# The valuation ring $R$ in $K(T)$, such that $K[T] \subsetneq R \subsetneq K(T)$

$K$ is an algebraically closed field, $K[T]$ is the ring of polynomials of one indeterminate over $K$, and $K(T)$ is its field of fractions. A valuation ring $R$ in $K(T)$ which includes $k[T]$ and is distinct from $K(T)$ must consist of all $f(T)/g(T)$, where $f(T),g(T) \in K[T]$ are relatively prime and where some fixed linear polynomial $T - a$ does not divide $g(T)$.

I think the proof for this statement can be divided into three steps:

1. to prove that for some $a \in K$, $\frac{1}{T-a} \notin R$;
2. to prove that for any $b \in K, b \neq a$, $\frac{1}{T - b}$ is contained in $R$;
3. to prove that $R$ consists of the polynomials described above.

First, as $K$ is algebraically closed, every irreducible polynomial is of degree $1$, and every polynomial $f(T)$ of positive degree can be written as $f(T) = k \prod_i(T - a_i)$, for certain $k, a_i \in K$.

1. If for any $a \in K$, $\frac{1}{T-a}$ is contained $R$, then for any $g(T) \in K[T]$ of positive degree, $\frac{1}{g(T)}$ must be in $R$. So $R$ contains all the fractions $\frac{f(T)}{g(T)}$, along with $K[T]$. $R$ must be equal to $K(T)$. A contradiction. So, there is an element $a$ in $K$, such that $\frac{1}{T-a} \notin R$.
2. This is where my question comes. What will happen if for $\frac{1}{T-b} \notin R$ for some $b \in K$ other than $a$?
3. For any relatively prime polynomials $f(T),g(T) \in K[T]$, if $g(T)$ is not divisible by $T-a$, $\frac{1}{g(T)}$ must be in $R$, so is $\frac{f(T)}{g(T)}$; Otherwise, $g(T)$ is divisible by $T-a$. So, $T-a$ is not a factor of $f(T)$. Thus, $\frac{1}{f(T)} \in R$. In this case, if $\frac{f(T)}{g(T)} \in R$, $\frac{1}{g(T)}$ must also be in $R$, so is $\frac{1}{T-a}$. Also a contradiction. This proves the statement.

I think I need some help on the second step... Or, some other method to prove the statement. Thanks very much.

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## 4 Answers

If you have another $b \neq a$ such that $(T - b)^{-1}$ is not in $R$, then consider the element $Q(T) = (T - b)/(T - a)$. As $R$ is a valuation ring, at least one of $Q$ and $Q^{-1}$ is in $R$. I claim that either of these will contradict our assumptions.

Suppose, for example, that $Q \in R$. Then surely $Q - 1 \in R$, and if you expand this it should become clear that $(T - a)^{-1} \in R$ as well.

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 Thank you very much for correcting mistakes in my question and answering it. I think I should have written in the question that I had "played around" with $(T-a)/(T-b)$ for quite a while without discovering "it"... Would you please give me some further hints on how I can find a contradiction if I suppose that $1/(T-a) \notin R$, $1/(T-b) \notin R$, $Q \in R$, and $Q^-1 \notin R$? Many thanks... – ShinyaSakai Sep 30 '11 at 3:38 @ShinyaSakai Oh, I don't think there were any mistakes. I just thought it could use another tag. The question was well-written and you showed your thought process. – Dylan Moreland Sep 30 '11 at 3:54 I know... I know... I forgot the the fact the valuation rings were local... If both $1/(T-a)$ and $1/ (T-b)$ are not in $R$, then $T-a$ and $T-b$ are both in the maximal ideal of $R$, $b-a = (T-a)-(T-b)$ is in the maximal ideal, but this is not right because it is nonzero and in $K$, thus invertible in $K$, also in $R$... – ShinyaSakai Sep 30 '11 at 5:43 Thanks again for editing and answering~ – ShinyaSakai Sep 30 '11 at 5:45

