The Stacks project

Lemma 42.5.1. The formula ( 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 ( Finally, we have $\text{length}_ B(C/\pi C) = \sum m_ k[\kappa _ k:\kappa ]$ by Algebra, Lemma 10.52.12. $\square$

Comments (0)

There are also:

  • 2 comment(s) on Section 42.5: Tame symbols

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0EAR. Beware of the difference between the letter 'O' and the digit '0'.