Download Independence Theorem and Flat Base Change

Independence Theorem and Flat Base Change
Notation 0
Throughout this script, let R and R0 be Noetherian rings, let f : R → R0 be a homomorphism
of rings, and let a ⊆ R be an ideal. If M 0 is an R0 -module, we write M 0 R for the restriction of
scalars by means of f .
Reminder and Exercise 1
A) Let x ∈ R. Let ηa,x : Γa (•) → Γa (•)x R denote the natural transformation of functors of RΓ (M )
modules arising from the canonical morphism ηx a
: Γa (M ) → Γa (M )xR for any R-module M ,
and let ιa,x : Γa+x (•) → Γa (•) denote the natural transformation of functors of R-modules arising
from the inclusion ιM
a,x : Γa+x (M ) → Γa (M ) for any R-module M . Then there is an admissible
triad of functors of R-modules
ιa,x
ηa,x
∆ : Γa+x (•) −→ Γa (•) −→ Γa (•)xR .
B) Let M be an R-module and let i ∈ N. Using ∆ it can be shown that there is a short exact
sequence of R-modules
1
i
0
0 → H(x)
(Hai−1 (M )) → Ha+(x)
(M ) → H(x)
(Hai (M )) → 0.
This short exact sequence is called the i-th comparison sequence of a and x associated to M .
Lemma 2
Let M be an R-module and let x ∈ R. The multiplication map x· : MxR → MxR is bijective.
m
. Then
Proof. Let u ∈ MxR . Then there are m ∈ M, n ∈ N0 such that u = xmn . Define v := xn+1
u = xv, and we see that x· is surjective.
m
Now assume that xu = 0. Then 0 = x · xmn = xn−1
. So there is a t ∈ N0 such that xt m = 0 and
t
m
thus u = xmn = xxn+t
= 0. Therefore x· is injective.
Lemma 3
Let M be an R-module and let x ∈ R. Let ηxM : M → MxR denote the canonical morphism. Then
a) ker(ηxM ) = Γ(x) (M ),
1
b) coker(ηxM ) ∼
(M ).
= H(x)
Proof. a) If m ∈ ker(ηxM ), then there is an integer n ∈ N0 such that xn · m = 0. This is the case
if and only if m ∈ (0 :M (x)n ) ⊆ Γ(x) (M ).
b) We write K := coker(ηxM ). Using a) we get a short exact sequence
0 → M/Γ(x) (M ) → MxR → K → 0.
M
Now let xmn + im(ηxM ) ∈ K. Then we have xn ( xmn + im(ηxM )) = m
1 + im(ηx ) = 0 ∈ K, so that
Γ(x) (K) = K. So, if we apply local cohomology to the above short exact sequence, we get an exact
sequence
1
1
Γ(x) (MxR ) → K → H(x)
(M/Γ(x) (M )) → H(x)
(MxR ).
1
By Lemma 2 the multiplication map x· : MxR → MxR is bijective. But then x· : Γ(x) (MxR ) →
1
1
Γ(x) (MxR ) and x· : H(x)
(MxR ) → H(x)
(MxR ) are bijective, too, and by 3.13 and 3.12 C) b) we
1
get Γ(x) (MxR ) = H(x) (MxR ) = 0. Therefore
1
1
(M ).
(M/Γ(x) (M )) ∼
coker(ηxM ) = K ∼
= H(x)
= H(x)
Proposition 4
Let M be an R-module and let x ∈ R. Then the following conditions are equivalent:
(i) The canonical morphism ηxM : M → MxR is surjective.
(ii) The multiplication map x· : M/Γ(x) (M ) → M/Γ(x) (M ) is surjective.
(iii) The multiplication map x· : M/Γ(x) (M ) → M/Γ(x) (M ) is bijective.
1
(iv) H(x)
(M ) = 0.
