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.

1. The annihilator $J$ of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is $J' \otimes _{B'} B$ where $J'$ is the annihilator of $\mathop{\mathrm{Ker}}(B' \otimes _ A B' \to B')$.

2. 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). $\square$

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