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    Zeros and poles of a rational function

    PR-DIVZEROPOLE

    Codimension-one valuations detect only finitely many zeros and poles; on a normal scheme, no poles means regular.Suggest a slogan

    Let \(X\) be an integral Noetherian scheme with function field \(K\). For a prime divisor \(Y \subseteq X\) with generic point \(\eta_Y\) such that \(\OO_{X, \eta_Y}\) is a discrete valuation ring with valuation \(v_Y\), a function \(t \in K^\times\) has a zero along \(Y\) if \(v_Y(t) > 0\) and a pole along \(Y\) if \(v_Y(t) < 0\).

    1. A Noetherian topological space has finitely many irreducible components [@Har10a, Proposition I.1.5].

    2. For \(t \in K^\times\), there are only finitely many codimension-one points \(y \in X\) with \(t \notin \OO_{X,y}^\times\).

    3. If \(X\) is regular in codimension one, then \(\div t = \sum_Y v_Y(t)\, Y\) is a finite sum, hence a Weil divisor [@Har10a, Lemma II.6.1].

    4. If \(X\) is normal, then \(t \in K\) has no poles if and only if \(t \in \OO_X(X)\); on an affine open \(\Spec A\) this is \(A = \bigcap_{\operatorname{ht} \mathfrak{p} = 1} A_{\mathfrak{p}}\) for a Noetherian normal domain \(A\) [@Har10a, Proposition II.6.3A].

    Why does a rational function on an integral Noetherian scheme have only finitely many zeros and poles?
    Why is a rational function with no poles on a normal Noetherian scheme regular?

    Dependencies

    • D-QJ5M9Normal domains and schemes, and their relation to regularityuses
    • D-5PQ5WWeil and Cartier divisors, and when they agreeuses

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