Proof. “(i) ⇒ (ii)”: As ker(ηxM ) = Γ(x) (M ), there is an injective morphism ϕ : M/Γ(x) (M ) MxR such that the following diagram commutes:
M
ηx
/ MxR
M KK
8
q
KK
q
ϕ qq
KK
q
KK
q
KK%
q8 qq
%
M/Γ(x) (M ).
As ηxM is surjective so is ϕ, and therefore M/Γ(x) (M ) ∼
= MxR . Now Lemma 2 yields (ii).
“(ii) ⇒ (iii)”: By 3.10 it is enough to show x ∈ NZDR (M/Γ(x) (M )). So let m + Γ(x) (M ) ∈
M/Γ(x) (M ) such that x·(m+M/Γ(x) (M )) = 0. Then xm ∈ Γ(x) (M ). So there is an integer n ∈ N0
such that 0 = xn (xm) = xn+1 m and therefore m ∈ Γ(x) (M ) whence x ∈ NZD(M/Γ(x) (M )).
1
1
“(iii) ⇒ (iv)”: As H(x)
(M ) ∼
(M/Γ(x) (M )), we may replace M by M/Γ(x) (M ) and assume
= H(x)
1
1
that Γ(x) (M ) = 0. As x· : M → M is bijective, we have an isomorphism x· : H(x)
(M ) → H(x)
(M ).
1
1
By 3.13 H(x) (M ) is (x)-torsion, and by 3.12 C) b) we get H(x) (M ) = 0.
“(iv) ⇒ (i)”: By Lemma 3 there is an exact sequence
ηM
x
1
M →
MxR → H(x)
(M ) → 0.
1
As H(x)
(M ) = 0, we see that ηxM is surjective.
Definition 5
A) Let x ∈ R and let M be an R-module. M is said to be x-quasi-divisible if the equivalent
conditions of Proposition 4 hold for M and x.
B) Let S ⊆ R. M is said to be S-quasi-divisible, if M is x-quasi-divisible for any x ∈ S.
C) Finally, M is said to be quasi-divisible, if M is R-quasi-divisible.
Remark 6
1
Let I be an injective R-module and let x ∈ R. By 2.15 b) H(x)
(I) = 0, and therefore injective
modules are quasi-divisible.
Remark and Exercise 7
A) Let n ∈ N0 . Show that (aR0 )n = an R0 .
B) Use A) to show: The functors Γa (•R ) and ΓaR0 (•)R from R0 -modules to R-modules are the
same.
C) Show that each R-module is 0-quasi-divisible.
Lemma 8
Let x ∈ R and let M 0 be an f (x)-quasi-divisible R0 -module. Then M 0R is x-quasi-divisible.
2
Proof. Let m ∈ M 0 . Then by the above Remark and Exercise
x · (m + Γ(x) (M 0R )) = x · (m + ΓxR0 (M 0 )R ) = f (x)m + Γ(f (x)) (M 0 )R .
As f (x)· : M 0 /Γ(f (x)) (M 0 ) → M 0 /Γ(f (x)) (M 0 ) is surjective, so must be x· : M 0R /Γ(x) (M 0R ) →
M 0R /Γ(x) (M 0R ).
Lemma 9
Let I 0 be an injective R0 -module. Then Γa (I 0R ) is a quasi-divisible R-module.
Proof. By 3.14 ΓaR0 (I 0 ) is injective and thus by Remark 6 a quasi-divisible R0 -module. So, by
Lemma 8, ΓaR0 (I 0 ) R is a quasi-divisible R-module, and by Remark and Exercise 7 we have
ΓaR0 (I 0 )R = Γa (I 0R ).
Proposition 10
Assume that f : R → R0 is flat. Let M be an R-module. Then there is a natural transformation
of functors from R-modules to R0 -modules
µ : HomR (•, M ) ⊗R R0 −→ HomR0 (• ⊗R R0 , M ⊗R R0 ),
which maps an R-module N to the morphism µN : HomR (N, M ) ⊗R R0 → HomR0 (N ⊗R R0 , M ⊗R
R0 ) given by
µN (g ⊗ r0 ) = r0 (g ⊗ idR0 ).
