A note on k-modules in an algebraic function field K/k of one variable

Let K be an algebraic function field of one variable with a constant field k. It is not necessary, in this note, that k is the exact constant field of K. We shall denote by M a finitely generated k-module in K. Moreover n(M) denotes the denominator divisor of M, i.e., n(M) is the divisor of K defined by ord p n(M)=max{ord p n(x)} x∈M xキO for every prime divisor p of K, where ord denotes the order at p and n(x) means the denominator divisor of x. Then it is well-known that there exists some element x in M such that n(M)=n(x), if k contains enough elements. (e.g., E. Artin 〔1〕; P.318, Lemma 2). The purpose of this note is to study it in minute detail. Under the condition d(n(M))≦ |k|, we shall prove that there exists an element x in M such that n(M)=n(x) in §2, where d(n(M)) means the degree of n(M) and |k| denotes the number of all the elements contained in k; if k is not finite, then we shall put |k|=∞. In §3, we shall show that the above inequality is the best condition in a sense