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Let $R$ be a ring with identity. I call a non-negative real valued function $N: R \to \mathbb R_{\geq 0}$ a ring norm, if it has the following properties:

  1. $N(r) = 0$ iff $r = 0$
  2. $N(r+s) \leq N(r) + N(s) $
  3. $N(rs) \leq N(r)N(s)$

(This is the notion usually used in Berkovich theory). We also might require the following property:

  1. $N(zs) = N(z\cdot 1_R) N(s) $ whenever $z \in \mathbb Z$.

What are the ring norms on $\mathbb Q$ up to strong equivalence and up to topological equivalence?

One thing I see is that any finite sum of ring norms is again a ring norm; the same for the maximum of finitely many ring norms. Furthermore, the supremum of infinitely many ring norms can be a ring norm (for example, $|q|_{sup} := \sup_{p\; \mathrm{ prime}} |q|_p$ is a ring norm).

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    $\begingroup$ The submultiplicative, power-multiplicative seminorms are precisely the infima of sets of mulitplicative seminorms. And I believe every submultiplicative seminorm $|-|$ can be "powermulitplicativified" by passing to $|x|_{pm} = \lim_{n \to \infty} | x^n |^{1/n}$. I don't know what the fibers of the map $|-| \mapsto |-|_{pm}$ look like. $\endgroup$
    – Tim Campion
    Commented May 4, 2023 at 12:25
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    $\begingroup$ Thank you for the comment, @TimCampion. I was not aware of the result about power-multiplicative seminorms. Does the "powermultiplicativified" norm still induce the same topology? $\endgroup$
    – Adelhart
    Commented May 4, 2023 at 18:12
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    $\begingroup$ Also, why is the infima of seminorms still a seminorm? I thought that one would lose the subadditivity. $\endgroup$
    – Adelhart
    Commented May 4, 2023 at 18:26
  • $\begingroup$ I think it should be the supremum, not the infimum. $\endgroup$
    – Oren Ben-Bassat
    Commented Nov 8 at 18:55

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