The Stacks project

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. $\square$