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In the book Algebraic Geometry by Hartshorne, page 74, it is said that the spectrum of a discrete valuation ring $R$ has only two points. How to show that a discrete valuation ring has only two prime ideals? In http://en.wikipedia.org/wiki/Discrete_valuation_ring, it is said that a discrete valuation ring has exactly one non-zero maximal ideal.

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  • $\begingroup$ A DVR is an integral domain by definition so the zero ideal is prime... $\endgroup$ – fretty Aug 27 '13 at 11:47
  • $\begingroup$ @fretty, thank you very much. But maybe it has more than two prime ideals. $\endgroup$ – LJR Aug 27 '13 at 11:49
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    $\begingroup$ Look at point 4 in the list at the top of your linked wiki article: "R is a noetherian local ring with Krull dimension one". With the maximal ideal and the zero ideal, there can thus be no prime ideals between them. $\endgroup$ – Arthur Aug 27 '13 at 11:50
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    $\begingroup$ @LJR While it's not a proof of any kind, I'm glad I could help. As far as I can tell from wikipedia, this can be seen as a consequence of a DVR being a (local) Dedekind domain that is not a field (those always have Krull dimension 1). However, you need to show care so as not to introduce circular arguments here (DVRs and Dedekind domains can be defined from eachother). $\endgroup$ – Arthur Aug 27 '13 at 12:05
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    $\begingroup$ Yes but consequence 6 suggests only one non-zero prime ideal. $\endgroup$ – fretty Aug 27 '13 at 12:10
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Here is an argument starting from the definition of a DVR as a ring admitting a discrete valuation:

if $P$ is a non-zero prime ideal, then it contains a non-zero element, which can be written in the form $u t^n$, where $u$ is a unit (i.e. has valuation $0$), and $t$ is a uniformizer (i.e. has valuation $1$). Then $t^n \in P$ (multiply by $u^{-1}$), and then $t \in P$ (using primality). Thus the unique non-zero ideal of $R$ is the (maximal) ideal of elements of valuation $\geq 1$.

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    $\begingroup$ I think saying a DVR is 'a ring with discrete valuation' is quite misleading, because a DVR is NOT a ring with a valuation having discrete values (take for example $\mathbb{Z}$ with a $p$-adic valuation), but rather the ring of integers of a field with discrete valuation. $\endgroup$ – user10676 Aug 27 '13 at 13:41
  • $\begingroup$ @user: Dear user, Fair enough. Regards, $\endgroup$ – Matt E Aug 28 '13 at 3:31
  • $\begingroup$ @user10676 : Why hasn't the $p$-adic valuation on $\Bbb Z$ discrete values? $\endgroup$ – Watson Jan 3 '17 at 17:53
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    $\begingroup$ @Watson: it does have discrete values (meaning that the valuation group is discrete). I am saying that $\mathbb{Z}$ equiped with the $p$-adic valuation is not a DVR (even if the $p$-adic valuation is a discrete valuation). It may be confusing because in the standard terminology a discrete valuation is only on a field and not ring, and maybe one should not talk about 'rings equipped with a discrete valuation' $\endgroup$ – user10676 Jan 6 '17 at 13:26
  • $\begingroup$ @user10676 : Thank you for your reply! If I understand well, $\Bbb Z$ equiped with the $p$-adic valuation $v_p$ is not a DVR because $$\{x \in \Bbb Q \mid v_p(x) \geq 0\}$$ is larger that $\Bbb Z$ (it contains $p/q$ for any prime $q \neq p$), am I right? $\endgroup$ – Watson Jan 6 '17 at 13:35

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