The Stacks project

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

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. $\square$

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. $\square$

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 $\omega _{C/A'} = \omega _{C_1/A'} \times \omega _{C_2/A'}$. $\square$

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. $\square$

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

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

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

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

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

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. $\square$

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

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

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