I've learned when you have a integral smooth scheme line bundles are the same as Cartier divisors are the same as Weil divisors. My question is to what extent does this continue to hold (if at all) when you are talking about ind objects. Since things become more subtle in positive characteristic lets say we are working over an algebraically closed field of characteristic 0.

Background

I think the situation can be quite different because smoothness, or algebraic smoothness is more complicated. In the non ind scheme case you can define smoothness of a point $p \in X$ as $S^q(m/m^2) \to m^q/m^{q+1}$ being an isomorphism for all $q\ge 0$, where $m$ is the maximal ideal of the local ring at $p$. In the ind case you have varieties $(X_n)_{n \ge 0}$ and $p \in X_n$ for all sufficiently large $n$ so you have $S_n^q(m_n/m_n^2) \to m^q_n/m^{q+1}_n$ for all sufficiently large $n$. Algebraic smoothness at $p$ means that the map

$\varprojlim S_n^q(m_n/m_n^2) \to \varprojlim m^q_n/m^{q+1}_n$

is an isomorphism for all $q\ge 0$

Thoughts

It is true that if $X$ is the union of $X_n$ and each $X_n$ is smooth then $X$ is algebraically smooth. However algebraic smoothness is not inductive in the sense that there are examples of $p \in X$ being algebraically smooth even when $p \in X_n$ is a singular point for all $X_n$ containing $p$. However line bundles are inductive in the sense that a line bundles on $X$ determines line bundles on each $X_n$ compatible under pull back and vice versa.

This suggests to me that you could have a collection of compatible Weil divisors in each $X_n$ which are not the zero section of any section of any line bundle. But I don't have an example.