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

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

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

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

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

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

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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