Residue Field of P-adic Norm on Rationals/Lemma 1

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

Let $\Z_{(p)}$ be the induced valuation ring on $\struct {\Q, \norm {\,\cdot\,}_p}$.

Let $p \Z_{(p)}$ be the induced valuation ideal on $\struct {\Q, \norm {\,\cdot\,}_p}$.

Let $\phi : \Z \to \Z_{(p)}/p\Z_{(p)}$ be the mapping defined by:


 * $\forall a \in \Z: \map \phi a = a + p\Z_{(p)}$

Then:
 * $\phi$ is a homomorphism.

Proof
Since $p \nmid 1$ then for all $a \in \Z$, $a = \dfrac a 1 \in \Z_{(p)}$.

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 ring epimorphism from $\Z_{(p)}$ to $\Z_{(p)} / p \Z_{(p)}$.

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

By definition, $\phi = q \circ i$ is the composition of $i$ with $q$.

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