## 49.6 The Noether different

There are many different differents available in the literature. We list some of them in this and the next sections; for more information we suggest the reader consult [Kunz].

Let $A \to B$ be a ring map. Denote

\[ \mu : B \otimes _ A B \longrightarrow B,\quad b \otimes b' \longmapsto bb' \]

the multiplication map. Let $I = \mathop{\mathrm{Ker}}(\mu )$. It is clear that $I$ is generated by the elements $b \otimes 1 - 1 \otimes b$ for $b \in B$. Hence the annihilator $J \subset B \otimes _ A B$ of $I$ is a $B$-module in a canonical manner. The *Noether different* of $B$ over $A$ is the image of $J$ under the map $\mu : B \otimes _ A B \to B$. Equivalently, the Noether different is the image of the map

\[ J = \mathop{\mathrm{Hom}}\nolimits _{B \otimes _ A B}(B, B \otimes _ A B) \longrightarrow B,\quad \varphi \longmapsto \mu (\varphi (1)) \]

We begin with some obligatory lemmas.

Lemma 49.6.1. Let $A \to B_ i$, $i = 1, 2$ be ring maps. Set $B = B_1 \times B_2$.

The annihilator $J$ of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$ is $J_1 \times J_2$ where $J_ i$ is the annihilator of $\mathop{\mathrm{Ker}}(B_ i \otimes _ A B_ i \to B_ i)$.

The Noether different $\mathfrak {D}$ of $B$ over $A$ is $\mathfrak {D}_1 \times \mathfrak {D}_2$, where $\mathfrak {D}_ i$ is the Noether different of $B_ i$ over $A$.

**Proof.**
Omitted.
$\square$

Lemma 49.6.2. Let $A \to B$ be a finite type ring map. Let $A \to A'$ be a flat ring map. Set $B' = B \otimes _ A A'$.

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

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.**
Choose generators $b_1, \ldots , b_ n$ of $B$ as an $A$-algebra. Then

\[ J = \mathop{\mathrm{Ker}}(B \otimes _ A B \xrightarrow {b_ i \otimes 1 - 1 \otimes b_ i} (B \otimes _ A B)^{\oplus n}) \]

Hence we see that the formation of $J$ commutes with flat base change. The result on the Noether different follows immediately from this.
$\square$

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.

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

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).
$\square$

Lemma 49.6.5. Let $A \to B$ be a quasi-finite homomorphism of Noetherian rings.

If $A \to A'$ is a flat map of Noetherian rings, then

\[ \xymatrix{ \omega _{B/A} \times J \ar[r] \ar[d] & B \ar[d] \\ \omega _{B'/A'} \times J' \ar[r] & B' } \]

is commutative where notation as in Lemma 49.6.2 and horizontal arrows are given by (49.6.4.1).

If $B = B_1 \times B_2$, then

\[ \xymatrix{ \omega _{B/A} \times J \ar[r] \ar[d] & B \ar[d] \\ \omega _{B_ i/A} \times J_ i \ar[r] & B_ i } \]

is commutative for $i = 1, 2$ where notation as in Lemma 49.6.1 and horizontal arrows are given by (49.6.4.1).

**Proof.**
Because of the construction of the pairing in Remark 49.6.4 both (1) and (2) reduce to the case where $A \to B$ is finite. Then (1) follows from the fact that the contraction map $\mathop{\mathrm{Hom}}\nolimits _ A(M, A) \otimes _ A M \otimes _ A M \to M$, $\lambda \otimes m \otimes m' \mapsto \lambda (m)m'$ commuted with base change. To see (2) use that $J = J_1 \times J_2$ is contained in the summands $B_1 \otimes _ A B_1$ and $B_2 \otimes _ A B_2$ of $B \otimes _ A B$.
$\square$

Lemma 49.6.6. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. The pairing of Remark 49.6.4 induces an isomorphism $J \to \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$.

