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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).