## 49.14 A generalization of the different

In this section we generalize Definition 49.9.1 to take into account all cases of ring maps $A \to B$ where the Dedekind different is defined and $1 \in \mathcal{L}_{B/A}$. First we explain the condition “$A \to B$ maps nonzerodivisors to nonzerodivisors and induces a flat map $Q(A) \to Q(A) \otimes _ A B$”.

Lemma 49.14.1. Let $A \to B$ be a map of Noetherian rings. Consider the conditions

nonzerodivisors of $A$ map to nonzerodivisors of $B$,

(1) holds and $Q(A) \to Q(A) \otimes _ A B$ is flat,

$A \to B_\mathfrak q$ is flat for every $\mathfrak q \in \text{Ass}(B)$,

(3) holds and $A \to B_\mathfrak q$ is flat for every $\mathfrak q$ lying over an element in $\text{Ass}(A)$.

Then we have the following implications

\[ \xymatrix{ (1) & (2) \ar@{=>}[l] \ar@{=>}[d] \\ (3) \ar@{=>}[u] & (4) \ar@{=>}[l] } \]

If going up holds for $A \to B$ then (2) and (4) are equivalent.

**Proof.**
The horizontal implications in the diagram are trivial. Let $S \subset A$ be the set of nonzerodivisors so that $Q(A) = S^{-1}A$ and $Q(A) \otimes _ A B = S^{-1}B$. Recall that $S = A \setminus \bigcup _{\mathfrak p \in \text{Ass}(A)} \mathfrak p$ by Algebra, Lemma 10.63.9. Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$.

Assume (2). If $\mathfrak q \in \text{Ass}(B)$ then $\mathfrak q$ consists of zerodivisors, hence (1) implies the same is true for $\mathfrak p$. Hence $\mathfrak p$ corresponds to a prime of $S^{-1}A$. Hence $A \to B_\mathfrak q$ is flat by our assumption (2). If $\mathfrak q$ lies over an associated prime $\mathfrak p$ of $A$, then certainly $\mathfrak p \in \mathop{\mathrm{Spec}}(S^{-1}A)$ and the same argument works.

Assume (3). Let $f \in A$ be a nonzerodivisor. If $f$ were a zerodivisor on $B$, then $f$ is contained in an associated prime $\mathfrak q$ of $B$. Since $A \to B_\mathfrak q$ is flat by assumption, we conclude that $\mathfrak p$ is an associated prime of $A$ by Algebra, Lemma 10.65.3. This would imply that $f$ is a zerodivisor on $A$, a contradiction.

Assume (4) and going up for $A \to B$. We already know (1) holds. If $\mathfrak q$ corresponds to a prime of $S^{-1}B$ then $\mathfrak p$ is contained in an associated prime $\mathfrak p'$ of $A$. By going up there exists a prime $\mathfrak q'$ containing $\mathfrak q$ and lying over $\mathfrak p$. Then $A \to B_{\mathfrak q'}$ is flat by (4). Hence $A \to B_{\mathfrak q}$ is flat as a localization. Thus $A \to S^{-1}B$ is flat and so is $S^{-1}A \to S^{-1}B$, see Algebra, Lemma 10.39.18.
$\square$

Lemma 49.14.3. Assume the Dedekind different is defined for $A \to B$. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. The generalization of Remark 49.14.2 applies to the morphism $f : Y \to X$ if and only if $1 \in \mathcal{L}_{B/A}$ (e.g., if $A$ is normal, see Lemma 49.8.1). In this case $\mathfrak {D}_{B/A}$ is an ideal of $B$ and we have

\[ \mathfrak {D}_ f = \widetilde{\mathfrak {D}_{B/A}} \]

as coherent ideal sheaves on $Y$.

**Proof.**
As the Dedekind different for $A \to B$ is defined we can apply Lemma 49.14.1 to see that $Y \to X$ satisfies condition (1) of Remark 49.14.2. Recall that there is a canonical isomorphism $c : \mathcal{L}_{B/A} \to \omega _{B/A}$, see Lemma 49.8.2. Let $K = Q(A)$ and $L = K \otimes _ A B$ as above. By construction the map $c$ fits into a commutative diagram

\[ \xymatrix{ \mathcal{L}_{B/A} \ar[r] \ar[d]_ c & L \ar[d] \\ \omega _{B/A} \ar[r] & \mathop{\mathrm{Hom}}\nolimits _ K(L, K) } \]

where the right vertical arrow sends $x \in L$ to the map $y \mapsto \text{Trace}_{L/K}(xy)$ and the lower horizontal arrow is the base change map (49.2.3.1) for $\omega _{B/A}$. We can factor the lower horizontal map as

\[ \omega _{B/A} = \Gamma (Y, \omega _{Y/X}) \to \Gamma (V, \omega _{V/X}) \to \mathop{\mathrm{Hom}}\nolimits _ K(L, K) \]

Since all associated points of $\omega _{V/X}$ map to associated primes of $A$ (Lemma 49.2.8) we see that the second map is injective. The element $\tau _{V/X}$ maps to $\text{Trace}_{L/K}$ in $\mathop{\mathrm{Hom}}\nolimits _ K(L, K)$ by the very definition of trace elements (Definition 49.4.1). Thus $\tau $ as in condition (2) of Remark 49.14.2 exists if and only if $1 \in \mathcal{L}_{B/A}$ and then $\tau = c(1)$. In this case, by Lemma 49.8.1 we see that $\mathfrak {D}_{B/A} \subset B$. Finally, the agreement of $\mathfrak {D}_ f$ with $\mathfrak {D}_{B/A}$ is immediate from the definitions and the fact $\tau = c(1)$ seen above.
$\square$

Example 49.14.4. Let $k$ be a field. Let $A = k[x, y]/(xy)$ and $B = k[u, v]/(uv)$ and let $A \to B$ be given by $x \mapsto u^ n$ and $y \mapsto v^ m$ for some $n, m \in \mathbf{N}$ prime to the characteristic of $k$. Then $A_{x + y} \to B_{x + y}$ is (finite) étale hence we are in the situation where the Dedekind different is defined. A computation shows that

\[ \text{Trace}_{L/K}(1) = (nx + my)/(x + y),\quad \text{Trace}_{L/K}(u^ i) = 0,\quad \text{Trace}_{L/K}(v^ j) = 0 \]

for $1 \leq i < n$ and $1 \leq j < m$. We conclude that $1 \in \mathcal{L}_{B/A}$ if and only if $n = m$. Moreover, a computation shows that if $n = m$, then $\mathcal{L}_{B/A} = B$ and the Dedekind different is $B$ as well. In other words, we find that the different of Remark 49.14.2 is defined for $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ if and only if $n = m$, and in this case the different is the unit ideal. Thus we see that in nonflat cases the nonvanishing of the different does not guarantee the morphism is étale or unramified.

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