**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.
$\square$

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