- Source: 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 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.
Definition
Let
θ
:
R
→
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}}
(where
⊗
{\displaystyle \otimes }
refers to
⊗
R
{\displaystyle \otimes _{R}}
, not
⊗
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}}
by inserting
1
{\displaystyle 1}
at the
i
{\displaystyle i}
th spot:
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}.}
Define the degeneracies
s
i
:
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:
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}.}
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:
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 }
where
δ
n
=
∑
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 }
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}}
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
⊗
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 }
is exact.
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
=
id
R
{\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
(
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}}}
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}
,
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}}
.
This is to say that
h
{\displaystyle h}
is a homotopy operator and so
id
S
⊗
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
⊗
R
S
,
x
↦
1
⊗
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}
. Thus, Step 1 applied to the split ring homomorphism
S
→
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 ,}
where
M
S
=
S
⊗
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}
, etc., by "faithfully flat", the original sequence is exact.
◻
{\displaystyle \square }
= Arc topology case
=Bhargav Bhatt and Peter 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).