The maximal ideal and residue field of Z_p

An element $x\in {\mathbb{Z}}_p$ is a unit exactly when $\mathrm{\mid }x{\mathrm{\mid }}_p=1$: if $\mathrm{\mid }x{\mathrm{\mid }}_p=1$ then $\mathrm{\mid }{x}^{-1}{\mathrm{\mid }}_p=1$ as well, so $x^{-1}\in {\mathbb{Z}}_p$ by [L1]; if $\mathrm{\mid }x{\mathrm{\mid }}_p<1$ then $\mathrm{\mid }{x}^{-1}{\mathrm{\mid }}_p>1$, so $x^{-1}\mathrm{\notin }{\mathbb{Z}}_p$. Therefore the nonunits are precisely the elements with $\mathrm{\mid }x{\mathrm{\mid }}_p<1$, which is the principal ideal $p{\mathbb{Z}}_p$.

In the compatible-residue model of [L2], multiplication by $p$ is exactly the condition that the first residue coordinate is $0$. Therefore the quotient by $p{\mathbb{Z}}_p$ remembers only the first residue class, giving a canonical map ${\mathbb{Z}}_p\mathrm{/}p{\mathbb{Z}}_p\to \mathbb{Z}\mathrm{/}p\mathbb{Z}={\mathbb{F}}_p\mathrm{.}$ This map is bijective because every residue class lifts to a compatible system and two systems differ by an element of $p{\mathbb{Z}}_p$ exactly when their first coordinates agree.

Since the quotient by $p{\mathbb{Z}}_p$ is the field ${\mathbb{F}}_p$, the ideal $p{\mathbb{Z}}_p$ is maximal, and step 1.1 shows it contains every nonunit, so it is the unique maximal ideal.