Showing surjectivity of trace map $B\to A$ from faithful flatness

Question

I've come across the claim if $A\hookrightarrow B$ is a finite etale extension of rings (commutative w/ $1$) with $A$ Noetherian then the trace map $\operatorname{Tr}_{B/A}:B\to A$ is surjective and am looking for some help to see why this is true.

Just to recall the definitions, if $M$ is a finite projective $A$-module then we can define the trace of an element $\varphi\in\operatorname{End}_A(M)$ by recalling that we can take some elements $f_i\in A$ generating the unit ideal, such that $M_{f_i}$ is free over $A_{f_i}$, and then defining the trace there via some choice of basis, glueing the results to get an element $\operatorname{Tr}(\varphi)\in A$ and showing this is independent of the choices we made. Then for $b\in B$ we get an associated $A$-endomorphism of $B$ via multiplication by $b$, and this gives us a trace element $\operatorname{Tr}_{B/A}(b)$; the resulting map $\operatorname{Tr}_{B/A}:B\to A$ is $A$-linear.

In our situation the assumptions imply that $B$ is finite projective over $A$, so the trace map makes sense. They also imply that $A\to B$ is faithfully flat, and my impression is that this is the key assumption for the surjectivity claim (but I've decided to leave the other assumptions in case they do matter). Moreover, something else I read when trying to figure this out made it sound like one could do as follows: tensor with $B$ and show the resulting map $B\otimes_A B\to A\otimes_A B\cong B$ is surjective instead, and note that this is actually just the multiplication map $b\otimes b'\mapsto bb'$. This doesn't seem correct though when I try to write it down, but maybe I'm being stupid or misunderstanding what they were trying to say.

If it helps to simplify things for a solution, I'd be happy with the case where $B$ is actually free as an $A$-module.

Answer 1

As in the comments, the "unramified" assumption is important, i.e. for finite faithfully flat ring maps, trace will not be surjective: examples are in the commments to the OP ($\mathbb{Z}_2 \subseteq \mathbb{Z}_2 [\sqrt{2}]$ is one of them).

The surjectivity of the $A$-module map $\mathrm{Tr}_{B/A}: B \rightarrow A$ can be checked stalk-locally, i.e. it is enough to show that $(\mathrm{Tr}_{B/A})_\mathfrak{p}: B_{\mathfrak{p}} \rightarrow A_{\mathfrak{p}}$ is surjective for all $\mathfrak{p}\in \mathrm{Spec}\,A$. Clearly we have $(\mathrm{Tr}_{B/A})_\mathfrak{p}=\mathrm{Tr}_{B_{\mathfrak{p}}/A_{\mathfrak{p}}}$. By Nakayama, it is enough to show surjectivity of $\overline{\mathrm{Tr}_{B_{\mathfrak{p}}/A_{\mathfrak{p}}}}=\mathrm{Tr}_{(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}})/{(A_{\mathfrak{p}}}/\mathfrak{p})}: B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}\rightarrow A_{\mathfrak{p}}/\mathfrak{p}=k(\mathfrak{p})$. Since $A \rightarrow B$ is finite, hence surjective on $\mathrm{Spec}$'s, there are primes $\mathfrak{q}_1, \mathfrak{q}_2, \dots, \mathfrak{q}_s \subseteq B$ such that $\mathfrak{q}_i\cap A=\mathfrak{p}$; then the unramified assumption shows that $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}=\prod_{i}k(\mathfrak{q}_i)$ (see this Lemma in Stackproject). It follows that the map $\mathrm{Tr}_{(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}})/k(\mathfrak{p})}$ is the sum of the maps $\mathrm{Tr}_{k(\mathfrak{q}_i)/k(\mathfrak{p})}$. These are the trace maps for finite separable field extensions by the unramified assumption again, hence surjective. This finishes the proof.