Residue Field of P-adic Norm on Rationals

Theorem
Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.

The induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is isomorphic to the field $\F_p$ of integers modulo $p$.

Proof
By Valuation Ring of P-adic Norm on Rationals,
 * $\Z_{(p)} = \set{ \dfrac a b \in \Q : p \nmid b }$

is the induced valuation ring on $\struct {\Q,\norm {\,\cdot\,}_p}$.

By Valuation Ideal of P-adic Norm on Rationals,
 * $p\Z_{(p)} = \set{ \dfrac a b \in \Q : p \nmid b, p \divides a}$

is the induced valuation ideal on $\struct {\Q,\norm {\,\cdot\,}_p}$.

By definition, the induced residue field on $\struct {\Q,\norm {\,\cdot\,}_p}$ is the quotient ring $\Z_{(p)}/p\Z_{(p)}$.

By Quotient Ring of Integers with Principal Ideal, $\F_p$ is isomorphic to $\Z/p\Z$, where $p\Z$ is the principal ideal of $\Z$ generated by $p$.

To complete the proof it is sufficient to show that $\Z/p\Z$ is isomorphic to $\Z_{(p)}/p\Z_{(p)}$.

Let $a \in \Z$.

Since $p \nmid 1$ then $a = \dfrac a 1 \in \Z_{(p)}$.

Hence $\Z \subset \Z_{(p)}$ is a subring of $\Z_{(p)}$.

Let $i : \Z \to \Z_{(p)}$ be the inclusion mapping defined by:
 * $\map i a = a$.

By Inclusion Mapping is Monomorphism then $i$ is a ring monomorphism.

Let $q : \Z_{(p)} \to \Z_{(p)}/p\Z_{(p)}$ be the quotient epimorphism from $\Z_{(p)}$ to $\Z_{(p)}/p\Z_{(p)}$.

By Quotient Epimorphism is Epimorphism then $q$ is a epimorphism.

Let $\phi = q \cdot i$ be the composition of $i$ with $q$.

By Composition of Ring Homomorphisms is Ring Homomorphism then $\phi$ is a homomorphism

Let $K = \ker \left({\phi}\right)$ be the kernel of $\phi$.

Then:

By Quotient Ring of Kernel of Ring Epimorphism