Lemma 49.6.3. Let $A \to B' \to B$ be ring maps with $A \to B'$ of finite type and $B' \to B$ inducing an open immersion of spectra.
The annihilator $J$ of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is $J' \otimes _{B'} B$ where $J'$ is the annihilator of $\mathop{\mathrm{Ker}}(B' \otimes _ A B' \to B')$.
The Noether different $\mathfrak {D}$ of $B$ over $A$ is $\mathfrak {D}'B$, where $\mathfrak {D}'$ is the Noether different of $B'$ over $A$.
Proof. Write $I = \mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ and $I' = \mathop{\mathrm{Ker}}(B' \otimes _ A B' \to B')$. As $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(B')$ is an open immersion, it follows that $B = (B \otimes _ A B) \otimes _{B' \otimes _ A B'} B'$. Thus we see that $I = I'(B \otimes _ A B)$. Since $I'$ is finitely generated and $B' \otimes _ A B' \to B \otimes _ A B$ is flat, we conclude that $J = J'(B \otimes _ A B)$, see Algebra, Lemma 10.40.4. Since the $B' \otimes _ A B'$-module structure of $J'$ factors through $B' \otimes _ A B' \to B'$ we conclude that (1) is true. Part (2) is a consequence of (1). $\square$
Your email address will not be published.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)