Lemma 42.5.4. Let (A, \mathfrak m, \kappa ) \to (A', \mathfrak m', \kappa ') be a local homomorphism of Noetherian local rings. Assume A \to A' is flat and \dim (A) = \dim (A') = 1. Set m = \text{length}_{A'}(A'/\mathfrak mA'). For a_1, a_2 \in A nonzerodivisors \partial _ A(a_1, a_2)^ m maps to \partial _{A'}(a_1, a_2) via \kappa \to \kappa '.
Proof. If a_1, a_2 are both units, then \partial _ A(a_1, a_2) = 1 and \partial _{A'}(a_1, a_2) = 1 and the result is true. If not, then we can choose a ring extension A \subset B and local factorizations as in Lemma 42.4.4. Denote \mathfrak m_1, \ldots , \mathfrak m_ m be the maximal ideals of B. Let \mathfrak m_1, \ldots , \mathfrak m_ m be the maximal ideals of B with residue fields \kappa _1, \ldots , \kappa _ m. For each j \in \{ 1, \ldots , m\} denote \pi _ j \in B_ j = B_{\mathfrak m_ j} a nonzerodivisor such that we have factorizations a_ i = u_{i, j}\pi _ j^{e_{i, j}} as in the lemma. By definition we have
\partial _ A(a_1, a_2) = \prod \nolimits _ j \text{Norm}_{\kappa _ j/\kappa } ((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}} \bmod \mathfrak m_ j)^{m_ j}
where m_ j = \text{length}_{B_ j}(B_ j/\pi _ j B_ j).
Set B' = A' \otimes _ A B. Since A' is flat over A we see that A' \subset B' is a ring extension with B'/A' annihilated by a power of \mathfrak m'. Let
\mathfrak m'_{j, l},\quad l = 1, \ldots , n_ j
be the maximal ideals of B' lying over \mathfrak m_ j. Denote \kappa '_{j, l} the residue field of \mathfrak m'_{j, l}. Denote B'_{j, l} the localization of B' at \mathfrak m'_{j, l}. As factorizations of a_1 and a_2 in B'_{j, l} we use the image of the factorizations a_ i = u_{i, j} \pi _ j^{e_{i, j}} given to us in B_ j. By definition we have
\partial _{A'}(a_1, a_2) = \prod \nolimits _{j, l} \text{Norm}_{\kappa '_{j, l}/\kappa '} ((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}} \bmod \mathfrak m'_{j, l})^{m'_{j, l}}
where m'_{j, l} = \text{length}_{B'_{j, l}}(B'_{j, l}/\pi _ j B'_{j, l}).
Comparing the formulae above we see that it suffices to show that for each j and for any unit u \in B_ j we have
42.5.4.1
\begin{equation} \label{chow-equation-to-prove} \left(\text{Norm}_{\kappa _ j/\kappa }(u \bmod \mathfrak m_ j)^{m_ j}\right)^ m = \prod \nolimits _ l \text{Norm}_{\kappa '_{j, l}/\kappa '}(u \bmod \mathfrak m'_{j, l})^{m'_{j, l}} \end{equation}
in \kappa '. We are going to use the construction of determinants of endomorphisms of finite length modules in More on Algebra, Section 15.120 to prove this. Set M = B_ j/\pi _ j B_ j. By More on Algebra, Lemma 15.120.2 we have
\text{Norm}_{\kappa _ j/\kappa }(u \bmod \mathfrak m_ j)^{m_ j} = \det \nolimits _\kappa (u : M \to M)
Thus, by More on Algebra, Lemma 15.120.3, the left hand side of (42.5.4.1) is equal to \det _{\kappa '}(u : M \otimes _ A A' \to M \otimes _ A A'). We have an isomorphism
M \otimes _ A A' = (B_ j/\pi _ j B_ j) \otimes _ A A' = \bigoplus \nolimits _ l B'_{j, l}/\pi _ j B'_{j, l}
of A'-modules. Setting M'_ l = B'_{j, l}/\pi _ j B'_{j, l} we see that \text{Norm}_{\kappa '_{j, l}/\kappa '}(u \bmod \mathfrak m'_{j, l})^{m'_{j, l}} = \det _{\kappa '}(u_ j : M'_ l \to M'_ l) by More on Algebra, Lemma 15.120.2 again. Hence (42.5.4.1) holds by multiplicativity of the determinant construction, see More on Algebra, Lemma 15.120.1. \square
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