We can construct $\mathbb{Q}_p$ purely algebraically by defining it as the field of fractions of $\varprojlim \mathbb{Z} / p \mathbb{Z}$. Then we can define the $p$-adic valuation on $\mathbb{Q}_p$ and recover $\mathbb{Z}_p$ as the ring of integers: $\mathbb{Z}_p = \{x \in \mathbb{Q}_p : v(x) \geq 0\}$. I'm curious whether we can recognize the ring of integers in $\mathbb{Q}_p$ without reference to the valuation $v_p$, or to the topology induced by this valuation.

I suspect that $\mathbb{Z}_p$ is either

  1. The only discrete valuation ring contained in $\mathbb{Q}_p$, or
  2. The unique maximal discrete valuation ring contained in $\mathbb{Q}_p$.

Are either of these correct?

Edit: I've found this post, and I think the answer there is showing that every non-trivial discrete valuation on $\mathbb{Q}_p$ must be equivalent to the $p$-adic valuation, but I'm struggling with the details in the proof.

  • $\begingroup$ I'm not sure how much can be said about $\mathbb Q_p$ as an abstract field. I would imagine one at least needs the topology to prove interesting things. $\endgroup$
    – Kenta S
    Sep 29, 2022 at 0:25
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    $\begingroup$ You can characterize the unit group $\mathbb Z_p^*$ inside $\mathbb Q_p$ purely algebraically as those elements which have $n$-th roots for infinitely many $n$. From there to the ring $\mathbb Z_p$ is not a long way. $\endgroup$ Sep 29, 2022 at 3:05
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    $\begingroup$ @TorstenSchoeneberg - thanks for the comment. I assume the argument is that if $a^n = x$ then $n v_p(a) = v(x)$, so $n \mid v(x)$, and if infinitely many $n$ divide $v(x)$ then $v(x) = 0$? Then from there you could take $\mathbb{Z}_p$ as $p^n u$ where $u$ is any unit? $\endgroup$ Sep 29, 2022 at 16:50
  • $\begingroup$ That is right, however don't forget the other direction that each $u \in \mathbb Z_p^\times$ indeed has $n$-th roots for infinitely many $n$, whose proof needs more $p$-adic theory (Hensel's lemma). Also, for that last step towards $\mathbb Z_p$ you have to identify $p$, but I think that's doable too ... $\endgroup$ Sep 29, 2022 at 17:11
  • $\begingroup$ @TorstenSchoeneberg once you have $\mathbf Z_p^\times$ described purely algebraically, you get $\mathbf Z_p$ as $\mathbf Z + \mathbf Z_p^\times$. $\endgroup$
    – KCd
    Oct 28, 2022 at 13:13

1 Answer 1


No, neither statement is true. $\mathbb{Q}_p$ contains $\mathbb{Q}$, so it contains the localizations $\mathbb{Z}_{(q)} = \mathbb{Q} \cap \mathbb{Z}_q$ (given by the rational numbers whose denominator is not divisible by $q$) for every prime $q$, not necessarily equal to $p$. For $q \neq p$ these subrings are not contained in $\mathbb{Z}_p$ because $p$ is invertible in them.

  • $\begingroup$ Wow very good point. I was so convinced there should be some characterization along the lines of what I gave that I missed this simple counterexample. I still wonder whether there is a good algebraic characterization of $\mathbb{Z}_p$ within $\mathbb{Q}_p$, but I see now that it would have to be at least much more complicated than what I had hoped for. $\endgroup$ Sep 29, 2022 at 2:55
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    $\begingroup$ One thing that comes to mind - the rings $\mathbb{Z}_{(q)}$ don't have $\mathbb{Q}_p$ as their field of fractions. Maybe adjusting number two to be "the unique maximal DVR in the set of DVRs whose field of fractions is $\mathbb{Q}_p$" would work? This is getting a little closer to the characterization of the ring of integers of a number field as the unique maximal order $\endgroup$ Sep 29, 2022 at 2:59
  • $\begingroup$ @stillconfused: that seems plausible to me. The question about algebraic characterizations of $\mathbb{Z}_p$ in $\mathbb{Q}_p$ is also interesting. Such characterizations are known for $\mathbb{Z}$ in $\mathbb{Q}$ but they're not too easy (see e.g. annals.math.princeton.edu/wp-content/uploads/Koenigsmann.pdf) but this case should be simpler. $\endgroup$ Sep 29, 2022 at 3:00
  • $\begingroup$ Fascinating, thanks for sharing! I would not have expected such a recent paper on this question. Perhaps this is generally more difficult than I thought $\endgroup$ Sep 29, 2022 at 16:48
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    $\begingroup$ @stillconfused as Qiaochu suggested, it is indeed much easier to "define" (in the sense of first-order logic) $\mathbb{Z}_p$ in $\mathbb{Q}_p$ than it is to define $\mathbb{Z}$ in $\mathbb{Q}$. for example, if $p\neq 2$, then $\mathbb{Z}_p$ is precisely the set of $a\in\mathbb{Q}_p$ such that $pa^2+1$ is a square in $\mathbb{Q}_p$. (hint for showing this: use Hensel's lemma.) whether this counts as a satisfactory algebraic characterization for you I am not sure :) $\endgroup$ May 3, 2023 at 12:36

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