For any variety $X$, we can find a birational equivalent model $X_1\times \ldots\times X_m$ such that every $X_i$ is indecomposable in the birational sense. For example, the case of surfaces is due to the Enriques–Kodaira classification: $\mathbb P^2$ decomposes as $\mathbb P^1 \times \mathbb P^1$, a ruled surfaces decomposes as $C\times \mathbb P^1$, and a general hypersurface in $\mathbb P^3$ is indecomposable. I want to know that:
(1) Is the decomposition unique? That is, if we have two decomposition $X_1\times \ldots\times X_m$ and $X'_1\times \ldots\times X'_n$, then it has to be $m=n$ and pairwise birational equivalent.
(2) For a smooth hypersurface $X\subset \mathbb P^n$ of degree $d$, is it true that for $d$ big enough, the $X$ is always indecomposable?
I guess (1) is false and (2) is true. Besides, I guess this is more or less the minimal model program, am I right?
Thanks in advance.
Answer to question 2) is yes. If $d\geq n+1$, then they are indecomposable. The argument is same as the one for surfaces. If $X$ is such a surface birational to $Y\times Z$, since $h^i(\mathcal{O})$ is a birational invariant, we get by Kunneth formula $h^r(\mathcal{O}_Y)=0$ where $r=\dim Y$, since it is a direct summand of $h^r(\mathcal{O}_X)$ which is zero since $0<r<\dim X$. Similarly $h^s(\mathcal{O}_Z)=0$ where $s=\dim Z$ and thus you get $h^{r+s}(\mathcal{O}_{Y\times Z})=0$. But $r+s=\dim X$ and $h^{r+s}(\mathcal{O}_X)\neq 0$ since $\deg X\geq n+1$.