**Proof.**
We first prove this when $A \to B$ is finite and flat. In this case we can localize on $A$ and assume $B$ is finite free as an $A$-module. Let $b_1, \ldots , b_ n$ be a basis of $B$ as an $A$-module and denote $b_1^\vee , \ldots , b_ n^\vee $ the dual basis of $\omega _{B/A}$. Note that $\sum b_ i \otimes c_ i \in J$ maps to the element of $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$ which sends $b_ i^\vee $ to $c_ i$. Suppose $\varphi : \omega _{B/A} \to B$ is $B$-linear. Then we claim that $\xi = \sum b_ i \otimes \varphi (b_ i^\vee )$ is an element of $J$. Namely, the $B$-linearity of $\varphi $ exactly implies that $(b \otimes 1)\xi = (1 \otimes b)\xi $ for all $b \in B$. Thus our map has an inverse and it is an isomorphism.

Let $\mathfrak q \subset B$ be a prime lying over $\mathfrak p \subset A$. We will show that the localization

\[ J_\mathfrak q \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ B(\omega _ B/A, B)_\mathfrak q \]

is an isomorphism. This suffices by Algebra, Lemma 10.23.1. By Algebra, Lemma 10.145.2 we can find an étale ring map $A \to A'$ and a prime ideal $\mathfrak p' \subset A'$ lying over $\mathfrak p$ such that $\kappa (\mathfrak p') = \kappa (\mathfrak p)$ and such that

\[ B' = B \otimes _ A A' = C \times D \]

with $A' \to C$ finite and such that the unique prime $\mathfrak q'$ of $B \otimes _ A A'$ lying over $\mathfrak q$ and $\mathfrak p'$ corresponds to a prime of $C$. Let $J'$ be the annihilator of $\mathop{\mathrm{Ker}}(B' \otimes _{A'} B' \to B')$. By Lemmas 49.2.5, 49.6.2, and 49.6.5 the map $J' \to \mathop{\mathrm{Hom}}\nolimits _{B'}(\omega _{B'/A'}, B')$ is gotten by applying the functor $- \otimes _ B B'$ to the map $J \to \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$. Since $B_\mathfrak q \to B'_{\mathfrak q'}$ is faithfully flat it suffices to prove the result for $(A' \to B', \mathfrak q')$. By Lemmas 49.2.7, 49.6.1, and 49.6.5 this reduces us to the case proved in the first paragraph of the proof.
$\square$

Lemma 49.6.7. Let $A \to B$ be a flat quasi-finite homomorphism of Noetherian rings. The diagram

\[ \xymatrix{ J \ar[rr] \ar[rd]_\mu & & \mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B) \ar[ld]^{\varphi \mapsto \varphi (\tau _{B/A})} \\ & B } \]

commutes where the horizontal arrow is the isomorphism of Lemma 49.6.6. Hence the Noether different of $B$ over $A$ is the image of the map $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B) \to B$.

**Proof.**
Exactly as in the proof of Lemma 49.6.6 this reduces to the case of a finite free map $A \to B$. In this case $\tau _{B/A} = \text{Trace}_{B/A}$. Choose a basis $b_1, \ldots , b_ n$ of $B$ as an $A$-module. Let $\xi = \sum b_ i \otimes c_ i \in J$. Then $\mu (\xi ) = \sum b_ i c_ i$. On the other hand, the image of $\xi $ in $\mathop{\mathrm{Hom}}\nolimits _ B(\omega _{B/A}, B)$ sends $\text{Trace}_{B/A}$ to $\sum \text{Trace}_{B/A}(b_ i)c_ i$. Thus we have to show

\[ \sum b_ ic_ i = \sum \text{Trace}_{B/A}(b_ i)c_ i \]

when $\xi = \sum b_ i \otimes c_ i \in J$. Write $b_ i b_ j = \sum _ k a_{ij}^ k b_ k$ for some $a_{ij}^ k \in A$. Then the right hand side is $\sum _{i, j} a_{ij}^ j c_ i$. On the other hand, $\xi \in J$ implies

\[ (b_ j \otimes 1)(\sum \nolimits _ i b_ i \otimes c_ i) = (1 \otimes b_ j)(\sum \nolimits _ i b_ i \otimes c_ i) \]

which implies that $b_ j c_ i = \sum _ k a_{jk}^ i c_ k$. Thus the left hand side is $\sum _{i, j} a_{ij}^ i c_ j$. Since $a_{ij}^ k = a_{ji}^ k$ the equality holds.
$\square$

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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