Let $X$ be a smooth projective variety over $\mathbb{C}$. For $L$ a line bundle, one can define the first Chern class in $H^2(X; \mathbb{Z})$ (for example using the exponential exact sequence, a generic smooth section, or the Euler class - I think they are all the same.)
Moreover, given a smooth closed subvariety $Y \subset X$, one can associate to a fundamental class in the cohomology of $X$, by extending the Thom class of a tubular neighborhood / Normal bundle. Such cycles are said to be algebraic. (I think.) (There is a similar procedure when this subvariety is singular, by stratifying it.)
Question: For $L$ an algebraic line bundle (the transition maps are polynomial), is $c_1(L)$ an algebraic cycle?
I would also like to know if the collection of algebraic cycles and cocycles is closed under cup and cap products. I guess so, but I could use a reference.
Partial answer:
Let $O(1)$ denote some very ample line bundle on X. Then $L(n))$ for some $n$ is globally generated, and hence defines a map to projective space $\phi : X \to P^N$, so that $\phi^*(O_{P^N}(1)) = L(n)$.
Provided that the pullback of an algebraic cycle is algebraic (which I don't know to prove still), we are done, because then the first chern of $L(n)$ is algberaic, being the pullback of the Chern class of $O_{P^N}(1))$ which is a hyperplane, and hence algebraic.
Now we just use that $c_1(O_X(1))$ is a hyperplane section of $X$, and hence algebraic, and that $c_1(L(n)) = c_1(L) + c_1(O_X(n))$ to conclude.
So now I need:
Lemma: Let $\phi : X \to Y$ be a regular map of smooth complex projective (algebraic) varieties. Let $Z \subset Y$ be a subvariety, and $[Z]$ the fundamental class in $H^*(Y)$. Then $[\phi^{-1}(Z)] = \phi^*([Z])$.
Rmk: In the case that $Z$ is a smooth subvariety, I understand reasonable well how to define this fundamental class - for example, it can be defined by pulling back forms to on $Y$ to $Z$, and then integrating them, and then using Poincare duality and universal coefficients. In the case of singular varieties, this is less clear -- I expect that one can pullback to the generic smooth open (or to a resolution of singularities) and use the same trick, though.