## 49.8 The Dedekind different

Let $A \to B$ be a ring map. We say *the Dedekind different is defined* if $A$ is Noetherian, $A \to B$ is finite, any nonzerodivisor on $A$ is a nonzerodivisor on $B$, and $K \to L$ is étale where $K = Q(A)$ and $L = B \otimes _ A K$. Then $K \subset L$ is finite étale and

\[ \mathcal{L}_{B/A} = \{ x \in L \mid \text{Trace}_{L/K}(bx) \in A \text{ for all }b \in B\} \]

is the Dedekind complementary module. In this situation the *Dedekind different* is

\[ \mathfrak {D}_{B/A} = \{ x \in L \mid x\mathcal{L}_{B/A} \subset B\} \]

viewed as a $B$-submodule of $L$. By Lemma 49.8.1 the Dedekind different is an ideal of $B$ either if $A$ is normal or if $B$ is flat over $A$.

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).
$\square$

Lemma 49.8.2. If the Dedekind different of $A \to B$ is defined, then there is a canonical isomorphism $\mathcal{L}_{B/A} \to \omega _{B/A}$.

**Proof.**
Recall that $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B, A)$ as $A \to B$ is finite. We send $x \in \mathcal{L}_{B/A}$ to the map $b \mapsto \text{Trace}_{L/K}(bx)$. Conversely, given an $A$-linear map $\varphi : B \to A$ we obtain a $K$-linear map $\varphi _ K : L \to K$. Since $K \to L$ is finite étale, we see that the trace pairing is nondegenerate (Lemma 49.3.1) and hence there exists a $x \in L$ such that $\varphi _ K(y) = \text{Trace}_{L/K}(xy)$ for all $y \in L$. Then $x \in \mathcal{L}_{B/A}$ maps to $\varphi $ in $\omega _{B/A}$.
$\square$

Lemma 49.8.3. If the Dedekind different of $A \to B$ is defined and $A \to B$ is flat, then

the canonical isomorphism $\mathcal{L}_{B/A} \to \omega _{B/A}$ sends $1 \in \mathcal{L}_{B/A}$ to the trace element $\tau _{B/A} \in \omega _{B/A}$, and

the Dedekind different is $\mathfrak {D}_{B/A} = \{ b \in B \mid b\omega _{B/A} \subset B\tau _{B/A}\} $.

**Proof.**
The first assertion follows from the proof of Lemma 49.8.1 and Lemma 49.4.3. The second assertion is immediate from the first and the definitions.
$\square$

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