Furthermore, if N is a finitely generated R-module, µN is an isomorphism.
Proof. Let N be an R-module. Then g ⊗ r0 7→ r0 (g ⊗ idR0 ) defines a morphism of R-modules
µN : HomR (N, M ) ⊗R R0 → HomR0 (N ⊗R R0 , M ⊗R R0 ).
Now let Ñ be another R-module and let ϕ : N → Ñ be a morphism of R-modules. Then there
are morphisms of R0 -modules
HomR (ϕ, M ) ⊗R R0 : HomR (Ñ , M ) ⊗R R0 → HomR (N, M ) ⊗R R0 ,
given by h 7→ h ◦ (ϕ ⊗R R0 ), and
HomR0 (ϕ ⊗R R0 , M ⊗R R0 ) : HomR0 (Ñ ⊗R R0 , M ⊗R R0 ) → HomR0 (N ⊗R R0 , M ⊗R R0 ),
given by g̃ ⊗ h 7→ (g̃ ◦ ϕ) ⊗ h.
It is left as an exercise to show that
HomR (ϕ ⊗R R0 , M ⊗R R0 ) ◦ µÑ = µN ◦ (HomR (ϕ, M ) ⊗R R0 ).
Now assume that N is finitely generated. As R is Noetherian, N is of finite presentation, meaning
that there is an exact sequence of R-modules
R⊕k → R⊕l → N → 0
for some k, l ∈ N0 . As both HomR (•, M ) ⊗R R0 and HomR0 (• ⊗R R0 , M ⊗R R0 ) are contravariant
and left exact, we get a commutative diagram with exact rows
0
/ HomR (N, M ) ⊗ R0
R
µN
0
R
µR⊕l
/ HomR0 (N ⊗ R0 , M ⊗ R0 )
R
/ HomR (R⊕l , M ) ⊗ R0
R
/ HomR0 (R⊕l ⊗ R0 , M ⊗ R0 )
3
R
µR⊕k
R
/ HomR (R⊕k , M ) ⊗ R0
R
/ HomR0 (R⊕k ⊗ R0 , M ⊗ R0 ).
R
R
Now for any n ∈ N0 there are isomorphisms
∼
=
αn : HomR (R⊕n , M ) ⊗R R0 −→ M ⊕n ⊗R R0 ,
given by h ⊗ r0 7→ (h(e1 ), . . . , h(en )) ⊗ r0 ,
∼
=
βn : HomR0 (R⊕n ⊗R R0 , M ⊗R R0 ) −→ (M ⊗R R0 )⊕n ,
given by h 7→ (h(e1 ⊗ 1R0 ), . . . , h(en ⊗ 1R0 )), and
∼
=
γn : M ⊕n ⊗R R0 −→ (M ⊗R R0 )⊕n ,
given by (m1 , . . . , mn ) ⊗ r0 7→ (m1 ⊗ r0 , . . . , mn ⊗ r0 ). It is easy to check that γn ◦ αn = βn ◦ µR⊕n ,
and therefore µR⊕n is an isomorphism. Now the above commutative diagram yields (ii).
Remark and Exercise 11
Let b ⊆ R be an ideal, and let M be an R-modul. Then there is an isomorphism of R-moduls
∼
=
b
νM
: (0 :M b) −→ HomR (R/b, M ),
given by m 7→ ((x + b) 7→ xm).
Lemma 12
Assume that f : R → R0 is flat. Then there is a natural equivalence of functors from R-modules
to R0 -modules
∼
=
% : Γa (•) ⊗R R0 −→ ΓaR0 (• ⊗R R0 ),
such that for each R-module M we have the commutative diagram
%M
/ ΓaR0 (M ⊗R R0 )
Γa (M ) ⊗R R0
'OOO
nnv
OOO
nnn
n
O
n
OOO
ιM
n 0
'
vnnn ιM
0
M ⊗R R ,
where ι0M is the inclusion ΓaR0 (M ⊗R R0 ) ,→ M ⊗R R0 , and ιM arises from the inclusion Γa (M ) ,→
M.
