## 49.4 Traces for flat quasi-finite ring maps

The trace referred to in the title of this section is of a completely different nature than the trace discussed in Duality for Schemes, Section 48.7. Namely, it is the trace as discussed in Fields, Section 9.20 and generalized in Exercises, Exercises 110.22.6 and 110.22.7.

Let $A \to B$ be a finite flat map of Noetherian rings. Then $B$ is finite flat as an $A$-module and hence finite locally free (Algebra, Lemma 10.78.2). Given $b \in B$ we can consider the trace $\text{Trace}_{B/A}(b)$ of the $A$-linear map $B \to B$ given by multiplication by $b$ on $B$. By the references above this defines an $A$-linear map $\text{Trace}_{B/A} : B \to A$. Since $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as $A \to B$ is finite, we see that $\text{Trace}_{B/A} \in \omega _{B/A}$.

For a general flat quasi-finite ring map we define the notion of a trace as follows.

Definition 49.4.1. Let $A \to B$ be a flat quasi-finite map of Noetherian rings. The trace element is the unique1 element $\tau _{B/A} \in \omega _{B/A}$ with the following property: for any Noetherian $A$-algebra $A_1$ such that $B_1 = B \otimes _ A A_1$ comes with a product decomposition $B_1 = C \times D$ with $A_1 \to C$ finite the image of $\tau _{B/A}$ in $\omega _{C/A_1}$ is $\text{Trace}_{C/A_1}$. Here we use the base change map (49.2.3.1) and Lemma 49.2.7 to get $\omega _{B/A} \to \omega _{B_1/A_1} \to \omega _{C/A_1}$.

We first prove that trace elements are unique and then we prove that they exist.

Lemma 49.4.2. Let $A \to B$ be a flat quasi-finite map of Noetherian rings. Then there is at most one trace element in $\omega _{B/A}$.

Proof. Let $\mathfrak q \subset B$ be a prime ideal lying over the prime $\mathfrak p \subset A$. By Algebra, Lemma 10.145.2 we can find an étale ring map $A \to A_1$ and a prime ideal $\mathfrak p_1 \subset A_1$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p_1) = \kappa (\mathfrak p)$ and such that

$B_1 = B \otimes _ A A_1 = C \times D$

with $A_1 \to C$ finite and such that the unique prime $\mathfrak q_1$ of $B \otimes _ A A_1$ lying over $\mathfrak q$ and $\mathfrak p_1$ corresponds to a prime of $C$. Observe that $\omega _{C/A_1} = \omega _{B/A} \otimes _ B C$ (combine Lemmas 49.2.5 and 49.2.7). Since the collection of ring maps $B \to C$ obtained in this manner is a jointly injective family of flat maps and since the image of $\tau _{B/A}$ in $\omega _{C/A_1}$ is prescribed the uniqueness follows. $\square$

Here is a sanity check.

Lemma 49.4.3. Let $A \to B$ be a finite flat map of Noetherian rings. Then $\text{Trace}_{B/A} \in \omega _{B/A}$ is the trace element.

Proof. Suppose we have $A \to A_1$ with $A_1$ Noetherian and a product decomposition $B \otimes _ A A_1 = C \times D$ with $A_1 \to C$ finite. Of course in this case $A_1 \to D$ is also finite. Set $B_1 = B \otimes _ A A_1$. Since the construction of traces commutes with base change we see that $\text{Trace}_{B/A}$ maps to $\text{Trace}_{B_1/A_1}$. Thus the proof is finished by noticing that $\text{Trace}_{B_1/A_1} = (\text{Trace}_{C/A_1}, \text{Trace}_{D/A_1})$ under the isomorphism $\omega _{B_1/A_1} = \omega _{C/A_1} \times \omega _{D/A_1}$ of Lemma 49.2.7. $\square$

Lemma 49.4.4. Let $A \to B$ be a flat quasi-finite map of Noetherian rings. Let $\tau \in \omega _{B/A}$ be a trace element.

1. If $A \to A_1$ is a map with $A_1$ Noetherian, then with $B_1 = A_1 \otimes _ A B$ the image of $\tau$ in $\omega _{B_1/A_1}$ is a trace element.

2. If $A = R_ f$, then $\tau$ is a trace element in $\omega _{B/R}$.

3. If $g \in B$, then the image of $\tau$ in $\omega _{B_ g/A}$ is a trace element.

4. If $B = B_1 \times B_2$, then $\tau$ maps to a trace element in both $\omega _{B_1/A}$ and $\omega _{B_2/A}$.

Proof. Part (1) is a formal consequence of the definition.

