Suppose that $X$ is a quasiprojective variety over an algebraically closed field $k$, thought of as a scheme. Is the Picard group of the $X$, in the sense of schemes, isomorphic to the Picard group, in the classical sense, of the underlying quasiprojective variety $X(k)$ whose points are the closed points of $X$?
I feel like the answer should be "yes", but I have not been able to find a reference. Here's how I think the argument should go. First, as $X(k)$ is very dense in $X$ (see this answer), the lattices of open sets on $X$ and $X(k)$ coincide. Since Picard groups can be computed from Cech cohomology, i.e. $Pic(X) = \check{H}^1(X, \mathcal{O}^*_X)$ and $Pic(X(k)) = \check{H}^1(X(k),\mathcal{O}^*_{X(k)})$, the fact that $\mathcal{O}_X^*(A) = \mathcal{O}^*_{X(k)}(A(k))$ for any open affine subscheme $A\subset X$, seems to imply that there is an isomorphism $$Pic(X)\rightarrow Pic(X(k))$$ obtained by sending a line bundle on $X$ to its restriction to $X(k)$.
