Congruence Modulo Equivalence for Integers in P-adic Integers/Lemma 1

Theorem

Let $\Z_p$ be the $p$-adic integers for some prime $p$.

Let $k \in \N_{>0}$.

Then:

$\forall a \in \Z: \dfrac a {p^k} \in \Z_p \iff \dfrac a {p^k} \in \Z$

Proof

Necessary Condition

Let $a \in \Z$.

We have:

\(\ds \dfrac a {p^k}\)

\(\in\)

\(\ds \Z_p\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac a {p^k}\)

\(\in\)

\(\ds \Z_p \cap \Q\)

\(\ds \leadsto \ \ \)

\(\ds \exists c, d \in \Z : p \nmid d: \, \)

\(\ds \dfrac a {p^k}\)

\(=\)

\(\ds \dfrac c d\)

Characterization of Rational P-adic Integer

\(\ds \leadsto \ \ \)

\(\ds ad\)

\(=\)

\(\ds cp^k\)

Multiply both sides by $dp^k$

\(\ds \leadsto \ \ \)

\(\ds p^k\)

\(\divides\)

\(\ds ad\)

Definition of Divisor

\(\ds \leadsto \ \ \)

\(\ds p^k\)

\(\divides\)

\(\ds a\)

Euclid's Lemma and $p \nmid d$

\(\ds \leadsto \ \ \)

\(\ds \dfrac a {p^k}\)

\(\in\)

\(\ds \Z\)

Definition of Divisor

$\Box$

Sufficient Condition

From Integers are Dense in P-adic Integers:

$\Z \subseteq \Z_p$

The result follows.

$\blacksquare$