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If $F$ is a field, then it is a well-known result that the following are equivalent for two norms $|\cdot|_1$ and $|\cdot|_2$ on $F$:

  1. Any sequence in $F$ is Cauchy with respect to $|\cdot|_1$ if and only if it is Cauchy with respect to $|\cdot|_2$.

  2. There exists some $\alpha > 0$ such that $|x|_1 = |x|_2^\alpha$ for all $x\in F$.

Is this still true in some sense for norms on rings? If so, could anyone give me a reference? I think the standard proof uses division at a couple of steps.

Thank you.


Just in case, from a norm on a ring I require that

  1. $|x| = 0 \iff x = 0$.
  2. $|xy| = |x|\cdot|y|$ (in particular, there are no zero divisors).
  3. $|x+y| \le |x| + |y|$.
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No. Take as ring the integers $\mathbb{Z}$ and as norms

$|\cdot|_1$ the trivial norm, i.e. $|x|_1 = 1$ if $x \neq 0$, $|0|_1 =0$, and

$|\cdot|_2$ = the restriction of the usual absolute value.

Both define the discrete topology on $\mathbb{Z}$; in particular, for both norms Cauchy sequences are exactly the sequences that become constant after finitely many steps, so your condition 1 is satisfied. But apparently condition 2 is not. (Although almost: One could argue that $|x|_1 = |x|_2^0$.)

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  • $\begingroup$ That is a nice example :-) Thank you! $\endgroup$ – AAA Jul 25 '17 at 12:02

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