# Constructing normal crossing varieties

Let $X_i$ be a smooth projective variety with a smooth divisor $D_i$ for $i=1,2$. Suppose that $D_1$ is isomorphic to $D_2$. Then does it make sense to construct a normal crossing variety $X=X_1 \cup_D X_2$ by 'pasting' $D_1$ and $D_2$, where '$\cup_D$' means pasting along $D_1$ and $D_2$?

If the answer is yes. Suppose that there is ample divisor $H_i$ on $X_i$ such that $H_1|_{D_1}$ is lineary equivalent to $H_2|_{D_2}$. Then is the normal crossing variety $X$ necessarily projective?

• For curves, this is true. The construction is standard, you can see e.g. geometry of algebraic curves Vol 2, especially for the construction of the line bundle – Giulio Mar 30 '15 at 14:42
• Which chapter (of "geometry of algebraic curves Vol 2") contains this? – Creg Mar 30 '15 at 15:20

Q2: Yes, and the proof is more or-less completely explicit. Choose a high enough $n$ such taht $\mathcal O(n H_i)$ is a projextive embedding of $X_i$ and $\mathcal I_{D_i} \otimes \mathcal O(n H_i)$ is globally generated with no higher cohomology, for both $i$.
Then in the embedding $X_i \to \mathbb P^{N_i}$, $D_i$ is contained in some linear subspace cut out by the functions in $H^0(X_i, \mathcal I_{D_i} \otimes \mathcal O(n H_i))$. Because they generate, in fact $D_i$ is the intersect of $X_i$ with that linear subspace. Because the map $H^0(X_i, \mathcal O(nH_i)) \to H^0(X_i, \mathcal O_{D_i} (nH_i))$ is surjective, the emedding of $D_i$ into that linear subspace is just the embedding coming from the very ample line bundle $nH_i$. This embedding is the same whether $i=1$ or $2$.
So $X_1$ and $X_2$ are expressed as two projective varieties with an isomorphism between their restrictions to two linear subspaces. We can easily embed the two projective spaces in a higher-dimensional projective space such that their intersection is exactly that linear subspace. Then the union of $X_1$ and $X_2$ in that higher space will clearly be $X_1 \cup_D X_2$, and hence it is a projective variety.