## 16.4 Intermezzo: Néron desingularization

We interrupt the attack on the general case of Popescu's theorem to an easier but already very interesting case, namely, when $R \to \Lambda $ is a homomorphism of discrete valuation rings. This is discussed in [Section 4, Artin-Algebraic-Approximation].

Situation 16.4.1. Here $R \subset \Lambda $ is an extension of discrete valuation rings with ramification index $1$ (More on Algebra, Definition 15.111.1). We assume given a factorization

\[ R \to A \xrightarrow {\varphi } \Lambda \]

with $R \to A$ flat and of finite type. Let $\mathfrak q = \mathop{\mathrm{Ker}}(\varphi )$ and $\mathfrak p = \varphi ^{-1}(\mathfrak m_\Lambda )$.

In Situation 16.4.1 let $\pi \in R$ be a uniformizer. Recall that flatness of $A$ over $R$ signifies that $\pi $ is a nonzerodivisor on $A$ (More on Algebra, Lemma 15.22.10). By our assumption on $R \subset \Lambda $ we see that $\pi $ maps to a uniformizer of $\Lambda $. Since $\pi \in \mathfrak p$ we can consider Néron's affine blowup algebra (see Algebra, Section 10.70)

\[ \varphi ' : A' = A[\textstyle {\frac{\mathfrak p}{\pi }}] \longrightarrow \Lambda \]

which comes endowed with an induced map to $\Lambda $ sending $a/\pi ^ n$, $a \in \mathfrak p^ n$ to $\pi ^{-n}\varphi (a)$ in $\Lambda $. We will denote $\mathfrak q' \subset \mathfrak p' \subset A'$ the corresponding prime ideals of $A'$. Observe that the isomorphism class of $A'$ does not depend on our choice of uniformizer. Repeating the construction we obtain a sequence

\[ A \to A' \to A'' \to \ldots \to \Lambda \]

Lemma 16.4.2. In Situation 16.4.1 Néron's blowup is functorial in the following sense

if $a \in A$, $a \not\in \mathfrak p$, then Néron's blowup of $A_ a$ is $A'_ a$, and

if $B \to A$ is a surjection of flat finite type $R$-algebras with kernel $I$, then $A'$ is the quotient of $B'/IB'$ by its $\pi $-power torsion.

**Proof.**
Both (1) and (2) are special cases of Algebra, Lemma 10.70.3. In fact, whenever we have $A_1 \to A_2 \to \Lambda $ such that $\mathfrak p_1 A_2 = \mathfrak p_2$, we have that $A_2'$ is the quotient of $A_1' \otimes _{A_1} A_2$ by its $\pi $-power torsion.
$\square$

Lemma 16.4.3. In Situation 16.4.1 assume that $R \to A$ is smooth at $\mathfrak p$ and that $R/\pi R \subset \Lambda /\pi \Lambda $ is a separable field extension. Then $R \to A'$ is smooth at $\mathfrak p'$ and there is a short exact sequence

\[ 0 \to \Omega _{A/R} \otimes _ A A'_{\mathfrak p'} \to \Omega _{A'/R, \mathfrak p'} \to (A'/\pi A')_{\mathfrak p'}^{\oplus c} \to 0 \]

where $c = \dim ((A/\pi A)_\mathfrak p)$.

**Proof.**
By Lemma 16.4.2 we may replace $A$ by a localization at an element not in $\mathfrak p$; we will use this without further mention. Write $\kappa = R/\pi R$. Since smoothness is stable under base change (Algebra, Lemma 10.137.4) we see that $A/\pi A$ is smooth over $\kappa $ at $\mathfrak p$. Hence $(A/\pi A)_\mathfrak p$ is a regular local ring (Algebra, Lemma 10.140.3). Choose $g_1, \ldots , g_ c \in \mathfrak p$ which map to a regular system of parameters in $(A/\pi A)_\mathfrak p$. Then we see that $\mathfrak p = (\pi , g_1, \ldots , g_ c)$ after possibly replacing $A$ by a localization. Note that $\pi , g_1, \ldots , g_ c$ is a regular sequence in $A_\mathfrak p$ (first $\pi $ is a nonzerodivisor and then Algebra, Lemma 10.106.3 for the rest of the sequence). After replacing $A$ by a localization we may assume that $\pi , g_1, \ldots , g_ c$ is a regular sequence in $A$ (Algebra, Lemma 10.68.6). It follows that

