The Stacks project

Lemma 54.5.3. Let $X$ be a scheme which is Noetherian, Nagata, and has dimension $2$. Let $f : Y \to X$ be a proper birational morphism. Then there exists a commutative diagram

\[ \xymatrix{ X_ n \ar[r] \ar[d] & X_{n - 1} \ar[r] & \ldots \ar[r] & X_1 \ar[r] & X_0 \ar[d] \\ Y \ar[rrrr] & & & & X } \]

where $X_0 \to X$ is the normalization and where $X_{i + 1} \to X_ i$ is the normalized blowing up of $X_ i$ at a closed point.

Proof. We will use the results of Morphisms, Sections 29.18, 29.52, and 29.54 without further mention. We may replace $Y$ by its normalization. Let $X_0 \to X$ be the normalization. The morphism $Y \to X$ factors through $X_0$. Thus we may assume that both $X$ and $Y$ are normal.

Assume $X$ and $Y$ are normal. The morphism $f : Y \to X$ is an isomorphism over an open which contains every point of codimension $0$ and $1$ in $Y$ and every point of $Y$ over which the fibre is finite, see Varieties, Lemma 33.17.3. Hence there is a finite set of closed points $T \subset X$ such that $f$ is an isomorphism over $X \setminus T$. For each $x \in T$ the fibre $Y_ x$ is a proper geometrically connected scheme of dimension $1$ over $\kappa (x)$, see More on Morphisms, Lemma 37.53.6. Thus

\[ BadCurves(f) = \{ C \subset Y\text{ closed} \mid \dim (C) = 1, f(C) = \text{a point}\} \]

is a finite set. We will prove the lemma by induction on the number of elements of $BadCurves(f)$. The base case is the case where $BadCurves(f)$ is empty, and in that case $f$ is an isomorphism.

Fix $x \in T$. Let $X' \to X$ be the normalized blowup of $X$ at $x$ and let $Y'$ be the normalization of $Y \times _ X X'$. Picture

\[ \xymatrix{ Y' \ar[r]_{f'} \ar[d] & X' \ar[d] \\ Y \ar[r]^ f & X } \]

Let $x' \in X'$ be a closed point lying over $x$ such that the fibre $Y'_{x'}$ has dimension $\geq 1$. Let $C' \subset Y'$ be an irreducible component of $Y'_{x'}$, i.e., $C' \in BadCurves(f')$. Since $Y' \to Y \times _ X X'$ is finite we see that $C'$ must map to an irreducible component $C \subset Y_ x$. If is clear that $C \in BadCurves(f)$. Since $Y' \to Y$ is birational and hence an isomorphism over points of codimension $1$ in $Y$, we see that we obtain an injective map

\[ BadCurves(f') \longrightarrow BadCurves(f) \]

Thus it suffices to show that after a finite number of these normalized blowups we get rid at of at least one of the bad curves, i.e., the displayed map is not surjective.

We will get rid of a bad curve using an argument due to Zariski. Pick $C \in BadCurves(f)$ lying over our $x$. Denote $\mathcal{O}_{Y, C}$ the local ring of $Y$ at the generic point of $C$. Choose an element $u \in \mathcal{O}_{X, C}$ whose image in the residue field $R(C)$ is transcendental over $\kappa (x)$ (we can do this because $R(C)$ has transcendence degree $1$ over $\kappa (x)$ by Varieties, Lemma 33.20.3). We can write $u = a/b$ with $a, b \in \mathcal{O}_{X, x}$ as $\mathcal{O}_{Y, C}$ and $\mathcal{O}_{X, x}$ have the same fraction fields. By our choice of $u$ it must be the case that $a, b \in \mathfrak m_ x$. Hence

\[ N_{u, a, b} = \min \{ \text{ord}_{\mathcal{O}_{Y, C}}(a), \text{ord}_{\mathcal{O}_{Y, C}}(b)\} > 0 \]

Thus we can do descending induction on this integer. Let $X' \to X$ be the normalized blowing up of $x$ and let $Y'$ be the normalization of $X' \times _ X Y$ as above. We will show that if $C$ is the image of some bad curve $C' \subset Y'$ lying over $x' \in X'$, then there exists a choice of $a', b' \mathcal{O}_{X', x'}$ such that $N_{u, a', b'} < N_{u, a, b}$. This will finish the proof. Namely, since $X' \to X$ factors through the blowing up, we see that there exists a nonzero element $d \in \mathfrak m_{x'}$ such that $a = a' d$ and $b = b' d$ (namely, take $d$ to be the local equation for the exceptional divisor of the blowup). Since $Y' \to Y$ is an isomorphism over an open containing the generic point of $C$ (seen above) we see that $\mathcal{O}_{Y', C'} = \mathcal{O}_{Y, C}$. Hence

\[ \text{ord}_{\mathcal{O}_{Y, C}}(a) = \text{ord}_{\mathcal{O}_{Y', C'}}(a' d) = \text{ord}_{\mathcal{O}_{Y', C'}}(a') + \text{ord}_{\mathcal{O}_{Y', C'}}(d) > \text{ord}_{\mathcal{O}_{Y', C'}}(a') \]

Similarly for $b$ and the proof is complete. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BBT. Beware of the difference between the letter 'O' and the digit '0'.