Lemma 54.15.1. Let $Y$ be a one dimensional integral Noetherian scheme. The following are equivalent

1. there exists an alteration $X \to Y$ with $X$ regular,

2. there exists a resolution of singularities of $Y$,

3. there exists a finite sequence $Y_ n \to Y_{n - 1} \to \ldots \to Y_1 \to Y$ of blowups in closed points with $Y_ n$ regular, and

4. the normalization $Y^\nu \to Y$ is finite.

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

$X \to \ldots \to Y_2 \to Y_1 \to Y$

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

$f_*\mathcal{O}_ X \supset \ldots \supset f_{2, *}\mathcal{O}_{Y_2} \supset f_{1, *}\mathcal{O}_{Y_1} \supset \mathcal{O}_ Y$

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. $\square$

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