Lemma 42.68.5. Let R be a local ring with maximal ideal \mathfrak m and residue field \kappa . Let M be a finite length R-module. The determinant \det _\kappa (M) defined above is a \kappa -vector space of dimension 1. It is generated by the symbol [f_1, \ldots , f_ l] for any admissible sequence such that \langle f_1, \ldots f_ l \rangle = M.
Proof. We know \det _\kappa (M) has dimension at most 1, and in fact that it is generated by [f_1, \ldots , f_ l], by Lemma 42.68.3 and its proof. We will show by induction on l = \text{length}(M) that it is nonzero. For l = 1 it follows from Lemma 42.68.4. Choose a nonzero element f \in M with \mathfrak m f = 0. Set \overline{M} = M /\langle f \rangle , and denote the quotient map x \mapsto \overline{x}. We will define a surjective map
which will prove the lemma since by induction the determinant of \overline{M} is nonzero.
We define \psi on symbols as follows. Let (e_1, \ldots , e_ l) be an admissible sequence. If f \not\in \langle e_1, \ldots , e_ l \rangle then we simply set \psi ([e_1, \ldots , e_ l]) = 0. If f \in \langle e_1, \ldots , e_ l \rangle then we choose an i minimal such that f \in \langle e_1, \ldots , e_ i \rangle . We may write e_ i = \lambda f + x for some unit \lambda \in R and x \in \langle e_1, \ldots , e_{i - 1} \rangle . In this case we set
Note that it is indeed the case that (\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l) is an admissible sequence in \overline{M}, so this makes sense. Let us show that extending this rule \kappa -linearly to linear combinations of symbols does indeed lead to a map on determinants. To do this we have to show that the admissible relations are mapped to zero.
Type (a) relations. Suppose we have (e_1, \ldots , e_ l) an admissible sequence and for some 1 \leq a \leq l we have e_ a \in \langle e_1, \ldots , e_{a - 1}\rangle . Suppose that f \in \langle e_1, \ldots , e_ i\rangle with i minimal. Then i \not= a and \overline{e}_ a \in \langle \overline{e}_1, \ldots , \hat{\overline{e}_ i}, \ldots , \overline{e}_{a - 1}\rangle if i < a or \overline{e}_ a \in \langle \overline{e}_1, \ldots , \overline{e}_{a - 1}\rangle if i > a. Thus the same admissible relation for \det _\kappa (\overline{M}) forces the symbol [\overline{e}_1, \ldots , \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots , \overline{e}_ l] to be zero as desired.
Type (b) relations. Suppose we have (e_1, \ldots , e_ l) an admissible sequence and for some 1 \leq a \leq l we have e_ a = \lambda e'_ a + x with \lambda \in R^*, and x \in \langle e_1, \ldots , e_{a - 1}\rangle . Suppose that f \in \langle e_1, \ldots , e_ i\rangle with i minimal. Say e_ i = \mu f + y with y \in \langle e_1, \ldots , e_{i - 1}\rangle . If i < a then the desired equality is
which follows from \overline{e}_ a = \lambda \overline{e}'_ a + \overline{x} and the corresponding admissible relation for \det _\kappa (\overline{M}). If i > a then the desired equality is
which follows from \overline{e}_ a = \lambda \overline{e}'_ a + \overline{x} and the corresponding admissible relation for \det _\kappa (\overline{M}). The interesting case is when i = a. In this case we have e_ a = \lambda e'_ a + x = \mu f + y. Hence also e'_ a = \lambda ^{-1}(\mu f + y - x). Thus we see that
as desired.
Type (c) relations. Suppose that (e_1, \ldots , e_ l) is an admissible sequence and \mathfrak m e_ a \subset \langle e_1, \ldots , e_{a - 2}\rangle . Suppose that f \in \langle e_1, \ldots , e_ i\rangle with i minimal. Say e_ i = \lambda f + x with x \in \langle e_1, \ldots , e_{i - 1}\rangle . We distinguish 4 cases:
Case 1: i < a - 1. The desired equality is
which follows from the type (c) admissible relation for \det _\kappa (\overline{M}).
Case 2: i > a. The desired equality is
which follows from the type (c) admissible relation for \det _\kappa (\overline{M}).
Case 3: i = a. We write e_ a = \lambda f + \mu e_{a - 1} + y with y \in \langle e_1, \ldots , e_{a - 2}\rangle . Then
by definition. If \overline{\mu } is nonzero, then we have e_{a - 1} = - \mu ^{-1} \lambda f + \mu ^{-1}e_ a - \mu ^{-1} y and we obtain
by definition. Since in \overline{M} we have \overline{e}_ a = \mu \overline{e}_{a - 1} + \overline{y} we see the two outcomes are equal by relation (a) for \det _\kappa (\overline{M}). If on the other hand \overline{\mu } is zero, then we can write e_ a = \lambda f + y with y \in \langle e_1, \ldots , e_{a - 2}\rangle and we have
which is equal to \psi ([e_1, \ldots , e_ l]).
Case 4: i = a - 1. Here we have
by definition. If f \not\in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle then
Since (-1)^{a - 1} = (-1)^{a + 1} the two expressions are the same. Finally, assume f \in \langle e_1, \ldots , e_{a - 2}, e_ a \rangle . In this case we see that e_{a - 1} = \lambda f + x with x \in \langle e_1, \ldots , e_{a - 2}\rangle and e_ a = \mu f + y with y \in \langle e_1, \ldots , e_{a - 2}\rangle for units \lambda , \mu \in R. We conclude that both e_ a \in \langle e_1, \ldots , e_{a - 1} \rangle and e_{a - 1} \in \langle e_1, \ldots , e_{a - 2}, e_ a\rangle . In this case a relation of type (a) applies to both [e_1, \ldots , e_ l] and [e_1, \ldots , e_{a - 2}, e_ a, e_{a - 1}, e_{a + 1}, \ldots , e_ l] and the compatibility of \psi with these shown above to see that both
are zero, as desired.
At this point we have shown that \psi is well defined, and all that remains is to show that it is surjective. To see this let (\overline{f}_2, \ldots , \overline{f}_ l) be an admissible sequence in \overline{M}. We can choose lifts f_2, \ldots , f_ l \in M, and then (f, f_2, \ldots , f_ l) is an admissible sequence in M. Since \psi ([f, f_2, \ldots , f_ l]) = [f_2, \ldots , f_ l] we win. \square
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