The Stacks project

Proof. The implications (3) $\Rightarrow $ (2) $\Rightarrow $ (1) are immediate. The implication (1) $\Rightarrow $ (4) follows from Lemma 54.13.1. Observe that a normal one dimensional scheme is regular hence the implication (4) $\Rightarrow $ (2) is clear as well. Thus it remains to show that the equivalent conditions (1), (2), and (4) imply (3).

Let $f : X \to Y$ be a resolution of singularities. Since the dimension of $Y$ is one we see that $f$ is finite by Varieties, Lemma 33.17.2. We will construct factorizations

where $Y_ i \to Y_{i - 1}$ is a blowing up of a closed point and not an isomorphism as long as $Y_{i - 1}$ is not regular. Each of these morphisms will be finite (by the same reason as above) and we will get a corresponding system

where $f_ i : Y_ i \to Y$ is the structure morphism. Since $Y$ is Noetherian, this increasing sequence of coherent submodules must stabilize (Cohomology of Schemes, Lemma 30.10.1) which proves that for some $n$ the scheme $Y_ n$ is regular as desired. To construct $Y_ i$ given $Y_{i - 1}$ we pick a singular closed point $y_{i - 1} \in Y_{i - 1}$ and we let $Y_ i \to Y_{i - 1}$ be the corresponding blowup. Since $X$ is regular of dimension $1$ (and hence the local rings at closed points are discrete valuation rings and in particular PIDs), the ideal sheaf $\mathfrak m_{y_{i - 1}} \cdot \mathcal{O}_ X$ is invertible. By the universal property of blowing up (Divisors, Lemma 31.32.5) this gives us a factorization $X \to Y_ i$. Finally, $Y_ i \to Y_{i - 1}$ is not an isomorphism as $\mathfrak m_{y_{i - 1}}$ is not an invertible ideal. $\square$

Proof. Let $Y_ n \to Y_{n - 1} \to \ldots \to Y_1 \to Y$ be the sequence of blowups given to us by Lemma 54.15.1. Let $X_ n \to X_{n - 1} \to \ldots \to X_1 \to X$ be the corresponding sequence of blowups of $X$. This works because the strict transform is the blowup by Divisors, Lemma 31.33.2. $\square$

Proof. Since $\mathcal{O}_{X, p} \to \mathcal{O}_{Y, p}$ is surjective and $\mathcal{O}_{Y, p}$ is a discrete valuation ring, we can pick an element $x_1 \in \mathfrak m_ p$ mapping to a uniformizer in $\mathcal{O}_{Y, p}$. Choose an affine open $U = \mathop{\mathrm{Spec}}(A)$ containing $p$ such that $x_1 \in A$. Let $\mathfrak m \subset A$ be the maximal ideal corresponding to $p$. Let $I, J \subset A$ be the ideals defining $Y, Z$ in $\mathop{\mathrm{Spec}}(A)$. After shrinking $U$ we may assume that $\mathfrak m = I + (x_1)$, in other words, that $V(x_1) \cap U \cap Y = \{ p\} $ scheme theoretically. We conclude that $p$ is an effective Cartier divisor on $Y$ and since $Y'$ is the blowing up of $Y$ in $p$ (Divisors, Lemma 31.33.2) we see that $Y' \to Y$ is an isomorphism by Divisors, Lemma 31.32.7. The relationship $\mathfrak m = I + (x_1)$ implies that $\mathfrak m^ n \subset I + (x_1^ n)$ hence we can define a map

by sending $y/x_1^ n \in A[\frac{\mathfrak m}{x_1}]$ to the class of $a$ in $A/I$ where $a$ is chosen such that $y \equiv ax_1^ n \bmod I$. Then $\psi $ corresponds to the morphism of $Y \cap U$ into $X'$ over $U$ given by $Y' \cong Y$. Since the image of $x_1$ in $A[\frac{\mathfrak m}{x_1}]$ cuts out the exceptional divisor we conclude that $m_ q(Y', E) = 1$. Finally, since $J \subset \mathfrak m$ implies that the ideal $J' \subset A[\frac{\mathfrak m}{x_1}]$ certainly contains the elements $f/x_1$ for $f \in J$. Thus if we choose $f \in J$ whose image $\overline{f}$ in $A/I$ has minimal valuation equal to $m_ p(Y \cap Z)$, then we see that $\psi (f/x_1) = \overline{f}/x_1$ in $A/I$ has valuation one less proving the last part of the lemma. $\square$

