Amitsur complex

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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]

Definition

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

d i ( x 0 ⊗ ⋯ ⊗ x n ) = x 0 ⊗ ⋯ ⊗ x i − 1 ⊗ 1 ⊗ x i ⊗ ⋯ ⊗ x n . {\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}.} {\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}.}

Define the degeneracies s i : S ⊗ n + 1 → S ⊗ n {\displaystyle s^{i}:S^{\otimes n+1}\to S^{\otimes n}} {\displaystyle s^{i}:S^{\otimes n+1}\to S^{\otimes n}} by multiplying out the i {\displaystyle i} {\displaystyle i}th and ( i + 1 ) {\displaystyle (i+1)} {\displaystyle (i+1)}th spots:

s i ( x 0 ⊗ ⋯ ⊗ x n ) = x 0 ⊗ ⋯ ⊗ x i x i + 1 ⊗ ⋯ ⊗ x n . {\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}.} {\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}.}

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

0 → R → θ S → δ 0 S ⊗ 2 → δ 1 S ⊗ 3 → ⋯ {\displaystyle 0\to R\,{\overset {\theta }{\to }}\,S\,{\overset {\delta ^{0}}{\to }}\,S^{\otimes 2}\,{\overset {\delta ^{1}}{\to }}\,S^{\otimes 3}\to \cdots } {\displaystyle 0\to R\,{\overset {\theta }{\to }}\,S\,{\overset {\delta ^{0}}{\to }}\,S^{\otimes 2}\,{\overset {\delta ^{1}}{\to }}\,S^{\otimes 3}\to \cdots }

where δ n = ∑ i = 0 n + 1 ( − 1 ) i d i . {\displaystyle \delta ^{n}=\sum _{i=0}^{n+1}(-1)^{i}d^{i}.} {\displaystyle \delta ^{n}=\sum _{i=0}^{n+1}(-1)^{i}d^{i}.}

Exactness of the Amitsur complex

Faithfully flat case

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

0 → M → S ⊗ R M → S ⊗ 2 ⊗ R M → S ⊗ 3 ⊗ R M → ⋯ {\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 } {\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 }

is exact.[3]

Proof:

Step 1: The statement is true if θ : R → S {\displaystyle \theta :R\to S} {\displaystyle \theta :R\to S} splits as a ring homomorphism.

That " θ {\displaystyle \theta } {\displaystyle \theta } splits" is to say ρ ∘ θ = id R {\displaystyle \rho \circ \theta =\operatorname {id} _{R}} {\displaystyle \rho \circ \theta =\operatorname {id} _{R}} for some homomorphism ρ : S → R {\displaystyle \rho :S\to R} {\displaystyle \rho :S\to R} ( ρ {\displaystyle \rho } {\displaystyle \rho } is a retraction and θ {\displaystyle \theta } {\displaystyle \theta } a section). Given such a ρ {\displaystyle \rho } {\displaystyle \rho }, define

h : S ⊗ n + 1 ⊗ M → S ⊗ n ⊗ M {\displaystyle h:S^{\otimes n+1}\otimes M\to S^{\otimes n}\otimes M} {\displaystyle h:S^{\otimes n+1}\otimes M\to S^{\otimes n}\otimes M}

by

h ( x 0 ⊗ m ) = ρ ( x 0 ) ⊗ m , h ( x 0 ⊗ ⋯ ⊗ x n ⊗ m ) = θ ( ρ ( x 0 ) ) x 1 ⊗ ⋯ ⊗ x n ⊗ m . {\displaystyle {\begin{aligned}&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{aligned}}} {\displaystyle {\begin{aligned}&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{aligned}}}

An easy computation shows the following identity: with δ − 1 = θ ⊗ id M : M → S ⊗ R M {\displaystyle \delta ^{-1}=\theta \otimes \operatorname {id} _{M}:M\to S\otimes _{R}M} {\displaystyle \delta ^{-1}=\theta \otimes \operatorname {id} _{M}:M\to S\otimes _{R}M},

h ∘ δ n + δ n − 1 ∘ h = id S ⊗ n + 1 ⊗ M {\displaystyle h\circ \delta ^{n}+\delta ^{n-1}\circ h=\operatorname {id} _{S^{\otimes n+1}\otimes M}} {\displaystyle h\circ \delta ^{n}+\delta ^{n-1}\circ h=\operatorname {id} _{S^{\otimes n+1}\otimes M}}.

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

Step 2: The statement is true in general.

We remark that S → T := S ⊗ R S , x ↦ 1 ⊗ x {\displaystyle S\to T:=S\otimes _{R}S,\,x\mapsto 1\otimes x} {\displaystyle S\to T:=S\otimes _{R}S,\,x\mapsto 1\otimes x} is a section of T → S , x ⊗ y ↦ x y {\displaystyle T\to S,\,x\otimes y\mapsto xy} {\displaystyle T\to S,\,x\otimes y\mapsto xy}. Thus, Step 1 applied to the split ring homomorphism S → T {\displaystyle S\to T} {\displaystyle S\to T} implies:

0 → M S → T ⊗ S M S → T ⊗ 2 ⊗ S M S → ⋯ , {\displaystyle 0\to M_{S}\to T\otimes _{S}M_{S}\to T^{\otimes 2}\otimes _{S}M_{S}\to \cdots ,} {\displaystyle 0\to M_{S}\to T\otimes _{S}M_{S}\to T^{\otimes 2}\otimes _{S}M_{S}\to \cdots ,}

where M S = S ⊗ R M {\displaystyle M_{S}=S\otimes _{R}M} {\displaystyle M_{S}=S\otimes _{R}M}, is exact. Since T ⊗ S M S ≃ S ⊗ 2 ⊗ R M {\displaystyle T\otimes _{S}M_{S}\simeq S^{\otimes 2}\otimes _{R}M} {\displaystyle T\otimes _{S}M_{S}\simeq S^{\otimes 2}\otimes _{R}M}, etc., by "faithfully flat", the original sequence is exact. ◻ {\displaystyle \square } {\displaystyle \square }

Arc topology case

Bhargav Bhatt and Peter Scholze (2019, §8) show that the Amitsur complex is exact if R {\displaystyle R} {\displaystyle R} and S {\displaystyle S} {\displaystyle 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).

Notes

  1. The reference (M. Artin) seems to have a typo, and this should be the correct formula; see the calculation of s 0 {\displaystyle s_{0}} {\displaystyle s_{0}} and d 2 {\displaystyle d^{2}} {\displaystyle d^{2}} in the note.

Citations

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

References