Let $X$ be a locally Noetherian scheme. In this section we say that a formal arc in $X$ is a morphism $a : T \to X$ where $T$ is the spectrum of a complete discrete valuation ring $R$ whose residue field $\kappa $ is identified with the residue field of the image $p$ of the closed point of $\mathop{\mathrm{Spec}}(R)$. Let us say that the formal arc $a$ is centered at $p$ in this case. We say the formal arc $T \to X$ is nonsingular if the induced map $\mathfrak m_ p/\mathfrak m_ p^2 \to \mathfrak m_ R/\mathfrak m_ R^2$ is surjective.
Let $a : T \to X$, $T = \mathop{\mathrm{Spec}}(R)$ be a nonsingular formal arc centered at a closed point $p$ of $X$. Assume $X$ is locally Noetherian. Let $b : X_1 \to X$ be the blowing up of $X$ at $x$. Since $a$ is nonsingular, we see that there is an element $f \in \mathfrak m_ p$ which maps to a uniformizer in $R$. In particular, we find that the generic point of $T$ maps to a point of $X$ not equal to $p$. In other words, with $K$ the fraction field of $R$, the restriction of $a$ defines a morphism $\mathop{\mathrm{Spec}}(K) \to X \setminus \{ p\} $. Since the morphism $b$ is proper and an isomorphism over $X \setminus \{ x\} $ we can apply the valuative criterion of properness to obtain a unique morphism $a_1$ making the following diagram commute
Let $p_1 \in X_1$ be the image of the closed point of $T$. Observe that $p_1$ is a closed point as it is a $\kappa = \kappa (p)$-rational point on the fibre of $X_1 \to X$ over $x$. Since we have a factorization
we see that $a_1$ is a nonsingular formal arc as well.
We can repeat the process and obtain a sequence of blowing ups
This kind of sequence of blowups can be characterized as follows.
Lemma 54.10.1. Let $X$ be a locally Noetherian scheme. Let be a sequence of blowups such that
$p_ i$ is closed, maps to $p_{i - 1}$, and $\kappa (p_ i) = \kappa (p_{i - 1})$,
there exists an $x_1 \in \mathfrak m_ p$ whose image in $\mathfrak m_{p_ i}$, $i > 0$ defines the exceptional divisor $E_ i \subset X_ i$.
\[ (X, p) = (X_0, p_0) \leftarrow (X_1, p_1) \leftarrow (X_2, p_2) \leftarrow (X_3, p_3) \leftarrow \ldots \]
Then the sequence is obtained from a nonsingular arc $a : T \to X$ as above.
Proof. Let us write $\mathcal{O}_ n = \mathcal{O}_{X_ n, p_ n}$ and $\mathcal{O} = \mathcal{O}_{X, p}$. Denote $\mathfrak m \subset \mathcal{O}$ and $\mathfrak m_ n \subset \mathcal{O}_ n$ the maximal ideals.
We claim that $x_1^ t \not\in \mathfrak m_ n^{t + 1}$. Namely, if this were the case, then in the local ring $\mathcal{O}_{n + 1}$ the element $x_1^ t$ would be in the ideal of $(t + 1)E_{n + 1}$. This contradicts the assumption that $x_1$ defines $E_{n + 1}$.
For every $n$ choose generators $y_{n, 1}, \ldots , y_{n, t_ n}$ for $\mathfrak m_ n$. As $\mathfrak m_ n \mathcal{O}_{n + 1} = x_1\mathcal{O}_{n + 1}$ by assumption (2), we can write $y_{n, i} = a_{n, i} x_1$ for some $a_{n, i} \in \mathcal{O}_{n + 1}$. Since the map $\mathcal{O}_ n \to \mathcal{O}_{n + 1}$ defines an isomorphism on residue fields by (1) we can choose $c_{n, i} \in \mathcal{O}_ n$ having the same residue class as $a_{n, i}$. Then we see that
and the elements $z_{n, i}$ map to elements of $\mathfrak m_{n + 1}^2$ in $\mathcal{O}_{n + 1}$.