Proof. It follows from Lemma 54.15.2 that we may assume $Y_ i$ is a regular curve for $i = 1, \ldots , n$. For every $i \not= j$ and $p \in Y_ i \cap Y_ j$ we have the invariant $m_ p(Y_ i \cap Y_ j)$ (54.15.2.1). If the maximum of these numbers is $> 1$, then we can decrease it (Lemma 54.15.3) by blowing up in all the points $p$ where the maximum is attained. If the maximum is $1$ then we can separate the curves using the same lemma by blowing up in all these points $p$. $\square$

Proof. Let $D \subset Z$ be the largest effective Cartier divisor contained in $Z$. Then $\mathcal{I}_ Z \subset \mathcal{I}_ D$ and the quotient is supported in closed points by Divisors, Lemma 31.15.8. Thus we can write $\mathcal{I}_ Z = \mathcal{I}_{Z'} \mathcal{I}_ D$ where $Z' \subset X$ is a closed subscheme which set theoretically consists of finitely many closed points. Applying Lemma 54.4.1 we find a sequence of blowups as in the statement of our lemma such that $\mathcal{I}_{Z'}\mathcal{O}_{X_ n}$ is invertible. This proves the lemma. $\square$

Proof. Let $X' \to X$ be a blowup in a closed point $p$. Then the inverse image $Z' \subset X'$ of $Z$ is supported on the strict transform of $Z$ and the exceptional divisor. The exceptional divisor is a regular curve (Lemma 54.3.1) and the strict transform $Y'$ of each irreducible component $Y$ is either equal to $Y$ or the blowup of $Y$ at $p$. Thus in this process we do not produce additional singular components of dimension $1$. Thus it follows from Lemmas 54.15.5 and 54.15.4 that we may assume $Z$ is an effective Cartier divisor and that all irreducible components $Y$ of $Z$ are regular. (Of course we cannot assume the irreducible components are pairwise disjoint because in each blowup of a point of $Z$ we add a new irreducible component to $Z$, namely the exceptional divisor.)

Assume $Z$ is an effective Cartier divisor whose irreducible components $Y_ i$ are regular. For every $i \not= j$ and $p \in Y_ i \cap Y_ j$ we have the invariant $m_ p(Y_ i \cap Y_ j)$ (54.15.2.1). If the maximum of these numbers is $> 1$, then we can decrease it (Lemma 54.15.3) by blowing up in all the points $p$ where the maximum is attained (note that the “new” invariants $m_{q_ i}(Y'_ i \cap E)$ are always $1$). If the maximum is $1$ then, if $p \in Y_1 \cap \ldots \cap Y_ r$ for some $r > 2$ and not any of the others (for example), then after blowing up $p$ we see that $Y'_1, \ldots , Y'_ r$ do not meet in points above $p$ and $m_{q_ i}(Y'_ i, E) = 1$ where $Y'_ i \cap E = \{ q_ i\} $. Thus continuing to blowup points where more than $3$ of the components of $Z$ meet, we reach the situation where for every closed point $p \in X$ there is either (a) no curves $Y_ i$ passing through $p$, (b) exactly one curve $Y_ i$ passing through $p$ and $\mathcal{O}_{Y_ i, p}$ is regular, or (c) exactly two curves $Y_ i$, $Y_ j$ passing through $p$, the local rings $\mathcal{O}_{Y_ i, p}$, $\mathcal{O}_{Y_ j, p}$ are regular and $m_ p(Y_ i \cap Y_ j) = 1$. This means that $\sum Y_ i$ is a strict normal crossings divisor on the regular surface $X$, see Étale Morphisms, Lemma 41.21.2. $\square$