## 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 15.99.1). 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 10.69)

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

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 10.69.3. 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

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

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 10.135.4) 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 10.138.3). 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 10.105.3 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 10.67.6). It follows that

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

by More on Algebra, Lemma 15.30.1. The exact sequence of Algebra, Lemma 10.130.9 for the surjection $A[y_1, \ldots , y_ c] \to A'$ produces an exact sequence

\[ (A')^{\oplus c} \longrightarrow \Omega _{A/R} \otimes _ A A' \oplus \bigoplus \nolimits _{i = 1, \ldots , c} A' \text{d}y_ i \longrightarrow \Omega _{A'/R} \to 0 \]

where the $i$the basis element in the first module is mapped to $- \text{d}g_ i + \pi \text{d}y_ i$ in the second. To finish the proof it therefore suffices to show that $\text{d}g_1, \ldots , \text{d}g_ c$ forms part of a basis for $\Omega _{A/R, \mathfrak p}$. Since $\Omega _{A/R, \mathfrak p}$ is a finite free $A_\mathfrak p$-module (part of the definition of smoothness) 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 10.130.12). Since $\kappa \subset \kappa (\mathfrak p) \subset \Lambda /\pi \Lambda $ we see that $\kappa (\mathfrak p)$ is separable over $\kappa $ (Algebra, Definition 10.41.1). The desired linear independence now follows from Algebra, Lemma 10.138.4.
$\square$

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 10.130.9. 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 10.124.1. Then $\Omega _{A/R, \mathfrak q}$ is free over $A_\mathfrak q$ of rank $d$ (Algebra, Lemma 10.138.3). 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 10.124.6. Then it follows that $R \to A$ is smooth at $\mathfrak p$ by Algebra, Lemmas 10.135.16 and 10.138.3.
$\square$

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 15.106.9) 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 10.138.3). 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 $\pi I'$. Hence if we look at the induced map of complexes we get

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

Let $c = \dim ((B/\pi B)_{\mathfrak p_ B})$. Observe that $M \subset M'$ are free $\Lambda $-modules of rank $r$. The quotient $M'/M$ has length at most $c$ by Lemma 16.4.3. Let $N \subset M$ and $N' \subset M'$ be the images of the horizontal maps. Then $N \subset N'$ are free $\Lambda $-modules of rank $r - d$. Since $I$ maps into $\pi I'$ we see that $N \subset \pi N'$. Hence $N'/N$ has length at least $r - d$. We conclude by a simple lemma with modules over discrete valuation rings that $e$ decreases by at least $r - d - c$ (we will see below this quantity is $\geq 0$).

Since $B$ is smooth over $R$ of relative dimension $r$ at $\mathfrak p_ B$ we see that $r = c + \text{trdeg}_\kappa (\kappa (\mathfrak p_ B))$ by Algebra, Lemma 10.115.3. Let $J = \mathop{\mathrm{Ker}}(A \to A_\mathfrak q)$ so that $A/J$ is a domain with $A_\mathfrak q = (A/J)_\mathfrak q$. It follows that $A_ g = (A/J)_ g$ for some $g \in A$, $g \not\in \mathfrak q$ and hence $\dim _\mathfrak q((A/J)/R)$ is $d$ as this is true for $A$. By the same lemma as before applied twice, the fraction field of $A/J$ has transcendence degree $d$ over the field $R[1/\pi ]$. Applying the dimension formula (Algebra, Lemma 10.112.1) to $R \to A/J$ we find

\[ 1 \leq \dim ((A/J)_\mathfrak p) \leq 1 + d - \text{trdeg}_\kappa (\kappa (\mathfrak p)) = 1 + d - r + c \]

First inequality as $(A/J)_\mathfrak p$ has at least two primes. Equality as $\kappa (\mathfrak p) = \kappa (\mathfrak p_ B)$. Thus we see that $r - d - c \geq 0$ and zero if and only if $r = d + c$.

To finish the proof we have to show that $N'$ is strictly bigger than $\pi ^{-1}N$; 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 10.130.9). Since we may assume that $R \to A$ 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 10.138.4; 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 $g_1, \ldots , g_ r \in I$ 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 10.126.4 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 10.138.9. 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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