Suppose $X$ is a reduced proper schemes over a noetherian ring, $X_i$ are its irreducible components，$L$ is a invertible sheaf on $X$. If $L|_{X_i}$ are all ample, how do we show $L$ is ample?
If $X=X_1\cup X_2$, $F$ be a coherent sheaf on $X$, then $0\to I_{X_1}\to O_{X_1\cup X_2}\to O_{X_1}\to 0$，using long exact sequence, we have
$\cdots\to H^i(FI_{X_1}\otimes L^n)\to H^i(F\otimes L^n)\to H^i(FO_{X_1}\otimes L^n)\to\cdots$
When $n$ is sufficiently large, the third terms would be $0$, but how to deal with the first terms?
The key point is that $I_{X_1}$ is actually a sheaf on $X_2$, i.e. its support is all on $X_2$. This is where reducedness comes in. Looking at $I_{X_1}$ restricted to any affine open $SpecA$ of $X - X_2$, it corresponds to an ideal which is contained in all the prime ideals of $A$. But $A$ has trivial nilradical, so $I_{X_1}|_{SpecA} = 0$. Thus, $H^i(FI_{X_1}\otimes L^n,X)=H^i(FI_{X_1}|_{X_2} \otimes L|_{X_2}^n,X_2)=0$ for $n>>0$.