$\newcommand\P{\mathbb P}$I read the following classic result which does not give a proof but refers to a very old Italian work (Gallarati) that I cannot get hold of (and even if I do, probably I wouldn't understand it as it is in Italian):
Suppose that $X$ is an irreducible surface nonsingular in codimension 1 in $\P^3$ and $Y,Y'\ne X$ are surfaces in $\P^3$ such that $$X\cap Y = C_1\cup \dots \cup C_k$$ and $X\cap Y' = C_1$ , where $C_i$ are irreducible curves. Then there exists another surface $Z$ such that $$X\cap Z = C_2 \cup \dots \cup C_k$$
By curve, surface etc. I mean in the context of algebraic geometry i.e. algebraic and over an algebraically closed field. Maybe this is trivial in modern algebraic geometry but I cannot see the proof. In fact I am more interested in the construction of $Z$ rather than the proof. Can anyone assist me in constructing $Z$ or at least give a proof of this (or maybe cite a reference that is accessible)?
The first equality gives a linear equivalence $$ dH \sim n_1C_1 + n_2C_2 + \dots + n_kC_k $$ of Weil divisors on $X$, where $H$ is the hyperplane class, $d$ is the degree of $Y$, and $n_i$ are some positive integers. Similarly, the second equality means $$ d'H \sim n'C_1. $$ Therefore $$ (n'd - d'n_1)H \sim n'n_2C_2 + \dots + n'n_kC_k. $$ The right side is effective, hence $n'd - d'n_1 > 0$. Since $X$ is a hypersurface, the linear system $\mathcal{O}_{\mathbb{P}^3}(n'd - d'n_1)$ cuts the complete linear system $|(n'd - d'n_1)H|$ on $X$, which implies the existence of a surface $Z$ as required.