Lemma 42.5.1. The formula (42.5.0.1) determines a well defined element of $\kappa (\mathfrak m)^*$. In other words, the right hand side does not depend on the choice of the local factorizations or the choice of $B$.

Proof. Independence of choice of factorizations. Suppose we have a Noetherian $1$-dimensional local ring $B$, elements $a_1, a_2 \in B$, and nonzerodivisors $\pi , \theta$ such that we can write

$a_1 = u_1 \pi ^{e_1} = v_1 \theta ^{f_1},\quad a_2 = u_2 \pi ^{e_2} = v_2 \theta ^{f_2}$

with $e_ i, f_ i \geq 0$ integers and $u_ i, v_ i$ units in $B$. Observe that this implies

$a_1^{e_2} = u_1^{e_2}u_2^{-e_1}a_2^{e_1},\quad a_1^{f_2} = v_1^{f_2}v_2^{-f_1}a_2^{f_1}$

On the other hand, setting $m = \text{length}_ B(B/\pi B)$ and $k = \text{length}_ B(B/\theta B)$ we find $e_2 m = \text{length}_ B(B/a_2 B) = f_2 k$. Expanding $a_1^{e_2m} = a_1^{f_2 k}$ using the above we find

$(u_1^{e_2}u_2^{-e_1})^ m = (v_1^{f_2}v_2^{-f_1})^ k$

This proves the desired equality up to signs. To see the signs work out we have to show $me_1e_2$ is even if and only if $kf_1f_2$ is even. This follows as both $me_2 = kf_2$ and $me_1 = kf_1$ (same argument as above).

Independence of choice of $B$. Suppose given two extensions $A \subset B$ and $A \subset B'$ as in Lemma 42.4.4. Then

$C = (B \otimes _ A B')/(\mathfrak m\text{-power torsion})$

will be a third one. Thus we may assume we have $A \subset B \subset C$ and factorizations over the local rings of $B$ and we have to show that using the same factorizations over the local rings of $C$ gives the same element of $\kappa (\mathfrak m)$. By transitivity of norms (Fields, Lemma 9.20.5) this comes down to the following problem: if $B$ is a Noetherian local ring of dimension $1$ and $\pi \in B$ is a nonzerodivisor, then

$\lambda ^ m = \prod \text{Norm}_{\kappa _ k/\kappa }(\lambda )^{m_ k}$

Here we have used the following notation: (1) $\kappa$ is the residue field of $B$, (2) $\lambda$ is an element of $\kappa$, (3) $\mathfrak m_ k \subset C$ are the maximal ideals of $C$, (4) $\kappa _ k = \kappa (\mathfrak m_ k)$ is the residue field of $C_ k = C_{\mathfrak m_ k}$, (5) $m = \text{length}_ B(B/\pi B)$, and (6) $m_ k = \text{length}_{C_ k}(C_ k/\pi C_ k)$. The displayed equality holds because $\text{Norm}_{\kappa _ k/\kappa }(\lambda ) = \lambda ^{[\kappa _ k : \kappa ]}$ as $\lambda \in \kappa$ and because $m = \sum m_ k[\kappa _ k:\kappa ]$. First, we have $m = \text{length}_ B(B/xB) = \text{length}_ B(C/\pi C)$ by Lemma 42.2.5 and (42.2.2.1). Finally, we have $\text{length}_ B(C/\pi C) = \sum m_ k[\kappa _ k:\kappa ]$ by Algebra, Lemma 10.52.12. $\square$

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