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60     Appendix to Chapter 2

                              +
                        0 ≤ ord ( f ) = ord ( f 1 ) +· · · + ord ( f r ).
                                                      +
                                        +
                              p         p             p
                           +
        We can assume that ord ( f i ) ≥ 0 for all i and that p is irreducible in R. This means
                           p
        that each f i is an irreducible in R[x].
           Let f be an irreducible in R[x] with f (x) dividing a(x)b(x) in R[x]. If deg( f ) =
        0, then f is an irreducible in R and
                            +
                                                       +
                                           +
                      0 < ord (a(x)b(x)) = ord (a(x)) + ord (b(x)).
                            f               f          f
                                                             +
                                   +
        Hence f divides a(x) when ord (a(x)) > 0or b(x) when ord (b(x)) > 0. If
                                   f                          f
                          +
        deg( f )> 0, then ord ( f ) = 0 for every irreducible p ∈ R and f is irreducible in
                          p
        F[x] by the argument in the previous paragraph. Since F[x] is factorial, f (x) divides
                                                                   +
        a(x) or b(x) in F[x]and so a(x) = f (x)q(x) for example. Then 0 ≤ ord (a(x)) =
                                                                   p
                                  +
                      +
           +
        ord ( f (x))+ord (q(x)) = ord (q(x)) and q(x) ∈ R[x]. Hence f (x) divides a(x)
                      p
           p
                                  p
        in R[x] with quotient q(x). This proves the theorem.
        §3. Remarks on Valuations and Algebraic Curves
        Let k be an algebraically closed field, and let V denote the set of valuations up to
        equivalence on the field k(x) of rational functions in one variable over k which are
        trivial on k. We map P 1 (k)  → V by
                     (1, a) in P 1 (k)  → v (1,a) = ord x−a  for k[x] ⊂ k(x),
                ∞= (0, 1) in P 1 (k)  → v (0,1) = ord 1/x  for k[1/x] ⊂ k(x).
        Then the function P 1 (k) → V given by P  → v p , is a bijection. For it is clearly
        an injection, and to show that each valuation v is of the form v p , consider the ideal
        M v ∩ k[x]in k[x]if x ∈ R v or the ideal M v ∩ k[1/x]if x ∈ R v . The remainder of
        the argument is left to the reader.
        (3.1) Remark. Let C f be an irreducible plane curve defined over k. Then the inte-
        gral domains

                         k[x, y]       k[w, y]      k[w, x]
                                ,             ,
                        f (1, x, y)   f (w, 1, y)   f (w, x, 1)
        all have naturally isomorphic fields of fractions, denoted k(C f ), and called the func-
        tion field of the curve. The subrings k[w], k[x], and k[y] all inject into k(C f ) except
        when f is w, x, or y, respectively. Thus the field k(C f ) is a finite extension of
        k(w), k(x), or k(y). The main assertion is: there is a natural map from C f (k) to val-
        uations on k(C f ) trivial over k up to equivalence which is a bijection when C f is
        nonsingular and k is algebraically closed. This map extends P 1 (k) → V considered
        above in the sense that for k(P 1 ) = k(t) ⊂ k(C f ) the points of C f mapping onto
        P ∈ P 1 under t : C f → P 1 correspond to the extensions to k(C f ) of the valuation
        v p on k(P 1 ).
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