So I've been learning about divisors on schemes lately and have a question about defining the sheaf $\mathcal{L}(D)$ for a Weil divisor $D$. Suppose $X$ is a noetherian separated integral scheme, regular in codimension $1$. Let $D = \sum n_{i} \cdot Y_{i}$ be a Weil divisor. Several different sources, for example here and here, say that if $X$ is normal, then we have a coherent sheaf $\mathcal{L}(D)$ by, $$ \Gamma(U, \mathcal{L}(D)) = \left\lbrace s \in K(X) : \text{div}(s)|_{U} + D|_{U} \geq 0 \right\rbrace $$ My problem is, I don't see where normality needs to be used to do this. I have my attempt at a proof, which I think is correct. Is someone able to tell me if what I've done is correct, and if so, is normality just not necessary?
It seems easy enough to show that $\mathcal{L}$ is a sheaf. My proof that it is coherent is as follows:
Let $U = \text{Spec } A$ be an affine open subset in $X$. I will construct an isomorphism $$ \psi: \Gamma(\text{Spec } A, \mathcal{L}(D))^{\sim} \stackrel{\simeq}{\longrightarrow} \mathcal{L}(D)|_{\text{Spec } A} $$ Since $X$ is qcqs, the sections of $\Gamma(\text{Spec } A, \mathcal{L}(D))^{\sim}$ over $D(f) \subseteq \text{Spec } A$ is precisely $\Gamma(\text{Spec } A, \mathcal{L}(D))_{f}$ and the corresponding restriction map is the canonical localization morphism for modules.
Identify $f \in A$ with its corresponding section in $\Gamma(\text{Spec } A, \mathcal{O}_{X})$. We first show that the section $f|_{D(f)} \in \Gamma(D(f), \mathcal{O}_{X})$ acts as an automorphism on the module $\Gamma(D(f), \mathcal{L}(D))$. But the section $f|_{D(f)}$ corresponds to $\frac{f}{1} \in A_{f}$ which is an invertible element, and since $\Gamma(D(f), \mathcal{L}(D))$ is an $A_{f}$-module this is clear. So let $\tau$ be the morphism $$ \tau: \Gamma(\text{Spec } A, \mathcal{L}(D))_{f} \longrightarrow \Gamma(D(f), \mathcal{L}(D)) $$ given by the universal property for localization.
This shows the existence of a morphism of sheaves in one direction. It remains to how that $\tau$ is an isomorphism. This suffices since we chose $f \in A$ arbitrarily and the distinguished affine open sets form a base in $\text{Spec }A$.
For surjectivity, let $s \in \Gamma(D(f), \mathcal{L}(D))$ be a section. So $s \in K(X)$ satisfying $\text{div}(s)|_{D(f)} + D|_{D(f)} \geq 0$. Let the divisor $D$ be given by $$ D = \sum n_{i} \cdot Y_{i} $$ where the sum is finitely supported and similarly $$ \text{div}(s) = \sum \nu_{Y_{i}}(s) \cdot Y_{i} $$ Restricted to $D(f)$, we have $$ \text{div}(s)|_{D(f)} + D|_{D(f)} = \sum ( n_{i} + \nu_{Y_{i}}(s)) \cdot (Y_{i} \cap D(f)) $$ where we only consider those divisors $Y_{i}$ that have non-empty intersection with $Y_{i}$. In order to obtain a section in $\Gamma(\text{Spec } A, \mathcal{L}(D))$ that restricts to this, we need only consider the coefficients of prime divisors meeting $\text{Spec } A$ but not meeting $D(f)$. Suppose $Y$ is a prime divisor in the support of $D$ which meets $\text{Spec } A$ but does not meet $D(f)$ so that $Y \subseteq V(f)$. Suppose $Y$ has integer coefficient $m$ in the representation of $D$. Let $\mathfrak{p}$ be the prime ideal of height $1$ in $A$ corresponding to the generic point of $Y \cap \text{Spec } A$ in $\text{Spec } A$. Since $Y \subseteq V(f)$ we have that $f \in \mathfrak{p}$ so that $\nu_{Y}(f) > 0 $. Raising $f$ to a sufficiently large power $N$, we have $$ \nu_{Y}(f^{N}s) + m = \nu_{Y}(f^{N}) + \nu_{Y}(s) + m \geq 0 $$ Since $X$ is noetherian, $V(f)$ can contain only finitely many prime divisors of $X$, and so we can choose an $N$ large enough to work for all such coefficients. Then we have the section $f^{N}s \in \Gamma(\text{Spec } A, \mathcal{L}(D))$ since $$ \text{div}(f^{N}s)|_{\text{Spec } A} + D|_{\text{Spec } A} = \sum (n_{i} + \nu_{Y_{i}}(f^{N}s)) \cdot (Y_{i} \cap \text{Spec } A) \geq 0 $$ This gives us that $s \in \Gamma(\text{Spec } A, \mathcal{L}(D))_{f}$.
Injectivity is clear from the fact that $\mathcal{L}(D)$ is a subsheaf of the constant sheaf assigning $K(X)$.
Is my proof correct? If so, have I implicitly used normality somewhere without realizing it?
What you are overlooking is that the associated line bundle $\mathcal{L}(D)$ is meant to be an invertible sheaf. The normality condition ensures this construction always results in an invertible sheaf.