P-adic Norms are Not Equivalent

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Theorem

Let $p_1$ and $p_2$ be prime numbers such that $p_1 \neq p_2$.

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 $\norm {\,\cdot\,}_{p_1}$:

\(\displaystyle \norm{p_1/p_2}_{p_1}\) \(=\) \(\displaystyle \norm{p_1}_{p_1} \norm{1/p_2}_{p_1}\) Norm axiom (M2) (Multiplicativity)
\(\displaystyle \) \(=\) \(\displaystyle \norm{p_1}_{p_1} \times 1\) $p_1$ does not divide $p_2$
\(\displaystyle \) \(=\) \(\displaystyle 1/{p_1}\)
\(\displaystyle \) \(\lt\) \(\displaystyle 1\)


On the other hand, with $\norm {\,\cdot\,}_{p_2}$:

\(\displaystyle \norm{p_1/p_2}_{p_2}\) \(=\) \(\displaystyle \norm{p_1}_{p_2} \norm{1/p_2}_{p_2}\) Norm axiom (M2) (Multiplicativity)
\(\displaystyle \) \(=\) \(\displaystyle 1 \times \norm{1/p_2}_{p_2}\) $p_2$ does not divide $p_1$
\(\displaystyle \) \(=\) \(\displaystyle p_2\)
\(\displaystyle \) \(\gt\) \(\displaystyle 1\)


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 $\norm {\,\cdot\,}_{p_2}$.

$\blacksquare$


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