How can we prove that every field of characteristic 0 has at least one Discrete Valuation Ring?

My effort: Let $K$ be an field of characteristic 0. Then $\mathbb{Z}$ is a subring of $K$. Let $p$ be a prime. By Theorem 10.2 in Matsumura, there exists a valuation ring $R$ of $K$ with $\mathbb{Z} \subset R$ and $m_R \cap \mathbb{Z}=p \mathbb{Z}$, where $m_R$ is the maximal ideal of $R$. If I could show that $R$ is Noetherian, or principal ideal domain, then I would be done by Theorem 11.1 of Matsumura. But I am having a hard time proving this and besides, it seems to me that this is not the right direction.

Edit: This question was motivated by the remark in Matsumura's Commutative Ring Theory p. 79, which mentions "Let $K$ be a field and $R$ a DVR of $K$..." As the answers point out, a field need not have a DVR. Then why would $K$ have a DVR in Matsumura's remark?

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    $\begingroup$ Does the algebraic closure of a finite field contain a discrete valuation ring? Perhaps by "infinite field" you really meant "field of characteristic 0"? $\endgroup$ – Jack Schmidt Apr 27 '13 at 19:33
  • $\begingroup$ @JackSchmidt: Yes, thank you, i mean characteristic zero. Could you please edit, for some reason my browser does not allow me to make edits. $\endgroup$ – Manos Apr 27 '13 at 19:39
  • $\begingroup$ I might miss something simple, but since $\mathbb Z$ can be embedded in $K$, don't you get $\mathbb Q \hookrightarrow K$? So it suffices to prove that $\mathbb Q$ has a DVR, which is trivial... $\endgroup$ – N. S. Apr 27 '13 at 20:01
  • $\begingroup$ DVRs are no fields (by definition). $\endgroup$ – Martin Brandenburg Apr 27 '13 at 22:14
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    $\begingroup$ When working with DVRs we need the uniformizer. Therefore fields are usually excluded. See also en.wikipedia.org/wiki/Discrete_valuation_ring, encyclopediaofmath.org/index.php/Discretely-normed_ring. See also Matsumura, Commutative Ring Theory, page 78 ff. There the definition of a DVR is correct and uniformizers are used extensively. Theorem 11.1 (the characterization DVR = noetherian valuation ring) is not quite correct, fields have to be excluded. Theorem 11.2 says explicitly that DVRs are not fields. $\endgroup$ – Martin Brandenburg Apr 27 '13 at 23:55

The statement is not true: $\mathbb{C}$ contains no discrete valuation ring having field of fractions $\mathbb{C}$, because a valuation of $\mathbb{C}$ must have a divisible value group. In particular this value group cannot be $\mathbb{Z}$.

The statement is true for example for every finitely generated extension field of $\mathbb{Q}$.

Sketch of the proof: let $L/K$ be a finite extension of fields, $v$ a valuation on $K$ and $w$ a prolongation of $v$ to $L$. Then $(w(L^\times ):v(K^\times ))\leq (L:K)$. In particular: if $v$ is discrete then $w$ is discrete.

Let $K/\mathbb{Q}$ be a finitely generated extension. Since all valuations on $\mathbb{Q}$ are discrete we are done if $K/\mathbb{Q}$ is algebraic.

If the extension is not algebraic it is a finite extension of a rational function field $\mathbb{Q}(T)$ in finitely many variables $T=\{t_1,\ldots ,t_n\}$. Thus it suffices to prove that the valuations of $\mathbb{Q}$ posses a discrete prolongation to $\mathbb{Q}(T)$. Such a prolongation is the Gauss prolongation of a valuation $v$. It assigns to a polynomial the minimum of the values of its coefficients.

Motivated by mr.bigproblem's answer I add the following:

A field $K$ contains a discrete valuation ring $O$ with field of fractions $K$ if and only if $K$ is the fraction field of a noetherian domain properly contained in $K$.

Sketch of the proof: the implication $\Rightarrow$ is obvious. If on the other hand $R$ is a noetherian domain, its integral closure $S$ in $K$ by the Mori-Nagata-theorem has the property that all localizations $S_p$ at primes of height $1$ are discrete valuation rings. Note that $S$ itself needs not be noetherian.

