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 15.121.9) 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. $\square$

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