Building a hypersurface from hyperplane sections Let $P_0,\ldots,P_n$ denote the coordinate hyperplanes in $\mathbb P^n$, and suppose that in each $P_i$ I have a degree $d$ hypersurface $V_i$. I am trying to understand what is the obstruction to the existence of a degree $d$ hypersurface $V \subset \mathbb P^n$ such that $V \cap P_i = V_i$.
There are some obvious obstructions: for example, my prescribed hyperplane sections $V_i$ must agree on intersections so that $V_i \cap P_j = V_j \cap P_i$.  I don't think this alone is quite sufficient.  In the case $n=3$ for example, let $p = P_1 \cap P_2 \cap P_3$ and suppose that $V_1,V_2,V_3$ all go through this point.  Then $T_p V_1 + T_p V_2 + T_p V_3$ had better have dimension $2$, not $3$.
I am sure there is some exact sequence I'm missing - what is it?
(note: of course $V$ does not have to be unique, at least if $d \geq n+1$.  You can add monomials containing all of the variables to the defining equation, and it doesn't change the intersection of the resulting surface with the coordinate hyperplanes)
 A: The condition $V_i \cap P_j = V_j \cap P_i$ (if understood properly) is the only obstruction. To see this denote by $Z$ the union of $P_i$. Then we have an exact sequnce
$$
0 \to \mathcal{O}_Z \to \bigoplus \mathcal{O}_{P_i} \to \bigoplus \mathcal{O}_{P_i \cap P_j} \to \bigoplus \mathcal{O}_{P_i \cap P_j \cap P_k} \to \dots
$$
(such an exact sequence exists for any union of transverse hypersurfaces and can be proved by induction on the number of components). Tensoring it by $\mathcal{O}(d)$ one deduces an isomorphism
$$
H^0(Z,\mathcal{O}_Z(d)) = \mathrm{Ker}\Big(\bigoplus H^0(P_i,\mathcal{O}_{P_i}(d)) \to \bigoplus H^0(P_i \cap P_j, \mathcal{O}_{P_i \cap P_j}(d)) \Big).
$$
Thus, to give a section of $\mathcal{O}_Z(d)$ one needs to give a collection of sections of $\mathcal{O}_{P_i}(d)$ that agree on pairwise intersections (this is the right way to state the condition).
On the other hand, $Z$ is a hypersurface of degree $n+1$, hence there is an exact sequence
$$
0 \to \mathcal{O}(d-n-1) \to \mathcal{O}(d) \to \mathcal{O}_Z(d) \to 0
$$
on $\mathbb{P}^n$, and since $H^1(\mathbb{P}^n,\mathcal{O}(d-n-1)) = 0$ (I assume that $n \ge 2$), it follows that the morphism
$$
H^0(\mathbb{P}^n, \mathcal{O}(d)) \to H^0(Z, \mathcal{O}_Z(d)) 
$$
is surjective, hence any such section lifts to an equation of a hypersurface in $\mathbb{P}^n$.
