
## 10.120 Orders of vanishing

Lemma 10.120.1. Let $R$ be a semi-local Noetherian ring of dimension $1$. If $a, b \in R$ are nonzerodivisors then

$\text{length}_ R(R/(ab)) = \text{length}_ R(R/(a)) + \text{length}_ R(R/(b))$

and these lengths are finite.

Proof. We saw the finiteness in Lemma 10.118.11. 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 10.51.3.) $\square$

Definition 10.120.2. Suppose that $K$ is a field, and $R \subset K$ is a local1 Noetherian subring of dimension $1$ with fraction field $K$. In this case we define the order of vanishing along $R$

$\text{ord}_ R : K^* \longrightarrow \mathbf{Z}$

by the rule

$\text{ord}_ R(x) = \text{length}_ R(R/(x))$

if $x \in R$ and we set $\text{ord}_ R(x/y) = \text{ord}_ R(x) - \text{ord}_ R(y)$ for $x, y \in R$ both nonzero.

We can use the order of vanishing to compare lattices in a vector space. Here is the definition.

Definition 10.120.3. Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. A lattice in $V$ is a finite $R$-submodule $M \subset V$ such that $V = K \otimes _ R M$.

The condition $V = K \otimes _ R M$ signifies that $M$ contains a basis for the vector space $K$. We remark that in many places in the literature the notion of a lattice may be defined only in case the ring $R$ is a discrete valuation ring. If $R$ is a discrete valuation ring then any lattice is a free $R$-module, and this may not be the case in general.

Lemma 10.120.4. Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space.

1. If $M$ is a lattice in $V$ and $M \subset M' \subset V$ is an $R$-submodule of $V$ containing $M$ then the following are equivalent

1. $M'$ is a lattice,

2. $\text{length}_ R(M'/M)$ is finite, and

3. $M'$ is finitely generated.

2. If $M$ is a lattice in $V$ and $M' \subset M$ is an $R$-submodule of $M$ then $M'$ is a lattice if and only if $\text{length}_ R(M/M')$ is finite.

3. If $M$, $M'$ are lattices in $V$, then so are $M \cap M'$ and $M + M'$.

4. If $M \subset M' \subset M'' \subset V$ are lattices in $V$ then

$\text{length}_ R(M''/M) = \text{length}_ R(M'/M) + \text{length}_ R(M''/M').$
5. If $M$, $M'$, $N$, $N'$ are lattices in $V$ and $N \subset M \cap M'$, $M + M' \subset N'$, then we have

\begin{eqnarray*} & & \text{length}_ R(M/M \cap M') - \text{length}_ R(M'/M \cap M')\\ & = & \text{length}_ R(M/N) - \text{length}_ R(M'/N) \\ & = & \text{length}_ R(M + M' / M') - \text{length}_ R(M + M'/M) \\ & = & \text{length}_ R(N' / M') - \text{length}_ R(N'/M) \end{eqnarray*}

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 10.59.12 and Proposition 10.59.6 or combine Lemmas 10.118.9 and 10.51.4). In other words $\mathfrak m^ nM' \subset M$ for some $n$ Hence $\text{length}(M'/M) < \infty$ by Lemma 10.51.8, in other words (1)(b) holds. Assume (1)(b). Then $M'/M$ is a finite $R$-module (see Lemma 10.51.2). 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 10.120.3.

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 10.51.4). 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 10.51.3) and the exact sequence

$0 \to M'/M \to M''/M \to M''/M' \to 0$

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

Definition 10.120.5. Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. Let $M$, $M'$ be two lattices in $V$. The distance between $M$ and $M'$ is the integer

$d(M, M') = \text{length}_ R(M/M \cap M') - \text{length}_ R(M'/M \cap M')$

of Lemma 10.120.4 part (5).

In particular, if $M' \subset M$, then $d(M, M') = \text{length}_ R(M/M')$.

Lemma 10.120.6. Let $R$ be a Noetherian local domain of dimension $1$ with fraction field $K$. Let $V$ be a finite dimensional $K$-vector space. This distance function has the property that

$d(M, M'') = d(M, M') + d(M', M'')$

whenever given three lattices $M$, $M'$, $M''$ of $V$. In particular we have $d(M, M') = - d(M', M)$.

Proof. Omitted. $\square$

Lemma 10.120.7. 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. $\square$

Lemma 10.120.8. Let $A \to B$ be a ring map. Assume

1. $A$ is a Noetherian local domain of dimension $1$,

2. $A \subset B$ is a finite extension of domains.

Let $L/K$ be the corresponding finite extension of fraction fields. Let $y \in L^*$ and $x = \text{Nm}_{L/K}(y)$. In this situation $B$ is semi-local. Let $\mathfrak m_ i$, $i = 1, \ldots , n$ be the maximal ideals of $B$. Then

$\text{ord}_ A(x) = \sum \nolimits _ i [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m_ A)] \text{ord}_{B_{\mathfrak m_ i}}(y)$

where $\text{ord}$ is defined as in Definition 10.120.2.

Proof. The ring $B$ is semi-local by Lemma 10.112.2. 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

$\sum [\kappa (\mathfrak m_ i) : \kappa (\mathfrak m_ A)] \text{length}_{B_{\mathfrak m_ i}}((B/yB)_{\mathfrak m_ i})$

By Lemma 10.51.12 this is equal to $\text{length}_ A(B/yB)$. By Lemma 10.120.7 we have

$\text{length}_ A(B/yB) = d(B, yB) = \text{ord}_ A(\det \nolimits _ K(L \xrightarrow {y} L)).$

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

[1] We could also define this when $R$ is only semi-local but this is probably never really what you want!

Comment #3509 by Dmitrii on

In the proof of Lemma 02MJ it says "In this case the left hand side of the formula"; I think this should be replaced by the "right hand side of the formula".

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