Hispanopedia Wiki
Articulo de referencia

Amitsur complex

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 f...

Hispanopedia WikiContenido en espanolLectura gratuita

In algebra, the Amitsur complex is a natural complex associated to a ring homomorphism. It was introduced by ShimshonAmitsur (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} be a homomorphism of (not-necessary-commutative) rings. First define the cosimplicial setC∙=S⊗∙+1{\displaystyle C^{\bullet }=S^{\otimes \bullet +1}} (where ⊗{\displaystyle \otimes } refers to ⊗R{\displaystyle \otimes _{R}}, not ⊗Z{\displaystyle \otimes _{\mathbb {Z} }}) as follows. Define the face maps di:S⊗n+1→S⊗n+2{\displaystyle d^{i}:S^{\otimes {n+1}}\to S^{\otimes n+2}} by inserting 1{\displaystyle 1} at the i{\displaystyle i}th spot:[a]

di(x0⊗⋯⊗xn)=x0⊗⋯⊗xi−1⊗1⊗xi⊗⋯⊗xn.{\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 si:S⊗n+1→S⊗n{\displaystyle s^{i}:S^{\otimes n+1}\to S^{\otimes n}} by multiplying out the i{\displaystyle i}th and (i+1){\displaystyle (i+1)}th spots:

si(x0⊗⋯⊗xn)=x0⊗⋯⊗xixi+1⊗⋯⊗xn.{\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}} is a cosimplicial set. It then determines the complex with the augumentation θ{\displaystyle \theta }, the Amitsur complex:[2]

0→R→θS→δ0S⊗2→δ1S⊗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 }

where δn=∑i=0n+1(−1)idi.{\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 } is right faithfully flat, then a theorem of Alexander Grothendieck states that the (augmented) complex 0→R→θS⊗∙+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 } is right faithfully flat, then, for each left R{\displaystyle R}-module M{\displaystyle M},

0→M→S⊗RM→S⊗2⊗RM→S⊗3⊗RM→⋯{\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} splits as a ring homomorphism.

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

by

h(x0⊗m)=ρ(x0)⊗m,h(x0⊗⋯⊗xn⊗m)=θ(ρ(x0))x1⊗⋯⊗xn⊗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}}}

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

h∘δn+δn−1∘h=idS⊗n+1⊗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} is a homotopy operator and so idS⊗n+1⊗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⊗RS,x↦1⊗x{\displaystyle S\to T:=S\otimes _{R}S,\,x\mapsto 1\otimes x} is a section of T→S,x⊗y↦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} implies:

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

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

Arc topology case

BhargavBhattandPeter Scholze (2019,§8) show that the Amitsur complex is exact if R{\displaystyle R} and 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 s0{\displaystyle s_{0}} and d2{\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