Statement (2) makes sense because $\omega _{B/R} = \omega _{B/A}$ by Lemma 49.2.3. Denote $\tau '$ the element $\tau$ but viewed as an element of $\omega _{B/R}$. To see that (2) is true suppose that we have $R \to R_1$ with $R_1$ Noetherian and a product decomposition $B \otimes _ R R_1 = C \times D$ with $R_1 \to C$ finite. Then with $A_1 = (R_1)_ f$ we see that $B \otimes _ A A_1 = C \times D$. Since $R_1 \to C$ is finite, a fortiori $A_1 \to C$ is finite. Hence we can use the defining property of $\tau$ to get the corresponding property of $\tau '$.

Statement (3) makes sense because $\omega _{B_ g/A} = (\omega _{B/A})_ g$ by Lemma 49.2.3. The proof is similar to the proof of (2). Suppose we have $A \to A_1$ with $A_1$ Noetherian and a product decomposition $B_ g \otimes _ A A_1 = C \times D$ with $A_1 \to C$ finite. Set $B_1 = B \otimes _ A A_1$. Then $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B_1)$ is an open immersion as $B_ g \otimes _ A A_1 = (B_1)_ g$ and the image is closed because $B_1 \to C$ is finite (as $A_1 \to C$ is finite). Thus we see that $B_1 = C \times D_1$ and $D = (D_1)_ g$. Then we can use the defining property of $\tau$ to get the corresponding property for the image of $\tau$ in $\omega _{B_ g/A}$.

Statement (4) makes sense because $\omega _{B/A} = \omega _{B_1/A} \times \omega _{B_2/A}$ by Lemma 49.2.7. Suppose we have $A \to A'$ with $A'$ Noetherian and a product decomposition $B \otimes _ A A' = C \times D$ with $A' \to C$ finite. Then it is clear that we can refine this product decomposition into $B \otimes _ A A' = C_1 \times C_2 \times D_1 \times D_2$ with $A' \to C_ i$ finite such that $B_ i \otimes _ A A' = C_ i \times D_ i$. Then we can use the defining property of $\tau$ to get the corresponding property for the image of $\tau$ in $\omega _{B_ i/A}$. This uses the obvious fact that $\text{Trace}_{C/A'} = (\text{Trace}_{C_1/A'}, \text{Trace}_{C_2/A'})$ under the decomposition $\omega _{C/A'} = \omega _{C_1/A'} \times \omega _{C_2/A'}$. $\square$

Lemma 49.4.5. Let $A \to B$ be a flat quasi-finite map of Noetherian rings. Let $g_1, \ldots , g_ m \in B$ be elements generating the unit ideal. Let $\tau \in \omega _{B/A}$ be an element whose image in $\omega _{B_{g_ i}/A}$ is a trace element for $A \to B_{g_ i}$. Then $\tau$ is a trace element.

Proof. Suppose we have $A \to A_1$ with $A_1$ Noetherian and a product decomposition $B \otimes _ A A_1 = C \times D$ with $A_1 \to C$ finite. We have to show that the image of $\tau$ in $\omega _{C/A_1}$ is $\text{Trace}_{C/A_1}$. Observe that $g_1, \ldots , g_ m$ generate the unit ideal in $B_1 = B \otimes _ A A_1$ and that $\tau$ maps to a trace element in $\omega _{(B_1)_{g_ i}/A_1}$ by Lemma 49.4.4. Hence we may replace $A$ by $A_1$ and $B$ by $B_1$ to get to the situation as described in the next paragraph.

Here we assume that $B = C \times D$ with $A \to C$ is finite. Let $\tau _ C$ be the image of $\tau$ in $\omega _{C/A}$. We have to prove that $\tau _ C = \text{Trace}_{C/A}$ in $\omega _{C/A}$. By the compatibility of trace elements with products (Lemma 49.4.4) we see that $\tau _ C$ maps to a trace element in $\omega _{C_{g_ i}/A}$. Hence, after replacing $B$ by $C$ we may assume that $A \to B$ is finite flat.

