Characterization of P-adic Unit has Square Root in P-adic Units/Condition 3 implies Condition 1
Theorem
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units.
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p \Z_p}$.
Then:
- $\exists x \in \Z_p^\times : x^2 = u$
Proof
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p\Z_p}$
Let $y = y_0 + y_1 p + y_2 p^2 + y_3 p^3 + \ldots$ be the $p$-adic expansion of $y$.
By definition of $p$-adic expansion:
- $y_0 \in {0, 1, \ldots, p - 1}$
From P-adic Expansion of P-adic Unit:
- $y_0 \ne 0$.
Hence:
- $y_0 \in {1, 2, \ldots, p - 1}$
It follows that:
- $p \nmid y_0$
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
- $y \equiv y_0 \pmod {p \Z_p}$
Then:
- $y_0^2 \equiv y^2 \equiv u \pmod {p \Z_p}$
Let $F \in \Z \sqbrk X$ be the polynomial:
- $\map F X = X^2 - u$
By definition of formal derivative the formal derivative of $F$ is:
- $\map {F'} X = 2X$
We have:
- $\map F {y_0} \equiv 0 \pmod p$
and
- $\map {F'} {y_0} = 2 y_0$
By hypothesis:
- $p \nmid 2$
From the contrapositive statement of Euclid's Lemma for Prime Divisors:
- $p \nmid 2 y_0$
Hence:
- $\map {F'} {y_0} = 2 y_0 \not \equiv 0 \pmod p$
From Congruence Modulo Equivalence for Integers in P-adic Integers:
- $\map F {y_0} \equiv 0 \pmod {p \Z}$
and
- $\map {F'} {y_0} \not \equiv 0 \pmod {p \Z}$
From Hensel's Lemma for P-adic Integers:
- $\exists x \in \Z_p : \map F x = 0$
That is:
- $\exists x \in \Z_p : x^2 = u$
It remains to show that $x \in \Z_p^\times$.
We have:
\(\ds \norm x_p^2\) | \(=\) | \(\ds \norm {x^2}_p\) | Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm u_p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1\) | P-adic Unit has Norm Equal to One | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \norm x_p\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \Z_p^\times\) | P-adic Unit has Norm Equal to One |
$\blacksquare$