It should be obvious from the question that I am not an algebraic geometer, and so I would really appreciate an answer without using schemes or functor.
Let $V$ be an (embedded) variety in a complex projective space, and let $L \subset V$ by a subvariety of $V$ of codimension 1 in $V$.
Question: is it possible to find a homogeneous polynomial $F$ (depending on the embedding of $V$, but that's OK) such that $x \in V$ and $F(x)=0$ if and only if $x \in L$ ? (here $x$ is to be understood in homogeneous coordinates).
If this is false, then can it always be done using more than one $F$ (ie, can we always find homogeneous pol. $F_1, \ldots, F_r$ such that $x \in V$ and for all $i \leq r$, $F_i(x)=0$ is equivalent to $x \in L$?
I would very vaguely think so, but that would require to somehow prove that one can choose a set of generators of $I(V)$ that could be completed into a set of generators of $I(L)$ by just adding $F$, and I'm not sure how to do that.
For the first, the answer is no and for the latter the answer is yes and your vague feeling is correct.
Let me give you a simple example to illustrate why the first is false. Take $V$ to be the quadric defined by $xy-zw=0$ in projective 3-space. Take $L$ to be a line, say given by $x=z=0$ contained in $V$. One can prove (and not difficult) that there is no single homogeneous polynomial in $x,y,z,w$ satisfying your requirement.