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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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- On the field of fractions of certain kind of integral domain
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- Is $\Bbb Q/\Bbb Z$ artinian as a $\Bbb Z$-module?
- Does $S = R \cap K$ of a field extension $K \subseteq L = Q(R)$ satisfy $Q(S) = K$?

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- faithfully flat ring extensions where primes extend to primes
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- Question regarding to Dedekind domain and PID
- About weakly associated primes
- $k$-algebra homomorphism of the polynomial ring $k$
- An integral domain $A$ is exactly the intersection of the localisations of $A$ at each maximal ideal
- When irreducible elements of a UFD remain irreducible in a ring extension
- Can a prime in a Dedekind domain be contained in the union of the other prime ideals?

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.

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 $K$.

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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