# Does a section having no zero locus divisor imply that it is nowhere vanishing?

Let $X$ be a normal variety over a field $k$, $L$ is a line bundle on it, and $s$ is a global section of $L$. If the Weil divisor associated to $s$ equals zero. Then can I conclude that $s$ is nonvanishing everwhere?

Equivalently, let $A$ be a integrally closed finitely generated $k$-algebra, and $s\in A$, if $s$ is not a unit, then can I conclude that there exits a prime idea $p$, such that

1. dim($A_p$)=1
2. the valuation of $s$ corresponding to $p$ is non-zero.

If it is not true in general, what extra conditions should I impose on the variety $X$?

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How about $X = {\rm Spec}\ k[z]$, $L = {\mathcal O}_X$, and $s=z$? Which is to say, let's start by asking $X$ complete. –  Allen Knutson Nov 6 '12 at 7:33

Yes, a normal domain equals the intersection of its localizations at height one primes inside its fraction field. Hartshorne ("Algebraic Geometry", Prop. II 6.3A) gives the reference: Matsumura "Commutative Algebra", Th. 38, p. 124. We apply this to $s$ and $1/s$.