Let us consider
We claim that $\mathcal{O}/J_ n$ has length $n + 1$ and that $\mathcal{O}/(x_1) + J_ n$ equals the residue field. For $n = 0$ this is immediate. Assume the statement holds for $n$. Let $f \in J_ n$. Then in $\mathcal{O}_ n$ we have
for some $a \in \mathcal{O}_ n$ and some $A_ i$ homogeneous of degree $i$ with coefficients in $\mathcal{O}_ n$. Since $\mathcal{O} \to \mathcal{O}_ n$ identifies residue fields, we may choose $a \in \mathcal{O}$ (argue as in the construction of $z_{n, i}$ above). Taking the image in $\mathcal{O}_{n + 1}$ we see that $f$ and $a x_1^{n + 1}$ have the same image modulo $\mathfrak m_{n + 1}^{n + 2}$. Since $x_ n^{n + 1} \not\in \mathfrak m_{n + 1}^{n + 2}$ it follows that $J_ n/J_{n + 1}$ has length $1$ and the claim is true.
Consider $R = \mathop{\mathrm{lim}}\nolimits \mathcal{O}/J_ n$. This is a quotient of the $\mathfrak m$-adic completion of $\mathcal{O}$ hence it is a complete Noetherian local ring. On the other hand, it is not finite length and $x_1$ generates the maximal ideal. Thus $R$ is a complete discrete valuation ring. The map $\mathcal{O} \to R$ lifts to a local homomorphism $\mathcal{O}_ n \to R$ for every $n$. There are two ways to show this: (1) for every $n$ one can use a similar procedure to construct $\mathcal{O}_ n \to R_ n$ and then one can show that $\mathcal{O} \to \mathcal{O}_ n \to R_ n$ factors through an isomorphism $R \to R_ n$, or (2) one can use Divisors, Lemma 31.32.6 to show that $\mathcal{O}_ n$ is a localization of a repeated affine blowup algebra to explicitly construct a map $\mathcal{O}_ n \to R$. Having said this it is clear that our sequence of blowups comes from the nonsingular arc $a : T = \mathop{\mathrm{Spec}}(R) \to X$. $\square$
The following lemma is a kind of Néron desingularization lemma.
Lemma 54.10.2. Let $(A, \mathfrak m, \kappa )$ be a Noetherian local domain of dimension $2$. Let $A \to R$ be a surjection onto a complete discrete valuation ring. This defines a nonsingular arc $a : T = \mathop{\mathrm{Spec}}(R) \to \mathop{\mathrm{Spec}}(A)$. Let be the sequence of blowing ups constructed from $a$. If $A_\mathfrak p$ is a regular local ring where $\mathfrak p = \mathop{\mathrm{Ker}}(A \to R)$, then for some $i$ the scheme $X_ i$ is regular at $x_ i$.
\[ \mathop{\mathrm{Spec}}(A) = X_0 \leftarrow X_1 \leftarrow X_2 \leftarrow X_3 \leftarrow \ldots \]
Proof. Let $x_1 \in \mathfrak m$ map to a uniformizer of $R$. Observe that $\kappa (\mathfrak p) = K$ is the fraction field of $R$. Write $\mathfrak p = (x_2, \ldots , x_ r)$ with $r$ minimal. If $r = 2$, then $\mathfrak m = (x_1, x_2)$ and $A$ is regular and the lemma is true. Assume $r > 2$. After renumbering if necessary, we may assume that $x_2$ maps to a uniformizer of $A_\mathfrak p$. Then $\mathfrak p/\mathfrak p^2 + (x_2)$ is annihilated by a power of $x_1$. For $i > 2$ we can find $n_ i \geq 0$ and $a_ i \in A$ such that
for some $a_{jk} \in A$. If $n_ i = 0$ for some $i$, then we can remove $x_ i$ from the list of generators of $\mathfrak p$ and we win by induction on $r$. If for some $i$ the element $a_ i$ is a unit, then we can remove $x_2$ from the list of generators of $\mathfrak p$ and we win in the same manner. Thus either $a_ i \in \mathfrak p$ or $a_ i = u_ i x_1^{m_1} \bmod \mathfrak p$ for some $m_1 > 0$ and unit $u_ i \in A$. Thus we have either
We will prove that after blowing up the integers $n_ i$, $m_ i$ decrease which will finish the proof.
Let us see what happens with these equations on the affine blowup algebra $A' = A[\mathfrak m/x_1]$. As $\mathfrak m = (x_1, \ldots , x_ r)$ we see that $A'$ is generated over $R$ by $y_ i = x_ i/x_1$ for $i \geq 2$. Clearly $A \to R$ extends to $A' \to R$ with kernel $(y_2, \ldots , y_ r)$. Then we see that either
and the proof is complete. $\square$
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