The Stacks project

Proof. Let $X$ be some minimal model of $C$. The numerical type associated to $X$ has genus $0$ and is minimal (Definition 55.11.4 and Lemma 55.11.5). Hence by Lemma 55.6.1 we see that $X_ k$ is reduced, irreducible, has $H^0(X_ k, \mathcal{O}_{X_ k}) = k$, and has genus $0$. Let $Y$ be a second minimal model for $C$ which is not isomorphic to $X$. By Resolution of Surfaces, Lemma 54.17.2 there exists a diagram of $S$-morphisms

where each morphism is a blowup in a closed point. We will prove the lemma by induction on $m$. The base case is $m = 0$; it is true in this case because we assumed that $Y$ is minimal hence this would mean $n = 0$, but $X$ is not isomorphic to $Y$, so this does not happen, i.e., there is nothing to check.

Before we continue, note that $n + 1 = m + 1$ is equal to the number of irreducible components of the special fibre of $X_ n = Y_ m$ because both $X_ k$ and $Y_ k$ are irreducible. Another observation we will use below is that if $X' \to X''$ is a morphism of regular proper models for $C$, then $X' \to X''$ is an isomorphism over an open set of $X''$ whose complement is a finite set of closed points of the special fibre of $X''$, see Varieties, Lemma 33.17.3. In fact, any such $X' \to X''$ is a sequence of blowing ups in closed points (Resolution of Surfaces, Lemma 54.17.1) and the number of blowups is the difference in the number of irreducible components of the special fibres of $X'$ and $X''$.

Let $E_ i \subset Y_ i$, $m \geq i \geq 1$ be the curve which is contracted by the morphism $Y_ i \to Y_{i - 1}$. Let $i$ be the biggest index such that $E_ i$ has multiplicity $> 1$ in the special fibre of $Y_ i$. Then the further blowups $Y_ m \to \ldots \to Y_{i + 1} \to Y_ i$ are isomorphisms over $E_ i$ since otherwise $E_ j$ for some $j > i$ would have multiplicity $> 1$. Let $E \subset Y_ m$ be the inverse image of $E_ i$. By what we just said $E \subset Y_ m$ is an exceptional curve of the first kind. Let $Y_ m \to Y'$ be the contraction of $E$ (which exists by Resolution of Surfaces, Lemma 54.16.9). The morphism $Y_ m \to X$ has to contract $E$, because $X_ k$ is reduced. Hence there are morphisms $Y' \to Y$ and $Y' \to X$ (by Resolution of Surfaces, Lemma 54.16.1) which are compositions of at most $n - 1 = m - 1$ contractions of exceptional curves (see discussion above). We win by induction on $m$. Upshot: we may assume that the special fibres of all of the curves $X_ i$ and $Y_ i$ are reduced.

Since the fibres of $X_ i$ and $Y_ i$ are reduced, it has to be the case that the blowups $X_ i \to X_{i - 1}$ and $Y_ i \to Y_{i - 1}$ happen in closed points which are regular points of the special fibres. Namely, if $X''$ is a regular model for $C$ and if $x \in X''$ is a closed point of the special fibre, and $\pi \in \mathfrak m_ x^2$, then the exceptional fibre $E$ of the blowup $X' \to X''$ at $x$ has multiplicity at least $2$ in the special fibre of $X'$ (local computation omitted). Hence $\mathcal{O}_{X''_ k, x} = \mathcal{O}_{X'', x}/\pi $ is regular (Algebra, Lemma 10.106.3) as claimed. In particular $x$ is a Cartier divisor on the unique irreducible component $Z'$ of $X''_ k$ it lies on (Varieties, Lemma 33.43.8). It follows that the strict transform $Z \subset X'$ of $Z'$ maps isomorphically to $Z'$ (use Divisors, Lemmas 31.33.2 and 31.32.7). In other words, if an irreducible component $Z$ of $X_ i$ is not contracted under the map $X_ i \to X_ j$ ($i > j$) then it maps isomorphically to its image.

Now we are ready to prove the lemma. Let $E \subset Y_ m$ be the exceptional curve of the first kind which is contracted by the morphism $Y_ m \to Y_{m - 1}$. If $E$ is contracted by the morphism $Y_ m = X_ n \to X$, then there is a factorization $Y_{m - 1} \to X$ (Resolution of Surfaces, Lemma 54.16.1) and moreover $Y_{m - 1} \to X$ is a sequence of blowups in closed points (Resolution of Surfaces, Lemma 54.17.1). In this case we lower $m$ and we win by induction. Finally, assume that $E$ is not contracted by the morphism $Y_ m \to X$. Then $E \to X_ k$ is surjective as $X_ k$ is irreducible and by the above this means it is an isomorphism. Hence $X_ k$ is isomorphic to a projective line as desired. $\square$