Here's a different approach. Let $R \subset k(x)$ be a valuation ring. Then the ideals of $R$ are totally ordered by inclusion. Therefore, there exists a unique maximal ideal $\mathfrak{m}$ of $R.$ Assuming $k[x] \subset R,$ it follows $\mathfrak{m} \cap k[x]$ is a maximal ideal of $k[x]$ and hence is principal generated by an irreducible polynomial $p(x).$

The elements of $k[x] \setminus (\mathfrak{m} \cap k[x])$ are not contained in any maximal ideal of $R$ and thus are units in $R.$ It follows that $R \supset k[x]_{(p(x))}.$

Note that $k[x]_{(p(x))}$ is a valuation ring with valuation group $\mathbb{Z}.$ It follows that the valuation group of $R$ i.e. $k(x)^{\times}/R^{\times}$ is a quotient of $\mathbb{Z}$ in the category of ordered Abelian groups and thus either trivial or equal to $\mathbb{Z}.$ In the former case, $R = k(x),$ whereas in the latter, $R = k[x]_{(p(x))}.$ We conclude as the former case was excluded by assumption, that $R = k[x]_{(p(x))}$ as desired.

Note that thus far the assumption that $k$ is algebraically closed has been superfluous. However, in the case that $k$ is algebraically closed $p(x) = x -a$ for some $a \in k.$ And therefore in this case the elements of $R = k[x]_{(p(x))}$ are of the form $f/g$ where $x-a\not|g.$

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 I see. This is a more concise proof. Thank you very much. – ShinyaSakai Sep 30 '11 at 5:51

This may be more than you asked for, but:

Theorem 15 in these notes classifies all valuations on a rational function field $K = k(t)$ which are trivial on $k$. Here $k$ may be any field of constants.

This result may be viewed as the "function field Ostrowski Theorem". In that there are no Archimedean norms to deal with, it is actually significantly easier than Ostrowski over $\mathbb{Q}$ (or over a number field).

You asked about valuation rings, not rank one valuations. Happily, since $k[T]$ is a PID, any ring intermediate between $k[T]$ and its fraction field is a localization of $k[T]$, hence Noetherian. So any valuation ring $k[T] \subset R \subset K(T)$ is indeed a discrete valuation ring, so is the valuation ring corresponding to some non-Archimedean $\mathbb{R}$-valued norm function.

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 Thank you very much for the answer and reference. This contains so much information on valuation rings, and I will study this carefully. Thanks again~ – ShinyaSakai Sep 30 '11 at 3:44

I have yet another strategy: $R$ is a valuation ring, and therefore it is local, with maximal ideal $\mathfrak{m}$. If $i: K[T] \to R$ is the inclusion, $\mathfrak{p} := i^{-1}(\mathfrak{m}) \subset K[T]$ will be a prime ideal. We would like to prove that $R = K[T]_{\mathfrak{p}}$. Once we have this, your claim follows directly from the definition of $K[T]_{\mathfrak{p}}$, as all prime ideals in $K[T]$ have the form $(T-a)$ for some $a \in K$.

So take some $x \in K(t)$ and assume $x \notin K[T]_{\mathfrak{p}}$. Then we can write $x = f/g$ where $g \in \mathfrak{p}$. But then, $g$ couldn't have been invertible in $R$ (as the invertible ones are exactly those not elements not lying in $\mathfrak{m}$). Therefore $x \notin R$. On the other hand, if $x \in K[T]_{\mathfrak{p}}$, then $x = f/g$ where $g \notin \mathfrak{p}$. As $R$ is a valuation ring we must either have $f/g \in R$ or $g/f \in R$. In the former case, we are done, in the latter case, it must then be true that $f \notin \mathfrak{m}$. But then, $x^{-1}$ is invertible in $R$ which means that $x \in R$ as well.

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 Thank you very much for showing me another way of proof. – ShinyaSakai Sep 30 '11 at 6:05