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
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