Corollary 2

*Let $K$ be a field, $K(X)$ the field of rational functions in one indeterminate over $K$ and $V$ a valuation ring of $K(X)$ containing $K$ and distinct from $K(X)$. If $X \in V$, there exists an irreducible polynomial $P \in K[X]$ such that $V = (K[X])_{(P)}$; otherwise $V$ is the local ring of $K[X^{-1}]$ at the prime ideal $X^{-1}K[X^{-1}]$ (in other words $V$ is composed of the fractions $A/B$, where $A \in K[X]$ and $B \in K[X]$, such that $\deg(A) \leq \deg(B)$).

If $X \in V$, then $K[X] \subset V$ and the assertion made follows from Proposition 3. If $X \notin V$, then $X^{-1} \in V$, whence $K[X^{-1}] \subset V$; then $V$ is the local ring of $K[X^{-1}]$ at a prime ideal $p$ (Proposition 3) and this ideal contains $X^{-1}$ since $X^{-1}$ is not invertible in V; then $p = X^{-1}K[X^{-1}]$ since the latter ideal is maximal. Finally let us consider a rational function $A(X)/B(X)$, where A and B are polynomials of respective degrees $a$ and $b$; then $A(X) = X^a A'(X^{-1})$ and $B(X) = X^b B'(X^{-1})$, where $A'$ and $B'$ are polynomials such that $A'(0) \neq 0$ and $B'(0) \neq 0$; hence, for $A(X)/B(X)$ to belong to the local ring of $K[X^{-1}]$ at $X^{-1}K[X^{-1}]$, it is necessary and sufficient that $a \leq b$.