P-adic Norms are Not Equivalent
Theorem
Let $p_1$ and $p_2$ be prime numbers such that $p_1 \ne p_2$.
Let $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ be the $p$-adic norms on the rationals $\Q$.
Then $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not equivalent norms.
That is, the topology induced by $\norm {\,\cdot\,}_{p_1}$ does not equal the topology induced by $\norm {\,\cdot\,}_{p_2}$.
Proof
Consider $p_1/p_2 \in \Q$.
With $\norm {\,\cdot\,}_{p_1}$:
\(\ds \norm {p_1/p_2}_{p_1}\) | \(=\) | \(\ds \norm {p_1}_{p_1} \norm {1/p_2}_{p_1}\) | Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {p_1}_{p_1} \times 1\) | $p_1$ does not divide $p_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 / {p_1}\) | ||||||||||||
\(\ds \) | \(\lt\) | \(\ds 1\) |
On the other hand, with $\norm {\,\cdot\,}_{p_2}$:
\(\ds \norm {p_1/p_2}_{p_2}\) | \(=\) | \(\ds \norm {p_1}_{p_2} \norm {1/p_2}_{p_2}\) | Non-Archimedean Norm Axiom $\text N 2$: Multiplicativity | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \norm {1/p_2}_{p_2}\) | $p_2$ does not divide $p_1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds p_2\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 1\) |
By open unit ball equivalence, $\norm {\,\cdot\,}_{p_1}$ and $\norm {\,\cdot\,}_{p_2}$ are not equivalent norms.
By Equivalence of Definitions of Equivalent Division Ring Norms and topological equivalence then the topology induced by $\norm {\,\cdot\,}_{p_1}$ does not equal the topology induced by $\norm {\,\cdot\,}_{p_2}$.
$\blacksquare$
Sources
- 1997: Fernando Q. Gouvea: p-adic Numbers: An Introduction: $\S 3.1.2$ Absolute Values on $\Q$, Problem $68$
- 2007: Svetlana Katok: p-adic Analysis Compared with Real ... (previous) ... (next): $\S 1.4$ The field of $p$-adic numbers $\Q_p$: Remark $1.28$