The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 15.97.1). 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

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

  2. 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 15.104.9) 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.


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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BJ1. Beware of the difference between the letter 'O' and the digit '0'.