Lemma 15.120.2. Let $(R, \mathfrak m, \kappa ) \to (R', \mathfrak m', \kappa ')$ be a local homomorphism of local rings. Assume that $\kappa '/\kappa $ is a finite extension. Let $u \in R'$. Then for any finite length $R'$-module $M'$ we have

\[ \det \nolimits _\kappa (u : M' \to M') = \text{Norm}_{\kappa '/\kappa }(u \bmod \mathfrak m')^ m \]

where $m = \text{length}_{R'}(M')$.

**Proof.**
Observe that the statement makes sense as $\text{length}_ R(M') = \text{length}_{R'}(M') [\kappa ' : \kappa ]$. If $M' = \kappa '$, then the equality holds by definition of the norm as the determinant of the linear operator given by multiplication by $u$. In general one reduces to this case by choosing a suitable filtration and using the multiplicativity of Lemma 15.120.1. Some details omitted.
$\square$

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