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Is the following statement (maybe with more or less conditions) true?

Given a sequence of proper birational morphisms of integral varieties $$X\leftarrow X_1\leftarrow X_2\leftarrow X_3\leftarrow... .$$ Suppose there is another morphism $X\leftarrow Y$ which is finite birational with $Y$ integral and is factorized through every $X_i$. Then the sequence is finally stationary.

This question arised from remark 8.1.27 in Liu's Algebraic geometry and arithmetic curves, in which he said this is true for curves (if I understand it correctly).

-------------- edits: a case with two more assumptions --------------------

Denote the morphisms $X_i\rightarrow X_0:=X$ by $g_i$, the morphisms $Y\rightarrow X_i$ by $f_i$, and the morphisms $X_{i+1}\rightarrow X_i$ by $\pi_i$.

As $f_0$ is finite birational (hence surjective), so is every $f_i$, which means $g_i$ are quasi-finite and therefore also finite (By EGA-IV-8.11.1). So now all $f_i, g_i, \pi_i$ are finite birational (hence of finite fibres and surjective).

Now make the first more assumption that the integral varieties $X_i$ have ample $O_{X_i}$-modules (I think this assumption makes the above finte morphisms become projective in the defintition used in [Liu] and [Hartshorne], right?).

Then [Liu, 8.1.24] implies all $\pi_i$ are blowing-ups along some closed subscheme $Z_i$ of $X_i$. Now we make the second more assumption that $\pi_i$ maps $Z_{i+1}$ into $Z_i$ (set-theoretically) and $Z_0 (\subseteq X_0=X)$ is of dimension $0$.

Then the facts that all morphisms in consideration are surjective and of finite fibres and [Liu, 8.1.12(d)] imply for sufficiently large $n$, $\pi_n$ induces set-theoretically bijective maps $Z_{n+1}\rightarrow Z_n$, and (schematic) isomorphisms $X_{n+1}\setminus Z_{n+1}\rightarrow X_n\setminus Z_n$.

Finally I want use [Liu, 7.2.20(b)] to conclude this case, i.e. I want to show for every $x\in Z_{n+1}$ we have an isomophism of stalks $O_{X_n,\pi_n(x)}\rightarrow O_{X_{n+1},x}$, and I am stucked here.

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As $X_i\to X$ is dominated by $Y\to X$, $X_i\to X$ is quasi-finite, hence finite. Now you should be able to complete the answer. –  user18119 Jun 13 '12 at 22:48
    
@QiL, thank you. I considered a more restricted case (editted in the post), but not successful.. –  Ch Zh Jun 16 '12 at 10:11

1 Answer 1

up vote 1 down vote accepted

We have an increasing sequence of coherent sheaves on $X$: $$ O_X\subseteq (g_1)_*O_{X_1} \subseteq (g_2)_*O_{X_2} \subseteq \cdots $$ all containd in $f_*O_Y $, where $f$ is the morphism $Y\to X$. As $f_*O_Y$ is coherent and $X$ is quasi-compact, the above sequence is stationary. Suppose $(g_r)_*O_{X_r}=(g_{n})_*O_{X_n}$ for all $n\ge r$. As the $g_i$'s are all finite, this implies that $X_n\to X_r$ are isomorphisms for all $n\ge r$.

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