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.
42.5 Tame symbols
Consider a Noetherian local ring (A, \mathfrak m) of dimension 1. We denote Q(A) the total ring of fractions of A, see Algebra, Example 10.9.8. The tame symbol will be a map
\partial _ A(-, -) : Q(A)^* \times Q(A)^* \longrightarrow \kappa (\mathfrak m)^*
satisfying the following properties:
\partial _ A(f, gh) = \partial _ A(f, g) \partial _ A(f, h) for f, g, h \in Q(A)^*,
\partial _ A(f, g) \partial _ A(g, f) = 1 for f, g \in Q(A)^*,
\partial _ A(f, 1 - f) = 1 for f \in Q(A)^* such that 1 - f \in Q(A)^*,
\partial _ A(aa', b) = \partial _ A(a, b)\partial _ A(a', b) and \partial _ A(a, bb') = \partial _ A(a, b)\partial _ A(a, b') for a, a', b, b' \in A nonzerodivisors,
\partial _ A(b, b) = (-1)^ m with m = \text{length}_ A(A/bA) for b \in A a nonzerodivisor,
\partial _ A(u, b) = u^ m \bmod \mathfrak m with m = \text{length}_ A(A/bA) for u \in A a unit and b \in A a nonzerodivisor, and
\partial _ A(a, b - a)\partial _ A(b, b) = \partial _ A(b, b - a)\partial _ A(a, b) for a, b \in A such that a, b, b - a are nonzerodivisors.
Since it is easier to work with elements of A we will often think of \partial _ A as a map defined on pairs of nonzerodivisors of A satisfying (4), (5), (6), (7). It is an exercise to see that setting
\partial _ A(\frac{a}{b}, \frac{c}{d}) = \partial _ A(a, c) \partial _ A(a, d)^{-1} \partial _ A(b, c)^{-1} \partial _ A(b, d)
we get a well defined map Q(A)^* \times Q(A)^* \to \kappa (\mathfrak m)^* satisfying (1), (2), (3) as well as the other properties.
We do not claim there is a unique map with these properties. Instead, we will give a recipe for constructing such a map. Namely, given a_1, a_2 \in A nonzerodivisors, we choose a ring extension A \subset B and local factorizations as in Lemma 42.4.4. Then we define
42.5.0.1
\begin{equation} \label{chow-equation-tame-symbol} \partial _ A(a_1, a_2) = \prod \nolimits _ j \text{Norm}_{\kappa (\mathfrak m_ j)/\kappa (\mathfrak m)} ((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}} \bmod \mathfrak m_ j)^{m_ j} \end{equation}
where m_ j = \text{length}_{B_ j}(B_ j/\pi _ j B_ j) and the product is taken over the maximal ideals \mathfrak m_1, \ldots , \mathfrak m_ r 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
Lemma 42.5.2. The tame symbol (42.5.0.1) satisfies (4), (5), (6), (7) and hence gives a map \partial _ A : Q(A)^* \times Q(A)^* \to \kappa (\mathfrak m)^* satisfying (1), (2), (3).
Proof. Let us prove (4). Let a_1, a_2, a_3 \in A be nonzerodivisors. Choose A \subset B as in Lemma 42.4.4 for a_1, a_2, a_3. Then the equality
\partial _ A(a_1a_2, a_3) = \partial _ A(a_1, a_3) \partial _ A(a_2, a_3)
follows from the equality
(-1)^{(e_{1, j} + e_{2, j})e_{3, j}} (u_{1, j}u_{2, j})^{e_{3, j}}u_{3, j}^{-e_{1, j} - e_{2, j}} = (-1)^{e_{1, j}e_{3, j}} u_{1, j}^{e_{3, j}}u_{3, j}^{-e_{1, j}} (-1)^{e_{2, j}e_{3, j}} u_{2, j}^{e_{3, j}}u_{3, j}^{-e_{2, j}}
in B_ j. Properties (5) and (6) are equally immediate.
Let us prove (7). Let a_1, a_2, a_1 - a_2 \in A be nonzerodivisors and set a_3 = a_1 - a_2. Choose A \subset B as in Lemma 42.4.4 for a_1, a_2, a_3. Then it suffices to show
(-1)^{e_{1, j}e_{2, j} + e_{1, j}e_{3, j} + e_{2, j}e_{3, j} + e_{2, j}} u_{1, j}^{e_{2, j} - e_{3, j}} u_{2, j}^{e_{3, j} - e_{1, j}} u_{3, j}^{e_{1, j} - e_{2, j}} \bmod \mathfrak m_ j = 1
This is clear if e_{1, j} = e_{2, j} = e_{3, j}. Say e_{1, j} > e_{2, j}. Then we see that e_{3, j} = e_{2, j} because a_3 = a_1 - a_2 and we see that u_{3, j} has the same residue class as -u_{2, j}. Hence the formula is true – the signs work out as well and this verification is the reason for the choice of signs in (42.5.0.1). The other cases are handled in exactly the same manner. \square
Lemma 42.5.3. Let (A, \mathfrak m) be a Noetherian local ring of dimension 1. Let A \subset B be a finite ring extension with B/A annihilated by a power of \mathfrak m and \mathfrak m not an associated prime of B. For a, b \in A nonzerodivisors we have
\partial _ A(a, b) = \prod \text{Norm}_{\kappa (\mathfrak m_ j)/\kappa (\mathfrak m)}(\partial _{B_ j}(a, b))
where the product is over the maximal ideals \mathfrak m_ j of B and B_ j = B_{\mathfrak m_ j}.
