The Stacks project

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 $B = B_1 \times B_2$.

  1. 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 $\mathop{\mathrm{Ker}}(B_ i \otimes _ A B_ i \to B_ i)$.

  2. 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 $B' = B \otimes _ A A'$.

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

  2. 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.

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

  2. 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). $\square$

Remark 49.6.4. Let $A \to B$ be a quasi-finite homomorphism of Noetherian rings. Let $J$ be the annihilator of $\mathop{\mathrm{Ker}}(B \otimes _ A B \to B)$. There is a canonical $B$-bilinear pairing

49.6.4.1
\begin{equation} \label{discriminant-equation-pairing-noether} \omega _{B/A} \times J \longrightarrow B \end{equation}

defined as follows. Choose a factorization $A \to B' \to B$ with $A \to B'$ finite and $B' \to B$ inducing an open immersion of spectra. Let $J'$ be the annihilator of $\mathop{\mathrm{Ker}}(B' \otimes _ A B' \to B')$. We first define

\[ \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \times J' \longrightarrow B',\quad (\lambda , \sum b_ i \otimes c_ i) \longmapsto \sum \lambda (b_ i)c_ i \]

This is $B'$-bilinear exactly because for $\xi \in J'$ and $b \in B'$ we have $(b \otimes 1)\xi = (1 \otimes b)\xi $. By Lemma 49.6.3 and the fact that $\omega _{B/A} = \mathop{\mathrm{Hom}}\nolimits _ A(B', A) \otimes _{B'} B$ we can extend this to a $B$-bilinear pairing as displayed above.

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

  1. 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).

  2. 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 ├ętale 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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