The Stacks project

Proof. We saw the finiteness in Lemma 10.119.11. Additivity holds since there is a short exact sequence $0 \to R/(a) \to R/(ab) \to R/(b) \to 0$ where the first map is given by multiplication by $b$. (Use length is additive, see Lemma 10.52.3.) $\square$

Proof. Proof of (1). Assume (1)(a). Say $y_1, \ldots , y_ m$ generate $M'$. Then each $y_ i = x_ i/f_ i$ for some $x_ i \in M$ and nonzero $f_ i \in R$. Hence we see that $f_1 \ldots f_ m M' \subset M$. Since $R$ is Noetherian local of dimension $1$ we see that $\mathfrak m^ n \subset (f_1 \ldots f_ m)$ for some $n$ (for example combine Lemmas 10.60.13 and Proposition 10.60.7 or combine Lemmas 10.119.9 and 10.52.4). In other words $\mathfrak m^ nM' \subset M$ for some $n$ Hence $\text{length}(M'/M) < \infty$ by Lemma 10.52.8, in other words (1)(b) holds. Assume (1)(b). Then $M'/M$ is a finite $R$-module (see Lemma 10.52.2). Hence $M'$ is a finite $R$-module as an extension of finite $R$-modules. Hence (1)(c). The implication (1)(c) $\Rightarrow$ (1)(a) follows from the remark following Definition 10.121.3.

Proof of (2). Suppose $M$ is a lattice in $V$ and $M' \subset M$ is an $R$-submodule. We have seen in (1) that if $M'$ is a lattice, then $\text{length}_ R(M/M') < \infty$. Conversely, assume that $\text{length}_ R(M/M') < \infty$. Then $M'$ is finitely generated as $R$ is Noetherian and for some $n$ we have $\mathfrak m^ n M \subset M'$ (Lemma 10.52.4). Hence it follows that $M'$ contains a basis for $V$, and $M'$ is a lattice.

Proof of (3). Assume $M$, $M'$ are lattices in $V$. Since $R$ is Noetherian the submodule $M \cap M'$ of $M$ is finite. As $M$ is a lattice we can find $x_1, \ldots , x_ n \in M$ which form a $K$-basis for $V$. Because $M'$ is a lattice we can write $x_ i = y_ i/f_ i$ with $y_ i \in M'$ and $f_ i \in R$. Hence $f_ ix_ i \in M \cap M'$. Hence $M \cap M'$ is a lattice also. The fact that $M + M'$ is a lattice follows from part (1).

Part (4) follows from additivity of lengths (Lemma 10.52.3) and the exact sequence

Part (5) follows from repeatedly applying part (4). $\square$

Proof. Omitted. $\square$

Proof. We can see that the integer $d(M, \varphi (M))$ does not depend on the lattice $M$ as follows. Suppose that $M'$ is a second such lattice. Then we see that

Since $\varphi$ is an isomorphism we see that $d(\varphi (M'), \varphi (M)) = d(M', M) = -d(M, M')$, and hence $d(M, \varphi (M)) = d(M', \varphi (M'))$. Moreover, both sides of the equation (of the lemma) are additive in $\varphi$, i.e.,

and also

by the independence shown above. Hence it suffices to prove the lemma for generators of $\text{GL}(V)$. Choose an isomorphism $K^{\oplus n} \cong V$. Then $\text{GL}(V) = \text{GL}_ n(K)$ is generated by elementary matrices $E$. The result is clear for $E$ equal to the identity matrix. If $E = E_{ij}(\lambda )$ with $i \not= j$, $\lambda \in K$, $\lambda \not= 0$, for example

then with respect to a different basis we get $E_{12}(1)$. The result is clear for $E = E_{12}(1)$ by taking as lattice $R^{\oplus n} \subset K^{\oplus n}$. Finally, if $E = E_ i(a)$, with $a \in K^*$ for example

then $E_1(a)(R^{\oplus b}) = aR \oplus R^{\oplus n - 1}$ and it is clear that $d(R^{\oplus n}, aR \oplus R^{\oplus n - 1}) = \text{ord}_ R(a)$ as desired. $\square$

Proof. The ring $B$ is semi-local by Lemma 10.113.2. Write $y = b/b'$ for some $b, b' \in B$. By the additivity of $\text{ord}$ and multiplicativity of $\text{Nm}$ it suffices to prove the lemma for $y = b$ or $y = b'$. In other words we may assume $y \in B$. In this case the right hand side of the formula is

By Lemma 10.52.12 this is equal to $\text{length}_ A(B/yB)$. By Lemma 10.121.7 we have

Since $x = \text{Nm}_{L/K}(y) = \det \nolimits _ K(L \xrightarrow {y} L)$ by definition the lemma is proved. $\square$