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. $\square$

Comment #3789 by Dario Weißmann on

Typos: $\pi,g_2,\dots,g_c$ is a regular sequence -> $\pi, g_1,\dots,g_c$

in the exact sequence: $\Omega_{A/R}\otimes_A A' \oplus A'dy_i$ -> $\Omega_{A/R}\otimes_A A' \oplus \bigoplus_{i=1}^c A'dy_i$

where the $i$the -> where the $i$th

$\Omega_{A/R,\mathfrak{p}}/\pi = ...$ -> $\Omega_{A/R,\mathfrak{p}}/\pi\Omega_{A/R,\mathfrak{p}}=...$

I didn't find the exactness of the sequence that obvious. I got it by using the Jacobi-Zariski sequence for $R\to A \to A'$ and the fact that $H^1(NL_{A'/A})=0$. Is there a faster way?

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