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 \ref{lemma-dimension-at-most-one} and its proof. We will show by induction on $l = \text{length}(M)$ that it is nonzero. For $l = 1$ it follows from Lemma \ref{lemma-compare-det}. 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 $$ \psi : \det\nolimits_k(M) \to \det\nolimits_\kappa(\overline{M}) $$ which will prove the lemma since by induction the determinant of $\overline{M}$ is nonzero. \medskip\noindent 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 $$ \psi([e_1, \ldots, e_l]) = (-1)^i \overline{\lambda}[\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l]. $$ 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. \medskip\noindent 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. \medskip\noindent 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 $$ (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] = (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_{a - 1}, \overline{e}'_a, \overline{e}_{a + 1}, \ldots, \overline{e}_l] $$ 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 $$ (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] = (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 1}, \overline{e}'_a, \overline{e}_{a + 1}, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] $$ 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 $$ \psi([e_1, \ldots, e_l]) = (-1)^i \overline{\mu} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] = \psi( \overline{\lambda} [e_1, \ldots, e_{a - 1}, e'_a, e_{a + 1}, \ldots, e_l] ) $$ as desired. \medskip\noindent 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: \medskip\noindent Case 1: $i < a - 1$. The desired equality is \begin{align*} & (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] \\ & = (-1)^{i + 1} \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_{a - 2}, \overline{e}_a, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots, \overline{e}_l] \end{align*} which follows from the type (c) admissible relation for $\det_\kappa(\overline{M})$. \medskip\noindent Case 2: $i > a$. The desired equality is \begin{align*} & (-1)^i \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] \\ & = (-1)^{i + 1} \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 2}, \overline{e}_a, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots, \overline{e}_{i - 1}, \overline{e}_{i + 1}, \ldots, \overline{e}_l] \end{align*} which follows from the type (c) admissible relation for $\det_\kappa(\overline{M})$. \medskip\noindent 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 $$ \psi([e_1, \ldots, e_l]) = (-1)^a \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots, \overline{e}_l] $$ 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 $$ \psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) = (-1)^a \overline{\mu^{-1}\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 2}, \overline{e}_a, \overline{e}_{a + 1}, \ldots, \overline{e}_l] $$ 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 $$ \psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) = (-1)^a \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 1}, \overline{e}_{a + 1}, \ldots, \overline{e}_l] $$ which is equal to $\psi([e_1, \ldots, e_l])$. \medskip\noindent Case 4: $i = a - 1$. Here we have $$ \psi([e_1, \ldots, e_l]) = (-1)^{a - 1} \overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 2}, \overline{e}_a, \ldots, \overline{e}_l] $$ by definition. If $f \not \in \langle e_1, \ldots, e_{a - 2}, e_a \rangle$ then $$ \psi(-[e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) = (-1)^{a + 1}\overline{\lambda} [\overline{e}_1, \ldots, \overline{e}_{a - 2}, \overline{e}_a, \ldots, \overline{e}_l] $$ 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 $$ \psi([e_1, \ldots, e_l]) \quad\text{and}\quad \psi([e_1, \ldots, e_{a - 2}, e_a, e_{a - 1}, e_{a + 1}, \ldots, e_l]) $$ are zero, as desired. \medskip\noindent 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.