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    $\begingroup$ Isn't $\mathbb Z_{(2)}$ a subring of $\mathbb C$? $\endgroup$ – N. S. Apr 27 '13 at 20:05
  • $\begingroup$ @N.S. yes, but I think they want the DVR to come from a valuation on the field, not on a subfield. In other words, they want a DVR whose field of fractions is the given field. $\endgroup$ – Jack Schmidt Apr 28 '13 at 5:34
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    $\begingroup$ In my answer I was assuming that the DVR has fraction field equal to $\mathbb{C}$. Otherwise the statement has a trivial proof because every field of characteristic $0$ contains the rationals. $\endgroup$ – Hagen Knaf Apr 28 '13 at 6:49
  • $\begingroup$ Thanks a lot +1 Please see my edit. $\endgroup$ – Manos Apr 30 '13 at 2:28

Nicely done Hagen! I would like to add one more fact. The reason why $\mathbb{C}$ has no (discrete) valuation ring whose quotient field is also $\mathbb{C}$ is (as Hagen pointed out for DVR, the value group is divisible) that it doesn't possess a (Noetherian) domain whose quotient field equals $\mathbb{C}$. In fact, one has the more general statement:

A field $K$ (not necessarily char 0) possesses a (discrete) valuation ring whose quotient field equals $K$ iff it contains at least a proper (Noetherian) subdomain $R$ whose quotient field equals $K$.

  • $\begingroup$ Sure. $(\Leftarrow)$ is clear as a valuation ring is a domain. $(\Rightarrow)$ is an immediate of the following nontrivial fact: If $R$ is a domain with quotient field $K$ and $\mathfrak{p}$ is a prime ideal of $R$ then there always exists a (discrete) valuation ring $V$ having center $\mathfrak{p}$ in $R$, i.e. $R \subset V$ and $\mathfrak{m}_V \cap R = \mathfrak{p}$. I'm sorry that I can't prove it here since the proof is too long, but you can look it up in Huneke & Swanson's text about integral closure. $\endgroup$ – mr.bigproblem Apr 29 '13 at 17:16
  • $\begingroup$ One more comment and one more question related to this. 1) The fact I stated above has two cases, discrete valuation ring and non-discrete valuation ring. The proof for the non-discrete case is a non-constructive proof and is not too hard. Amuzingly, the only proof (at least to me), for the discrete case is constructive and deep (using Akizuki-Krull Theorem). 2) It yields from Hagen's observation and the fact I provided that any closed $K$ field doesn't contain a domain whose quotient field equals $K$. The question is whether if the converse is true? I haven't thought much of it. $\endgroup$ – mr.bigproblem Apr 29 '13 at 17:20
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    $\begingroup$ I do not agree with your statement: every field $K$ has a proper subdomain $R$ such that $K$ is the fraction field of $R$. Take a transcendence basis $T$ of $K$ over the prime field $P$ and consider the integral closure $R$ of the polynomial ring $P[T]$ in $K$. $\endgroup$ – Hagen Knaf Apr 29 '13 at 18:32
  • $\begingroup$ Thanks Hagen! Your counter example is right. I obviously omitted the Noetherian condition for my subdomain in the DVR case. I edited my statement. This also fits your example too. Take $K = \mathbb{C}$, $P$ is just $\mathbb{Q}$ so $T$ is infinite so that $P[T]$ is not Noetherian. Also, my question should be modified as: if $K$ doesn't contain any Noetherian proper subdomain whose quotient field equals $K$, is it algebraically closed? $\endgroup$ – mr.bigproblem Apr 29 '13 at 19:34
  • $\begingroup$ The answer to your question is "No". The reals do not carry discrete valuations for almost the same reason as for the complex numbers: one can take $n$-th roots of positve elements for every $n\in\mathbb{N}$. $\endgroup$ – Hagen Knaf Apr 30 '13 at 7:10

I don't have the text so I can't be sure, but when I read that phrase, I take the entire phrase as being a hypothesis for what comes next. So whatever comes next is only meant to apply to discretely valued fields, rather than to arbitrary fields.


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