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

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

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

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

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

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

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

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

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

follows from the equality

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

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

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

be the local factorizations which exist by our choice of $C_ j \cong C_{\mathfrak m_ j}$. By definition we have

and

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

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

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

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

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

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

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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