Assume $A \to B$ is finite flat. In this case $\text{Trace}_{B/A}$ is a trace element by Lemma 49.4.3. Hence $\text{Trace}_{B/A}$ maps to a trace element in $\omega _{B_{g_ i}/A}$ by Lemma 49.4.4. Since trace elements are unique (Lemma 49.4.2) we find that $\text{Trace}_{B/A}$ and $\tau$ map to the same elements in $\omega _{B_{g_ i}/A} = (\omega _{B/A})_{g_ i}$. As $g_1, \ldots , g_ m$ generate the unit ideal of $B$ the map $\omega _{B/A} \to \prod \omega _{B_{g_ i}/A}$ is injective and we conclude that $\tau _ C = \text{Trace}_{B/A}$ as desired. $\square$

Lemma 49.4.6. Let $A \to B$ be a flat quasi-finite map of Noetherian rings. There exists a trace element $\tau \in \omega _{B/A}$.

Proof. Choose a factorization $A \to B' \to B$ with $A \to B'$ finite and $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ an open immersion. Let $g_1, \ldots , g_ n \in B'$ be elements such that $\mathop{\mathrm{Spec}}(B) = \bigcup D(g_ i)$ as opens of $\mathop{\mathrm{Spec}}(B')$. Suppose that we can prove the existence of trace elements $\tau _ i$ for the quasi-finite flat ring maps $A \to B_{g_ i}$. Then for all $i, j$ the elements $\tau _ i$ and $\tau _ j$ map to trace elements of $\omega _{B_{g_ ig_ j}/A}$ by Lemma 49.4.4. By uniqueness of trace elements (Lemma 49.4.2) they map to the same element. Hence the sheaf condition for the quasi-coherent module associated to $\omega _{B/A}$ (see Algebra, Lemma 10.24.1) produces an element $\tau \in \omega _{B/A}$. Then $\tau$ is a trace element by Lemma 49.4.5. In this way we reduce to the case treated in the next paragraph.

Assume we have $A \to B'$ finite and $g \in B'$ with $B = B'_ g$ flat over $A$. It is our task to construct a trace element in $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B$. Choose a resolution $F_1 \to F_0 \to B' \to 0$ of $B'$ by finite free $A$-modules $F_0$ and $F_1$. Then we have an exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \to F_0^\vee \to F_1^\vee$

where $F_ i^\vee = \mathop{\mathrm{Hom}}\nolimits _ A(F_ i, A)$ is the dual finite free module. Similarly we have the exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _ A(B', B') \to F_0^\vee \otimes _ A B' \to F_1^\vee \otimes _ A B'$

The idea of the construction of $\tau$ is to use the diagram

$B' \xrightarrow {\mu } \mathop{\mathrm{Hom}}\nolimits _ A(B', B') \leftarrow \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _ A B' \xrightarrow {ev} A$

where the first arrow sends $b' \in B'$ to the $A$-linear operator given by multiplication by $b'$ and the last arrow is the evaluation map. The problem is that the middle arrow, which sends $\lambda ' \otimes b'$ to the map $b'' \mapsto \lambda '(b'')b'$, is not an isomorphism. If $B'$ is flat over $A$, the exact sequences above show that it is an isomorphism and the composition from left to right is the usual trace $\text{Trace}_{B'/A}$. In the general case, we consider the diagram

$\xymatrix{ & \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _ A B' \ar[r] \ar[d] & \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _ A B'_ g \ar[d] \\ B' \ar[r]_-\mu \ar@{..>}[rru] \ar@{..>}[ru]^\psi & \mathop{\mathrm{Hom}}\nolimits _ A(B', B') \ar[r] & \mathop{\mathrm{Ker}}(F_0^\vee \otimes _ A B'_ g \to F_1^\vee \otimes _ A B'_ g) }$

By flatness of $A \to B'_ g$ we see that the right vertical arrow is an isomorphism. Hence we obtain the unadorned dotted arrow. Since $B'_ g = \mathop{\mathrm{colim}}\nolimits \frac{1}{g^ n}B'$, since colimits commute with tensor products, and since $B'$ is a finitely presented $A$-module we can find an $n \geq 0$ and a $B'$-linear (for right $B'$-module structure) map $\psi : B' \to \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _ A B'$ whose composition with the left vertical arrow is $g^ n\mu$. Composing with $ev$ we obtain an element $ev \circ \psi \in \mathop{\mathrm{Hom}}\nolimits _ A(B', A)$. Then we set

$\tau = (ev \circ \psi ) \otimes g^{-n} \in \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B'_ g = \omega _{B'_ g/A} = \omega _{B/A}$

We omit the easy verification that this element does not depend on the choice of $n$ and $\psi$ above.