Proof. Since f is flat, the inclusion map Γa (M ) ,→ M for any R-module M induces a natural
transformation of functors from R-modules to R0 -modules
ι : Γa (•) ⊗R R0 −→ • ⊗R R0
sucht that for all R-modules M the R0 -morphism ιM : Γa (M ) ⊗R R0 M ⊗R R0 is injective.
Now, let M be an R-module, and let m ∈ Γa (M ) and let r0 ∈ R0 . Then there is an integer n ∈ N0
such that an m = 0. By Remark and Exercise 7 A) we get
(aR0 )n (m ⊗ r0 ) = (an R0 )(m ⊗ r0 ) = (an m) ⊗ r0 = 0.
So we have ιM (Γa (M ) ⊗R R0 ) ⊆ ΓaR0 (M ⊗R R0 ) for every R-module M , and therefore ι induces
a natural transformation
% : Γa (•) ⊗R R0 −→ ΓaR0 (• ⊗R R0 ),
such that for all R-modules M the morphism %M : Γa (M ) ⊗R R0 −→ ΓaR0 (M ⊗R R0 ) is injective
and for all m ∈ Γa (M ), r0 ∈ R0 it holds %M (m ⊗ r0 ) = m ⊗ r0 .
Let us show that %M is surjective for any R-module M . For this, let n ∈ N0 . Since R/an is finitely
generated, by Proposition 10 there is an isomorphism of R0 -modules
µR/an : HomR (R/an , M ) ⊗R R0 −→ HomR0 (R/an ⊗R R0 , M ⊗R R0 ).
4
Using the above Remark and Exercise we get a chain of isomorphisms of R0 -modules
n
(0 :M an ) ⊗R R0
a
νM
⊗R R0
−→
−→
∼
=
−→
µR/an
(aR0 )n −1
0)
RR
(νM ⊗
−→
HomR (R/an , M ) ⊗R R0
HomR0 (R/an ⊗R R0 , M ⊗R R0 )
HomR0 (R0 /(aR0 )n , M ⊗R R0 )
(0 :M ⊗R R0 (aR0 )n ).
Let εn denote the composition of these isomorphisms. For simplicity’s sake we write ι for any
inclusion if source and target are obvious. Then it is easy to check that we have the following
commutative diagram:
εn
/ (0 :M ⊗ R0 (aR0 )n )
(0 :M an ) ⊗R R0
R
P
(
mv
PPP
ιmmmm
PPP
m
m
0 PPP
m
ι⊗R R
P(
vmmm
ι
ι⊗R R0
M ⊗R RhQ0
QQQ 0
n6
n
n
ι
QQQM
nnn
QQQ
nnnιM
n
Qh
6
n
/ ΓaR0 (M ⊗R R0 )
Γa (M ) ⊗R R0
%M
Now, let u ∈ ΓaR0 (M ⊗R R0 ). Then there exists a n ∈ N0 such that u ∈ (0 :M ⊗R R0 (aR0 )n ), and
0
therefore we can define y := (ι ⊗R R0 ) ◦ ε−1
n (u) ∈ Γa (M ) ⊗R R . Then %M (y) = u whence %M is
surjective.
Finally, the above commutative diagram yields the commutative diagram in the claim, too.
Lemma 13
Assume that f : R → R0 is flat. Let S ⊆ R, and let M be an S-quasi-divisible R-module. Then
M ⊗R R0 is f (S)-quasi-divisible.
Proof. Let x ∈ S. We have to show that the multiplication map
f (x)· : (M ⊗R R0 )/Γ(f (x)) (M ⊗R R0 ) −→ (M ⊗R R0 )/Γ(f (x)) (M ⊗R R0 )
is surjective. By Lemma 12, we have the following equality of submodules of M ⊗R R0
Γ(f (x)) (M ⊗R R0 ) = ΓxR0 (M ⊗R R0 ) = Γ(f (x)) (M ) ⊗R R0 .
