The Stacks project

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. $\square$