The Stacks project

Lemma 10.121.7. Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. Let $\varphi : V \to V$ be a $K$-linear isomorphism. For any lattice $M \subset V$ we have

\[ d(M, \varphi (M)) = \text{ord}_ R(\det (\varphi )) \]

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

\begin{eqnarray*} d(M, \varphi (M)) & = & d(M, M') + d(M', \varphi (M)) \\ & = & d(M, M') + d(\varphi (M'), \varphi (M)) + d(M', \varphi (M')) \end{eqnarray*}

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.,

\[ \text{ord}_ R(\det (\varphi \circ \psi )) = \text{ord}_ R(\det (\varphi )) + \text{ord}_ R(\det (\psi )) \]

and also

\begin{eqnarray*} d(M, \varphi (\psi ((M))) & = & d(M, \psi (M)) + d(\psi (M), \varphi (\psi (M))) \\ & = & d(M, \psi (M)) + d(M, \varphi (M)) \end{eqnarray*}

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

\[ E_{12}(\lambda ) = \left( \begin{matrix} 1 & \lambda & \ldots \\ 0 & 1 & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right) \]

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

\[ E_1(a) = \left( \begin{matrix} a & 0 & \ldots \\ 0 & 1 & \ldots \\ \ldots & \ldots & \ldots \end{matrix} \right) \]

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. $\square$

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