\[ A' = A[y_1, \ldots , y_ c]/(\pi y_1 - g_1, \ldots , \pi y_ c - g_ c) = A[y_1, \ldots , y_ c]/I \]

by More on Algebra, Lemma 15.31.2. In the following we will use the definition of smoothness using the naive cotangent complex (Algebra, Definition 10.137.1) and the criterion of Algebra, Lemma 10.137.12 without further mention. The exact sequence of Algebra, Lemma 10.134.4 for $R \to A[y_1, \ldots , y_ c] \to A'$ looks like this

\[ 0 \to H_1(\mathop{N\! L}\nolimits _{A'/R}) \to I/I^2 \to \Omega _{A/R} \otimes _ A A' \oplus \bigoplus \nolimits _{i = 1, \ldots , c} A' \text{d}y_ i \to \Omega _{A'/R} \to 0 \]

where the class of $\pi y_ i - g_ i$ in $I/I^2$ is mapped to $- \text{d}g_ i + \pi \text{d}y_ i$ in the next term. Here we have used Algebra, Lemma 10.134.6 to compute $\mathop{N\! L}\nolimits _{A'/A[y_1, \ldots , y_ c]}$ and we have used that $R \to A[y_1, \ldots , y_ c]$ is smooth, so $H_1(\mathop{N\! L}\nolimits _{A[y_1, \ldots , y_ c]/R}) = 0$ and $\Omega _{A[y_1, \ldots , y_ c]/R}$ is a finite projective (a fortiori flat) $A[y_1, \ldots , y_ c]$-module which is in fact the direct sum of $\Omega _{A/R} \otimes _ A A[y_1, \ldots , y_ c]$ and a free module with basis $\text{d}y_ i$. To finish the proof it suffices to show that $\text{d}g_1, \ldots , \text{d}g_ c$ forms part of a basis for the finite free module $\Omega _{A/R, \mathfrak p}$. Namely, this will show $(I/I^2)_\mathfrak p$ is free on $\pi y_ i - g_ i$, the localization at $\mathfrak p$ of the middle map in the sequence is injective, so $H_1(\mathop{N\! L}\nolimits _{A'/R})_\mathfrak p = 0$, and that the cokernel $\Omega _{A'/R, \mathfrak p}$ is finite free. To do this it suffices to show that the images of $\text{d}g_ i$ are $\kappa (\mathfrak p)$-linearly independent in $\Omega _{A/R, \mathfrak p}/\pi = \Omega _{(A/\pi A)/\kappa , \mathfrak p}$ (equality by Algebra, Lemma 10.131.12). Since $\kappa \subset \kappa (\mathfrak p) \subset \Lambda /\pi \Lambda $ we see that $\kappa (\mathfrak p)$ is separable over $\kappa $ (Algebra, Definition 10.42.1). The desired linear independence now follows from Algebra, Lemma 10.140.4.
$\square$

Lemma 16.4.4. In Situation 16.4.1 assume that $R \to A$ is smooth at $\mathfrak q$ and that we have a surjection of $R$-algebras $B \to A$ with kernel $I$. Assume $R \to B$ smooth at $\mathfrak p_ B = (B \to A)^{-1}\mathfrak p$. If the cokernel of

\[ I/I^2 \otimes _ A \Lambda \to \Omega _{B/R} \otimes _ B \Lambda \]

is a free $\Lambda $-module, then $R \to A$ is smooth at $\mathfrak p$.

