Amitsur complex

From HandWiki

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by Shimshon Amitsur (1959). When the homomorphism is faithfully flat, the Amitsur complex is exact (thus determining a resolution), which is the basis of the theory of faithfully flat descent. The notion should be thought of as a mechanism to go beyond the conventional localization of rings and modules.[1]


Let [math]\displaystyle{ \theta\colon R \to S }[/math] be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial set [math]\displaystyle{ C^\bullet = S^{\otimes \bullet+1} }[/math] (where [math]\displaystyle{ \otimes }[/math] refers to [math]\displaystyle{ \otimes_R }[/math], not [math]\displaystyle{ \otimes_{\Z} }[/math]) as follows. Define the face maps [math]\displaystyle{ d^i\colon S^{\otimes {n+1}} \to S^{\otimes n+2} }[/math] by inserting 1 at the i-th spot:[lower-alpha 1]

[math]\displaystyle{ d^i(x_0 \otimes \cdots \otimes x_n) = x_0 \otimes \cdots \otimes x_{i-1} \otimes 1 \otimes x_i \otimes \cdots \otimes x_n. }[/math]

Define the degeneracies [math]\displaystyle{ s^i\colon S^{\otimes n+1} \to S^{\otimes n} }[/math] by multiplying out the i-th and (i + 1)-th spots:

[math]\displaystyle{ s^i(x_0 \otimes \cdots \otimes x_n) = x_0 \otimes \cdots \otimes x_i x_{i+1} \otimes \cdots \otimes x_n. }[/math]

They satisfy the "obvious" cosimplicial identities and thus [math]\displaystyle{ S^{\otimes \bullet + 1} }[/math] is a cosimplicial set. It then determines the complex with the augumentation [math]\displaystyle{ \theta }[/math], the Amitsur complex:[2]

[math]\displaystyle{ 0 \to R \,\overset{\theta}\to\, S \,\overset{\delta^0}\to\, S^{\otimes 2} \,\overset{\delta^1}\to\, S^{\otimes 3} \to \cdots }[/math]

where [math]\displaystyle{ \delta^n = \sum_{i=0}^{n+1} (-1)^i d^i. }[/math]

Exactness of the Amitsur complex

Faithfully flat case

In the above notations, if [math]\displaystyle{ \theta }[/math] is right faithfully flat, then a theorem of Alexander Grothendieck states that the (augmented) complex [math]\displaystyle{ 0 \to R \overset{\theta}\to S^{\otimes \bullet + 1} }[/math] is exact and thus is a resolution. More generally, if [math]\displaystyle{ \theta }[/math] is right faithfully flat, then, for each left R-module M,

[math]\displaystyle{ 0 \to M \to S \otimes_R M \to S^{\otimes 2} \otimes_R M \to S^{\otimes 3} \otimes_R M \to \cdots }[/math]

is exact.[3]


Step 1: The statement is true if [math]\displaystyle{ \theta\colon R \to S }[/math] splits as a ring homomorphism.

That "[math]\displaystyle{ \theta }[/math] splits" is to say [math]\displaystyle{ \rho \circ \theta = \operatorname{id}_R }[/math] for some homomorphism [math]\displaystyle{ \rho\colon S \to R }[/math] ([math]\displaystyle{ \rho }[/math] is a retraction and [math]\displaystyle{ \theta }[/math] a section). Given such a [math]\displaystyle{ \rho }[/math], define

[math]\displaystyle{ h\colon S^{\otimes n+1} \otimes M \to S^{\otimes n} \otimes M }[/math]


[math]\displaystyle{ \begin{align} & h(x_0 \otimes m) = \rho(x_0) \otimes m, \\ & h(x_0 \otimes \cdots \otimes x_n \otimes m) = \theta(\rho(x_0)) x_1 \otimes \cdots \otimes x_n \otimes m. \end{align} }[/math]

An easy computation shows the following identity: with [math]\displaystyle{ \delta^{-1}=\theta \otimes \operatorname{id}_M\colon M \to S \otimes_R M }[/math],

[math]\displaystyle{ h \circ \delta^n + \delta^{n-1} \circ h = \operatorname{id}_{S^{\otimes n+1} \otimes M} }[/math].

This is to say that h is a homotopy operator and so [math]\displaystyle{ \operatorname{id}_{S^{\otimes n+1} \otimes M} }[/math] determines the zero map on cohomology: i.e., the complex is exact.

Step 2: The statement is true in general.

We remark that [math]\displaystyle{ S \to T := S \otimes_R S, \, x \mapsto 1 \otimes x }[/math] is a section of [math]\displaystyle{ T \to S, \, x \otimes y \mapsto xy }[/math]. Thus, Step 1 applied to the split ring homomorphism [math]\displaystyle{ S \to T }[/math] implies:

[math]\displaystyle{ 0 \to M_S \to T \otimes_S M_S \to T^{\otimes 2} \otimes_S M_S \to \cdots, }[/math]

where [math]\displaystyle{ M_S = S \otimes_R M }[/math], is exact. Since [math]\displaystyle{ T \otimes_S M_S \simeq S^{\otimes 2} \otimes_R M }[/math], etc., by "faithfully flat", the original sequence is exact. [math]\displaystyle{ \square }[/math]

The case of the arc topology

Bhargav Bhatt and Peter Scholze (2019, §8) show that the Amitsur complex is exact if R and S are (commutative) perfect rings, and the map is required to be a covering in the arc topology (which is a weaker condition than being a cover in the flat topology).


  1. Note the reference (M. Artin) seems to have a typo, and this should be the correct formula; see the calculation of [math]\displaystyle{ s_0 }[/math] and [math]\displaystyle{ d^2 }[/math] in the note.


  1. Artin 1999, III.7.
  2. Artin 1999, III.6.
  3. Artin 1999, Theorem III.6.6.