$K$ 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:
- to prove that for some $a \in K$, $\frac{1}{T-a} \notin R$;
- to prove that for any $b \in K, b \neq a$, $\frac{1}{T - b}$ is contained in $R$;
- 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 $k, a_i \in K$.
- 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 $\frac{1}{T-a} \notin R$.
- This is where my question comes. What will happen if for $\frac{1}{T-b} \notin R$ for some $b \in K$ other than $a$?
- 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.