Lemma 49.8.1. Assume the Dedekind different of $A \to B$ is defined. Consider the statements

$A \to B$ is flat,

$A$ is a normal ring,

$\text{Trace}_{L/K}(B) \subset A$,

$1 \in \mathcal{L}_{B/A}$, and

the Dedekind different $\mathfrak {D}_{B/A}$ is an ideal of $B$.

Then we have (1) $\Rightarrow $ (3), (2) $\Rightarrow $ (3), (3) $\Leftrightarrow $ (4), and (4) $\Rightarrow $ (5).

**Proof.**
The equivalence of (3) and (4) and the implication (4) $\Rightarrow $ (5) are immediate.

If $A \to B$ is flat, then we see that $\text{Trace}_{B/A} : B \to A$ is defined and that $\text{Trace}_{L/K}$ is the base change. Hence (3) holds.

If $A$ is normal, then $A$ is a finite product of normal domains, hence we reduce to the case of a normal domain. Then $K$ is the fraction field of $A$ and $L = \prod L_ i$ is a finite product of finite separable field extensions of $K$. Then $\text{Trace}_{L/K}(b) = \sum \text{Trace}_{L_ i/K}(b_ i)$ where $b_ i \in L_ i$ is the image of $b$. Since $b$ is integral over $A$ as $B$ is finite over $A$, these traces are in $A$. This is true because the minimal polynomial of $b_ i$ over $K$ has coefficients in $A$ (Algebra, Lemma 10.38.6) and because $\text{Trace}_{L_ i/K}(b_ i)$ is an integer multiple of one of these coefficients (Fields, Lemma 9.20.3).
$\square$

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