Let us prove that $\tau$ as constructed in the previous paragraph has the desired property in a special case. Namely, say $B' = C' \times D'$ and $g = (f, h)$ where $A \to C'$ flat, $D'_ h$ is flat, and $f$ is a unit in $C'$. To show: $\tau$ maps to $\text{Trace}_{C'/A}$ in $\omega _{C'/A}$. In this case we first choose $n_ D$ and $\psi _ D : D' \to \mathop{\mathrm{Hom}}\nolimits _ A(D', A) \otimes _ A D'$ as above for the pair $(D', h)$ and we can let $\psi _ C : C' \to \mathop{\mathrm{Hom}}\nolimits _ A(C', A) \otimes _ A C' = \mathop{\mathrm{Hom}}\nolimits _ A(C', C')$ be the map seconding $c' \in C'$ to multiplication by $c'$. Then we take $n = n_ D$ and $\psi = (f^{n_ D} \psi _ C, \psi _ D)$ and the desired compatibility is clear because $\text{Trace}_{C'/A} = ev \circ \psi _ C$ as remarked above.

To prove the desired property in general, suppose given $A \to A_1$ with $A_1$ Noetherian and a product decomposition $B'_ g \otimes _ A A_1 = C \times D$ with $A_1 \to C$ finite. Set $B'_1 = B' \otimes _ A A_1$. Then $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(B'_1)$ is an open immersion as $B'_ g \otimes _ A A_1 = (B'_1)_ g$ and the image is closed as $B'_1 \to C$ is finite (since $A_1 \to C$ is finite). Thus $B'_1 = C \times D'$ and $D'_ g = D$. We conclude that $B'_1 = C \times D'$ and $g$ over $A_1$ are as in the previous paragraph. Since formation of the displayed diagram above commutes with base change, the formation of $\tau$ commutes with the base change $A \to A_1$ (details omitted; use the resolution $F_1 \otimes _ A A_1 \to F_0 \otimes _ A A_1 \to B'_1 \to 0$ to see this). Thus the desired compatibility follows from the result of the previous paragraph. $\square$

Remark 49.4.7. Let $f : Y \to X$ be a flat locally quasi-finite morphism of locally Noetherian schemes. Let $\omega _{Y/X}$ be as in Remark 49.2.11. It is clear from the uniqueness, existence, and compatibility with localization of trace elements (Lemmas 49.4.2, 49.4.6, and 49.4.4) that there exists a global section

$\tau _{Y/X} \in \Gamma (Y, \omega _{Y/X})$

such that for every pair of affine opens $\mathop{\mathrm{Spec}}(B) = V \subset Y$, $\mathop{\mathrm{Spec}}(A) = U \subset X$ with $f(V) \subset U$ that element $\tau _{Y/X}$ maps to $\tau _{B/A}$ under the canonical isomorphism $H^0(V, \omega _{Y/X}) = \omega _{B/A}$.

Lemma 49.4.8. Let $k$ be a field and let $A$ be a finite $k$-algebra. Assume $A$ is local with residue field $k'$. The following are equivalent

1. $\text{Trace}_{A/k}$ is nonzero,

2. $\tau _{A/k} \in \omega _{A/k}$ is nonzero, and

3. $k'/k$ is separable and $\text{length}_ A(A)$ is prime to the characteristic of $k$.

Proof. Conditions (1) and (2) are equivalent by Lemma 49.4.3. Let $\mathfrak m \subset A$. Since $\dim _ k(A) < \infty$ it is clear that $A$ has finite length over $A$. Choose a filtration

$A = I_0 \supset \mathfrak m = I_1 \supset I_2 \supset \ldots I_ n = 0$

by ideals such that $I_ i/I_{i + 1} \cong k'$ as $A$-modules. See Algebra, Lemma 10.52.11 which also shows that $n = \text{length}_ A(A)$. If $a \in \mathfrak m$ then $aI_ i \subset I_{i + 1}$ and it is immediate that $\text{Trace}_{A/k}(a) = 0$. If $a \not\in \mathfrak m$ with image $\lambda \in k'$, then we conclude

$\text{Trace}_{A/k}(a) = \sum \nolimits _{i = 0, \ldots , n - 1} \text{Trace}_ k(a : I_ i/I_{i - 1} \to I_ i/I_{i - 1}) = n \text{Trace}_{k'/k}(\lambda )$

The proof of the lemma is finished by applying Fields, Lemma 9.20.7. $\square$

[1] Uniqueness and existence will be justified in Lemmas 49.4.2 and 49.4.6.

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