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This is related to Hartshorne's book Algebraic Geometry, Lemma II.7.8, page 158.

Let $X$ be a nonsingular projective variety over an algebraically closed field $k$ and let $\mathcal{L}$ be an invertible sheaf on $X$. For any $s \in \Gamma(X, \mathcal{L})$, we have $\mathrm{Supp}(s)_{0} = (X_{s})^{c}$, where $\mathrm{Supp}(s)_{0}$ means the union of the prime divisors of the divisor of zeros $(s)_{0}$ of $s$ and $X_{s} = \{ P \in X \mid s_{p} \not\in \mathfrak{m}_{p}\mathcal{L}_{p} \}$.


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up vote 3 down vote accepted

Since checking that two subsets of $X$ are equal can be done locally, we may assume that $X=Spec(A)$ is affine and that $\mathcal L=\mathcal O_X$, since $\mathcal L$ is an invertible sheaf i.e. a locally free sheaf of rank one.

But then $s\in \Gamma(X,\mathcal O)=A$ is just an ordinary regular function on $X$ and to say that $s_P\notin \mathfrak m_P \mathcal L_P=\mathfrak m_P \mathcal O_P= \mathfrak m_P$ means that $s(P)=s_P \:\text{mod }\mathfrak m_P\neq0\in \kappa(P)=\mathcal O_P/\mathfrak m_P=k$.
So we see that $X_s=D(s)$ and $(X_s)^c=V(s)$.

On the other hand the prime divisors of $(s)_0$ are the irreducible components of $V(s)$ so that their union $(s)_0$ is also equal to $V(s)$ .

Thus we have the required equality $\mathrm{Supp}(s)_{0} = (X_{s})^{c} $.

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