Proof. Choose B_ j \subset C_ j as in Lemma 42.4.4 for a, b. By Lemma 42.4.1 we can choose a finite ring extension B \subset C with C_ j \cong C_{\mathfrak m_ j} for all j. Let \mathfrak m_{j, k} \subset C be the maximal ideals of C lying over \mathfrak m_ j. Let
a = u_{j, k}\pi _{j, k}^{f_{j, k}},\quad b = v_{j, k}\pi _{j, k}^{g_{j, k}}
be the local factorizations which exist by our choice of C_ j \cong C_{\mathfrak m_ j}. By definition we have
\partial _ A(a, b) = \prod \nolimits _{j, k} \text{Norm}_{\kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m)} ((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}} \bmod \mathfrak m_{j, k})^{m_{j, k}}
and
\partial _{B_ j}(a, b) = \prod \nolimits _ k \text{Norm}_{\kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m_ j)} ((-1)^{f_{j, k}g_{j, k}}u_{j, k}^{g_{j, k}}v_{j, k}^{-f_{j, k}} \bmod \mathfrak m_{j, k})^{m_{j, k}}
The result follows by transitivity of norms for \kappa (\mathfrak m_{j, k})/\kappa (\mathfrak m_ j)/\kappa (\mathfrak m), see Fields, Lemma 9.20.5. \square
Lemma 42.5.4. Let (A, \mathfrak m, \kappa ) \to (A', \mathfrak m', \kappa ') be a local homomorphism of Noetherian local rings. Assume A \to A' is flat and \dim (A) = \dim (A') = 1. Set m = \text{length}_{A'}(A'/\mathfrak mA'). For a_1, a_2 \in A nonzerodivisors \partial _ A(a_1, a_2)^ m maps to \partial _{A'}(a_1, a_2) via \kappa \to \kappa '.
Proof. If a_1, a_2 are both units, then \partial _ A(a_1, a_2) = 1 and \partial _{A'}(a_1, a_2) = 1 and the result is true. If not, then we can choose a ring extension A \subset B and local factorizations as in Lemma 42.4.4. Denote \mathfrak m_1, \ldots , \mathfrak m_ m be the maximal ideals of B. Let \mathfrak m_1, \ldots , \mathfrak m_ m be the maximal ideals of B with residue fields \kappa _1, \ldots , \kappa _ m. For each j \in \{ 1, \ldots , m\} denote \pi _ j \in B_ j = B_{\mathfrak m_ j} a nonzerodivisor such that we have factorizations a_ i = u_{i, j}\pi _ j^{e_{i, j}} as in the lemma. By definition we have
\partial _ A(a_1, a_2) = \prod \nolimits _ j \text{Norm}_{\kappa _ j/\kappa } ((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}} \bmod \mathfrak m_ j)^{m_ j}
where m_ j = \text{length}_{B_ j}(B_ j/\pi _ j B_ j).
Set B' = A' \otimes _ A B. Since A' is flat over A we see that A' \subset B' is a ring extension with B'/A' annihilated by a power of \mathfrak m'. Let
\mathfrak m'_{j, l},\quad l = 1, \ldots , n_ j
be the maximal ideals of B' lying over \mathfrak m_ j. Denote \kappa '_{j, l} the residue field of \mathfrak m'_{j, l}. Denote B'_{j, l} the localization of B' at \mathfrak m'_{j, l}. As factorizations of a_1 and a_2 in B'_{j, l} we use the image of the factorizations a_ i = u_{i, j} \pi _ j^{e_{i, j}} given to us in B_ j. By definition we have
\partial _{A'}(a_1, a_2) = \prod \nolimits _{j, l} \text{Norm}_{\kappa '_{j, l}/\kappa '} ((-1)^{e_{1, j}e_{2, j}}u_{1, j}^{e_{2, j}}u_{2, j}^{-e_{1, j}} \bmod \mathfrak m'_{j, l})^{m'_{j, l}}
where m'_{j, l} = \text{length}_{B'_{j, l}}(B'_{j, l}/\pi _ j B'_{j, l}).
Comparing the formulae above we see that it suffices to show that for each j and for any unit u \in B_ j we have
42.5.4.1
\begin{equation} \label{chow-equation-to-prove} \left(\text{Norm}_{\kappa _ j/\kappa }(u \bmod \mathfrak m_ j)^{m_ j}\right)^ m = \prod \nolimits _ l \text{Norm}_{\kappa '_{j, l}/\kappa '}(u \bmod \mathfrak m'_{j, l})^{m'_{j, l}} \end{equation}
in \kappa '. We are going to use the construction of determinants of endomorphisms of finite length modules in More on Algebra, Section 15.120 to prove this. Set M = B_ j/\pi _ j B_ j. By More on Algebra, Lemma 15.120.2 we have
\text{Norm}_{\kappa _ j/\kappa }(u \bmod \mathfrak m_ j)^{m_ j} = \det \nolimits _\kappa (u : M \to M)
Thus, by More on Algebra, Lemma 15.120.3, the left hand side of (42.5.4.1) is equal to \det _{\kappa '}(u : M \otimes _ A A' \to M \otimes _ A A'). We have an isomorphism
M \otimes _ A A' = (B_ j/\pi _ j B_ j) \otimes _ A A' = \bigoplus \nolimits _ l B'_{j, l}/\pi _ j B'_{j, l}
of A'-modules. Setting M'_ l = B'_{j, l}/\pi _ j B'_{j, l} we see that \text{Norm}_{\kappa '_{j, l}/\kappa '}(u \bmod \mathfrak m'_{j, l})^{m'_{j, l}} = \det _{\kappa '}(u_ j : M'_ l \to M'_ l) by More on Algebra, Lemma 15.120.2 again. Hence (42.5.4.1) holds by multiplicativity of the determinant construction, see More on Algebra, Lemma 15.120.1. \square
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