every field of characteristic 0 has a discrete valuation ring? 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?
 A: 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.
A: 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 $K$.
A: 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.