**Proof.**
The cokernel of the map $I/I^2 \to \Omega _{B/R} \otimes _ B A$ is $\Omega _{A/R}$, see Algebra, Lemma 10.131.9. Let $d = \dim _\mathfrak q(A/R)$ be the relative dimension of $R \to A$ at $\mathfrak q$, i.e., the dimension of $\mathop{\mathrm{Spec}}(A[1/\pi ])$ at $\mathfrak q$. See Algebra, Definition 10.125.1. Then $\Omega _{A/R, \mathfrak q}$ is free over $A_\mathfrak q$ of rank $d$ (Algebra, Lemma 10.140.3). Thus if the hypothesis of the lemma holds, then $\Omega _{A/R} \otimes _ A \Lambda $ is free of rank $d$. It follows that $\Omega _{A/R} \otimes _ A \kappa (\mathfrak p)$ has dimension $d$ (as it is true upon tensoring with $\Lambda /\pi \Lambda $). Since $R \to A$ is flat and since $\mathfrak p$ is a specialization of $\mathfrak q$, we see that $\dim _\mathfrak p(A/R) \geq d$ by Algebra, Lemma 10.125.6. Then it follows that $R \to A$ is smooth at $\mathfrak p$ by Algebra, Lemmas 10.137.17 and 10.140.3.
$\square$

Lemma 16.4.5. In Situation 16.4.1 assume that $R \to A$ is smooth at $\mathfrak q$ and that $R/\pi R \subset \Lambda /\pi \Lambda $ is a separable extension of fields. Then after a finite number of affine Néron blowups the algebra $A$ becomes smooth over $R$ at $\mathfrak p$.

**Proof.**
We choose an $R$-algebra $B$ and a surjection $B \to A$. Set $\mathfrak p_ B = (B \to A)^{-1}(\mathfrak p)$ and denote $r$ the relative dimension of $R \to B$ at $\mathfrak p_ B$. We choose $B$ such that $R \to B$ is smooth at $\mathfrak p_ B$. For example we can take $B$ to be a polynomial algebra in $r$ variables over $R$. Consider the complex

\[ I/I^2 \otimes _ A \Lambda \longrightarrow \Omega _{B/R} \otimes _ B \Lambda \]

of Lemma 16.4.4. By the structure of finite modules over $\Lambda $ (More on Algebra, Lemma 15.124.9) we see that the cokernel looks like

\[ \Lambda ^{\oplus d} \oplus \bigoplus \nolimits _{i = 1, \ldots , n} \Lambda /\pi ^{e_ i} \Lambda \]

for some $d \geq 0$, $n \geq 0$, and $e_ i \geq 1$. Observe that $d$ is the relative dimension of $A/R$ at $\mathfrak q$ (Algebra, Lemma 10.140.3). If the defect $e = \sum _{i = 1, \ldots , n} e_ i$ is zero, then we are done by Lemma 16.4.4.

Next, we consider what happens when we perform the Néron blowup. Recall that $A'$ is the quotient of $B'/IB'$ by its $\pi $-power torsion (Lemma 16.4.2) and that $R \to B'$ is smooth at $\mathfrak p_{B'}$ (Lemma 16.4.3). Thus after blowup we have exactly the same setup. Picture

\[ \xymatrix{ 0 \ar[r] & I' \ar[r] & B' \ar[r] & A' \ar[r] & 0 \\ 0 \ar[r] & I \ar[u] \ar[r] & B \ar[u] \ar[r] & A \ar[r] \ar[u] & 0 } \]

Since $I \subset \mathfrak p_ B$, we see that $I \to I'$ factors through $\pi I'$. Looking at the induced map of complexes we get

\[ \xymatrix{ I'/(I')^2 \otimes _{A'} \Lambda \ar[r] & \Omega _{B'/R} \otimes _{B'} \Lambda \ar@{=}[r] & M' \\ I/I^2 \otimes _ A \Lambda \ar[r] \ar[u] & \Omega _{B/R} \otimes _ B \Lambda \ar[u] \ar@{=}[r] & M } \]

Then $M \subset M'$ are finite free $\Lambda $-modules with quotient $M'/M$ annihilated by $\pi $, see Lemma 16.4.3. Let $N \subset M$ and $N' \subset M'$ be the images of the horizontal maps and denote $Q = M/N$ and $Q' = M'/N'$. We obtain a commutative diagram

\[ \xymatrix{ 0 \ar[r] & N' \ar[r] & M' \ar[r] & Q' \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[u] & M \ar[r] \ar[u] & Q \ar[r] \ar[u] & 0 } \]

Then $N \subset N'$ are free $\Lambda $-modules of rank $r - d$. Since $I$ maps into $\pi I'$ we see that $N \subset \pi N'$.

