From here: can someone provide a proof of the true implication (linear implies algebraic)? I am taking definitions from Griffiths-Harris, 461.
I'm afraid I really fail to see how the existence of a (meromorphic) section whose principal divisor is $D$ implies the "deformation" property at the base of the definition of algebraic equivalence.
Thank you in advance.
Write out the definitions. Suppose $C,D$ are divisors on $X$. They're linearly equivalent if there is a meromorphic function $f$ so that $C=D+(f)$, or in other words, the map $\mathcal{O}(C)\to \mathcal{O}(D)$ given by $1\mapsto f$ is an isomorphism. They are algebraically equivalent if there exists a connected scheme $T$ with points $t_1,t_2\in T$ and a line bundle $L$ on $X\times T$ so that $L|_{X\times t_1}\cong \mathcal{O}(C)$ and $L|_{X\times t_2}\cong\mathcal{O}(D)$.
Do you see a natural choice of $T$ for the case of two linearly equivalent divisors?