How to find the set of valuations of the field $k(x)$ that satisfy certain properties? This is a problem from Borevich and Shafarevich (p.180).  
If $\mathfrak{O} = k[x]$, the problem is to find the set of valuations $A$ of the field $K = k(x)$ which satisfies the conditions (1),(2),(3) below.  I think I want to find the largest such set.  I also want to do the same when $\mathfrak{O} = k[1/x]$.
Can someone please give me some guidance on how to do this?
Here is the relevant terminology.
A function $v$ on a field $K$ is called a valuation of $K$ if it satisfies the following:
(i) $v(\alpha)$ takes on all rational integral values as $\alpha$ ranges through the nonzero elements of $K$, $v(0)=\infty$
(ii) $v(\alpha \beta) = v(\alpha) + v(\beta)$
(iii) $v(\alpha + \beta) \geq \min(v(\alpha),v(\beta))$
Let $\mathfrak{O}$ be a integral domain with quotient field $K$ and let $A$ be a set of valuations of $K$.  Consider the following properties that $A$ may possess.
(1) For any $\alpha \neq 0$ of $\mathfrak{O}$, $v(\alpha) = 0$ for all but finitely many valuations $v \in A$
(2) An element $\alpha$ of $K$ belongs to $\mathfrak{O}$ if and only if $v(\alpha) \geq 0$ for all $v \in A$
(3) For any finite set of distinct valuations $v_1, \ldots, v_m$ of $A$ and for any set of nonnegative integers $k_1,\ldots,k_m$, there is an element $\alpha \in \mathfrak{O}$ for which $v_i(\alpha) = k_i$, $(i=1,\ldots,m)$.
I know that $A$ satisfying the conditions (1),(2),(3) is equivalent to $A$ inducing a theory of divisors on $\mathfrak{O}$.
A theory of divisors on a domain $\mathfrak{O}$ consists of a commutative semigroup $D$ (semigroup = associative multiplication with identity) having the property of unique factorization into irreducibles along with a homomorphism $\alpha \rightarrow (\alpha)$ of the semigroup $\mathfrak{O}^{\times}$ into $D$ satisfying
(i) An element $\alpha \in \mathfrak{O}^{\times}$ is divisible by $\beta \in \mathfrak{O}^{\times}$ in the ring $\mathfrak{O}$ if and only if $(\alpha)$ is divisible by $(\beta)$ in $D$.
(ii) If $\alpha$ and $\beta$ of $\mathfrak{O}$ are divisible by $a \in D$, then $\alpha \pm \beta$ are also divisible by $a$.
(iii) If $a$ and $b$ are two elements of $D$ and the set of all elements $\alpha \in \mathfrak{O}$ which are divisible by $a$ coincides with the set of all elements $\beta \in \mathfrak{O}$ which are divisible by $b$, then $a = b$.
EDIT: I have one clarification question about condition (i) in the definition of a valuation.  Is the intention that $v$ should assume all integers values and also only integer values?  
I ask because another problem is to show that an algebraically closed field $K$ has no valuations, and I have only been able to prove this when $v$ assumes only integer values.  My proof is that for any given $\alpha \in K$ with valuation $v(\alpha) = m \neq 0$ and every positive integer $n$, the equation $x^n = \alpha$ has a solution $x \in K$, and so the equation $ny = m$ has a solution $y = v(\alpha) \in \mathbb{Z}$, which says that every integer $n$ divides the given nonzero integer $m$.  Contradiction.
 A: In the sequel let $v$ be a valuation of $k(x)$, let $R$ be the valuation ring of $v$ and $M$ its maximal ideal.
If $v$ satisfies (2) then $k[x]\subseteq R$; in particular $v$ is trivial on $k$.
The prime ideal $P:=M\cap k[x]$ is non-zero, thus has the form $P=pk[x]$ with a monic prime polynomial $p$ (monic means that the coefficient of the leading term of $p$ is $1$, not that $p$ has degree $1$).
The localication $k[x]_P$ is contained in $R$. Since $k[x]$ is factorial we can write every element of $k(x)$ in the form $f/g$, $f,g\in k[x]$ and not both divisible by $p$. Condition (2) implies that $f/g\in R$ is equivalent to $p$ does not divide $g$. The latter implies $f/g\in k[x]_P$, hence $k[x]_P=R$. Moreover one sees that the valuation $v$ is the $p$-adic valuation of $k(x)$.
For the polynomial ring $k[x]$ these valuations are in bijection with the monic prime polynomials of $k[x]$. The valuation $v$ associated to such a polynomial $p$ is defined in the same way in which the p-adic valuations of the rationals are defined.
The set $A$ of valuations you get actually coincides with the set of all valuations of $K(x)$ that are non-negative on $k[x]$. So it remains to prove that (1) and (3) are satisfied. 
Proof of (1) for the family $A$ of all $p$-adic valuations of $k(x)$: an element $\alpha\in k[x]$ has a positive value at a $p$-adic valuation if and only if $\alpha$ is divisible by $p$. Thus the relevant $p$ are the prime factors of $\alpha$.
Proof of (2): take the prime elements $p_i$ corresponding to the $v_i$, raise them to the powers $k_i$ and multiply these powers to get $\alpha$.
