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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.

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up vote 1 down vote accepted

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 $\alpha$.

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What you write is the answer to the question "What are the valuations $v$ on $k(x)$ which are trivial on $k$ such that $v(x) \geq 0$?" I don't see the first condition in the OP's question, although perhaps s/he means for it to be there. – Pete L. Clark Feb 23 '11 at 13:59
I don't see how this answers the question. I have verified that the question is stated correctly. – admchrch Feb 23 '11 at 14:21
Also, why do the valuations corresponding to irreducible polynomials of degree > 1 not appear? – admchrch Feb 23 '11 at 14:40
@Pete I think condition (2) implies that the valuations in $A$ are trivial on $k$. – Plop Feb 23 '11 at 16:33
@Plop: thanks -- you're right. If $x$ is a nonzero element of $k$ then both $x$ and $x^{-1}$ have non-negative valuation, and since $v(x^{-1}) = -v(x)$, this implies $v(x) = 0$. – Pete L. Clark Feb 24 '11 at 0:45

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