# P-adic Norms are Not Equivalent

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

$\blacksquare$