Lemma 42.68.3. With notations as above we have $\dim _\kappa (\det _\kappa (M)) \leq 1$.

Proof. Fix an admissible sequence $(f_1, \ldots , f_ l)$ of $M$ such that

$\text{length}_ R(\langle f_1, \ldots , f_ i\rangle ) = i$

for $i = 1, \ldots , l$. Such an admissible sequence exists exactly because $M$ has length $l$. We will show that any element of $\det _\kappa (M)$ is a $\kappa$-multiple of the symbol $[f_1, \ldots , f_ l]$. This will prove the lemma.

Let $(e_1, \ldots , e_ l)$ be an admissible sequence of $M$. It suffices to show that $[e_1, \ldots , e_ l]$ is a multiple of $[f_1, \ldots , f_ l]$. First assume that $\langle e_1, \ldots , e_ l\rangle \not= M$. Then there exists an $i \in [1, \ldots , l]$ such that $e_ i \in \langle e_1, \ldots , e_{i - 1}\rangle$. It immediately follows from the first admissible relation that $[e_1, \ldots , e_ n] = 0$ in $\det _\kappa (M)$. Hence we may assume that $\langle e_1, \ldots , e_ l\rangle = M$. In particular there exists a smallest index $i \in \{ 1, \ldots , l\}$ such that $f_1 \in \langle e_1, \ldots , e_ i\rangle$. This means that $e_ i = \lambda f_1 + x$ with $x \in \langle e_1, \ldots , e_{i - 1}\rangle$ and $\lambda \in R^*$. By the second admissible relation this means that $[e_1, \ldots , e_ l] = \overline{\lambda }[e_1, \ldots , e_{i - 1}, f_1, e_{i + 1}, \ldots , e_ l]$. Note that $\mathfrak m f_1 = 0$. Hence by applying the third admissible relation $i - 1$ times we see that

$[e_1, \ldots , e_ l] = (-1)^{i - 1}\overline{\lambda } [f_1, e_1, \ldots , e_{i - 1}, e_{i + 1}, \ldots , e_ l].$

Note that it is also the case that $\langle f_1, e_1, \ldots , e_{i - 1}, e_{i + 1}, \ldots , e_ l\rangle = M$. By induction suppose we have proven that our original symbol is equal to a scalar times

$[f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l]$

for some admissible sequence $(f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l)$ whose elements generate $M$, i.e., with $\langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l\rangle = M$. Then we find the smallest $i$ such that $f_{j + 1} \in \langle f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ i\rangle$ and we go through the same process as above to see that

$[f_1, \ldots , f_ j, e_{j + 1}, \ldots , e_ l] = (\text{scalar}) [f_1, \ldots , f_ j, f_{j + 1}, e_{j + 1}, \ldots , \hat{e_ i}, \ldots , e_ l]$

Continuing in this vein we obtain the desired result. $\square$

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