Let $K = \Lambda _\pi $ be the fraction field of $\Lambda $. We have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & N' \ar[r] & N'_ K \cap M' \ar[r] & Q'_{tor} \ar[r] & 0 \\ 0 \ar[r] & N \ar[r] \ar[u] & N_ K \cap M \ar[r] \ar[u] & Q_{tor} \ar[r] \ar[u] & 0 } \]

whose rows are short exact sequences. This shows that the change in defect is given by

\[ e - e' = \text{length}(Q_{tor}) - \text{length}(Q'_{tor}) = \text{length}(N'/N) - \text{length}(N'_ K \cap M' / N_ K \cap M) \]

Since $M'/M$ is annihilated by $\pi $, so is $N'_ K \cap M' / N_ K \cap M$, and its length is at most $\dim _ K(N_ K)$. Since $N \subset \pi N'$ we get $\text{length}(N'/N) \ge \dim _ K(N_ K)$, with equality if and only if $N = \pi N'$.

To finish the proof we have to show that $N$ is strictly smaller than $\pi N'$ when $A$ is not smooth at $\mathfrak p$; this is the key computation one has to do in Néron's argument. To do this, we consider the exact sequence

\[ I/I^2 \otimes _ B \kappa (\mathfrak p_ B) \to \Omega _{B/R} \otimes _ B \kappa (\mathfrak p_ B) \to \Omega _{A/R} \otimes _ A \kappa (\mathfrak p) \to 0 \]

(follows from Algebra, Lemma 10.131.9). Since $R \to A$ is not smooth at $\mathfrak p$ we see that the dimension $s$ of $\Omega _{A/R} \otimes _ A \kappa (\mathfrak p)$ is bigger than $d$. On the other hand the first arrow factors through the injective map

\[ \mathfrak p B_\mathfrak p/\mathfrak p^2 B_\mathfrak p \to \Omega _{B/R} \otimes _ B \kappa (\mathfrak p_ B) \]

of Algebra, Lemma 10.140.4; note that $\kappa (\mathfrak p)$ is separable over $k$ by our assumption on $R/\pi R \subset \Lambda /\pi \Lambda $. Hence we conclude that we can find generators $g_1, \ldots , g_ t \in I$ such that $g_ j \in \mathfrak p^2$ for $j > r - s$. Then the images of $g_ j$ in $A'$ are in $\pi ^2 I'$ for $j > r - s$. Since $r - s < r - d$ we find that at least one of the minimal generators of $N$ becomes divisible by $\pi ^2$ in $N'$. Thus we see that $e$ decreases by at least $1$ and we win.
$\square$

If $R \to \Lambda $ is an extension of discrete valuation rings, then $R \to \Lambda $ is regular if and only if (a) the ramification index is $1$, (b) the extension of fraction fields is separable, and (c) $R/\mathfrak m_ R \subset \Lambda /\mathfrak m_\Lambda $ is separable. Thus the following result is a special case of general Néron desingularization in Theorem 16.12.1.

slogan
Lemma 16.4.6. Let $R \subset \Lambda $ be an extension of discrete valuation rings which has ramification index $1$ and induces a separable extension of residue fields and of fraction fields. Then $\Lambda $ is a filtered colimit of smooth $R$-algebras.

**Proof.**
By Algebra, Lemma 10.127.4 it suffices to show that any $R \to A \to \Lambda $ as in Situation 16.4.1 can be factored as $A \to B \to \Lambda $ with $B$ a smooth $R$-algebra. After replacing $A$ by its image in $\Lambda $ we may assume that $A$ is a domain whose fraction field $K$ is a subfield of the fraction field of $\Lambda $. In particular, $A$ is separable over the fraction field of $R$ by our assumptions. Then $R \to A$ is smooth at $\mathfrak q = (0)$ by Algebra, Lemma 10.140.9. After a finite number of Néron blowups, we may assume $R \to A$ is smooth at $\mathfrak p$, see Lemma 16.4.5. Then, after replacing $A$ by a localization at an element $a \in A$, $a \not\in \mathfrak p$ it becomes smooth over $R$ and the lemma is proved.
$\square$

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