Lemma 49.6.7. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. The diagram
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
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
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)