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

$\blacksquare$

## 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 $1.28$