Lemma 49.6.7. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. The diagram

\[ \xymatrix{ J \ar[rr] \ar[rd]_\mu & & \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B) \ar[ld]^{\varphi \mapsto \varphi (\tau _{B/A})} \\ & B } \]

commutes where the horizontal arrow is the isomorphism of Lemma 49.6.6. Hence the Noether different of $B$ over $A$ is the image of the map $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B) \to B$.

**Proof.**
Exactly as in the proof of Lemma 49.6.6 this reduces to the case of a finite free map $A \to B$. In this case $\tau _{B/A} = \text{Trace}_{B/A}$. Choose a basis $b_1, \ldots , b_ n$ of $B$ as an $A$-module. Let $\xi = \sum b_ i \otimes c_ i \in J$. Then $\mu (\xi ) = \sum b_ i c_ i$. On the other hand, the image of $\xi $ in $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$ sends $\text{Trace}_{B/A}$ to $\sum \text{Trace}_{B/A}(b_ i)c_ i$. Thus we have to show

\[ \sum b_ ic_ i = \sum \text{Trace}_{B/A}(b_ i)c_ i \]

when $\xi = \sum b_ i \otimes c_ i \in J$. Write $b_ i b_ j = \sum _ k a_{ij}^ k b_ k$ for some $a_{ij}^ k \in A$. Then the right hand side is $\sum _{i, j} a_{ij}^ j c_ i$. On the other hand, $\xi \in J$ implies

\[ (b_ j \otimes 1)(\sum \nolimits _ i b_ i \otimes c_ i) = (1 \otimes b_ j)(\sum \nolimits _ i b_ i \otimes c_ i) \]

which implies that $b_ j c_ i = \sum _ k a_{jk}^ i c_ k$. Thus the left hand side is $\sum _{i, j} a_{ij}^ i c_ j$. Since $a_{ij}^ k = a_{ji}^ k$ the equality holds.
$\square$

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