field and hence a domain, so [L2] shows $(p)$ is prime. Finally $(0)$ is prime because $\mathbb{Z}$ is an integral domain.

Steps 1.1, 1.2, and 1.3 prove the listed prime ideals of $F$ and $\mathbb{Z}$, and the inclusion order is immediate because every nonzero prime ideal of $\mathbb{Z}$ is maximal.