42.68.45 Application to the key lemma
In this section we apply the results above to show the analogue of the key lemma (Lemma 42.6.3) with the tame symbol d_ A constructed above. Please see Remark 42.6.4 for the relationship with Milnor K-theory.
Lemma 42.68.46 (Key Lemma).reference Let A be a 2-dimensional Noetherian local domain with fraction field K. Let f, g \in K^*. Let \mathfrak q_1, \ldots , \mathfrak q_ t be the height 1 primes \mathfrak q of A such that either f or g is not an element of A^*_{\mathfrak q}. Then we have
\sum \nolimits _{i = 1, \ldots , t} \text{ord}_{A/\mathfrak q_ i}(d_{A_{\mathfrak q_ i}}(f, g)) = 0
We can also write this as
\sum \nolimits _{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(d_{A_{\mathfrak q}}(f, g)) = 0
since at any height one prime \mathfrak q of A where f, g \in A^*_{\mathfrak q} we have d_{A_{\mathfrak q}}(f, g) = 1 by Lemma 42.68.33.
Proof.
Since the tame symbols d_{A_{\mathfrak q}}(f, g) are additive (Lemma 42.68.30) and the order functions \text{ord}_{A/\mathfrak q} are additive (Algebra, Lemma 10.121.1) it suffices to prove the formula when f = a \in A and g = b \in A. In this case we see that we have to show
\sum \nolimits _{\text{height}(\mathfrak q) = 1} \text{ord}_{A/\mathfrak q}(\det \nolimits _\kappa (A_{\mathfrak q}/(ab), a, b)) = 0
By Proposition 42.68.43 this is equivalent to showing that
e_ A(A/(ab), a, b) = 0.
Since the complex A/(ab) \xrightarrow {a} A/(ab) \xrightarrow {b} A/(ab) \xrightarrow {a} A/(ab) is exact we win.
\square
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