Lemma 49.6.8. Let $A \to B$ be a finite type ring map. Let $\mathfrak {D} \subset B$ be the Noether different. Then $V(\mathfrak {D})$ is the set of primes $\mathfrak q \subset B$ such that $A \to B$ is not unramified at $\mathfrak q$.
Proof. Assume $A \to B$ is unramified at $\mathfrak q$. After replacing $B$ by $B_ g$ for some $g \in B$, $g \not\in \mathfrak q$ we may assume $A \to B$ is unramified (Algebra, Definition 10.151.1 and Lemma 49.6.3). In this case $\Omega _{B/A} = 0$. Hence if $I = \mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$, then $I/I^2 = 0$ by Algebra, Lemma 10.131.13. Since $A \to B$ is of finite type, we see that $I$ is finitely generated. Hence by Nakayama's lemma (Algebra, Lemma 10.20.1) there exists an element of the form $1 + i$ annihilating $I$. It follows that $\mathfrak {D} = B$.
Conversely, assume that $\mathfrak {D} \not\subset \mathfrak q$. Then after replacing $B$ by a principal localization as above we may assume $\mathfrak {D} = B$. This means there exists an element of the form $1 + i$ in the annihilator of $I$. Conversely this implies that $I/I^2 = \Omega _{B/A}$ is zero and we conclude. $\square$
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