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

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