As f is flat, there is a short exact sequence
0 → Γ(x) (M ) ⊗R R0 → M ⊗R R0 → (M/Γ(x) (M )) ⊗R R0 → 0.
It follows by means of Lemma 12
(M/Γ(x) (M )) ⊗R R0 ∼
= (M ⊗R R0 )/(Γ(x) (M ) ⊗R R0 ) ∼
= (M ⊗R R0 )/Γ(x) (M ⊗R R0 ).
Thus it suffices to show that the multiplication map
f (x)· : (M/Γ(x) (M )) ⊗R R0 −→ (M/Γ(x) (M )) ⊗R R0
is surjective. Now, for every m ∈ M/Γ(x) (M ) and every r0 ∈ R0
f (x)(m ⊗ r0 ) = (xm) ⊗ r0 = ((x·) ⊗R R0 )(m ⊗ r0 ).
Since M is S-quasi-divisible, the multiplication map x· : M/Γ(x) (M ) → M/Γ(x) (M ) is surjective,
and as the tensor product functor is right exact, f (x)· = (x·) ⊗R R0 : (M/Γ(x) (M )) ⊗R R0 →
(M/Γ(x) (M )) ⊗R R0 is surjective, too.
5
Definition 14
Let F be an additive functor from R-modules to R0 -modules. An R-module M is said to be
F-acyclic, if Ri F (M ) = 0 for all i > 0.
Lemma 15
Let M be an R-module which is Γa -acyclic, and let x ∈ R such that Γa (M ) is x-quasi-divisible.
Then M is Γa+(x) ()-acyclic.
Proof. Let i ∈ N. Then by Reminder and Exercise 1 there is a short exact sequence
1
i
0
(Hai−1 (M )) → Ha+(x)
(M ) → H(x)
(Hai (M )) → 0.
0 → H(x)
As Hai (M ) = 0, we get an isomorphism
1
i
H(x)
(Hai−1 (M )) ∼
(M ).
= Ha+(x)
1
For i > 1, we have Hai−1 (M ) = 0, and since Γa (M ) is x-quasi-divisible, it holds H(x)
(Ha0 (M )) =
1
i
H(x)
(Γa (M )) = 0. So Ha+(x)
(M ) = 0 for all i > 0.
Lemma 16
Let M be a Γa -acyclic R-module. Let t ∈ N0 and let x1 , . . . , xt ∈ R such that Γa+(x1 ,...,xk−1 ) (M )
is xk -quasi-divisible for any k ∈ {1, . . . , t}. Then M is Γa+(x1 ,...,xk ) -acyclic for any k ∈ {0, . . . , t}.
Proof. We use induction on k. For k = 0 the claim is trivial. So let k > 0. By induction
M is Γa+(x1 ,...,xk−1 ) -acyclic, and as Γa+(x1 ,...,xk−1 ) (M ) is xk -quasi-divisible, Lemma 15 yields our
claim.
Definition and Exercise 17
A) Let S ⊆ R. a is said to be an S-ideal if a can be generated by elements of S.
B) Show that every S-ideal can be generated by finitely many elements of S.
Proposition 18
Let S ⊆ R with 0 ∈ S, and let M be an R-module such that Γb (M ) is S-quasi-divisible for every
S-ideal b ⊆ R. Then M is Γb -acyclic for every S-ideal b ⊆ R.
Proof. Let b be an S-ideal, and let x1 , . . . , xt ∈ S such that b = (x1 , . . . , xt ). As Γ0 (•) = id(•), it
holds H0i (M ) = 0 for all i > 0 and M is Γ0 -acyclic. Therefore and by assumption Γ(x1 ,...,xk−1 ) (M )
is xk -quasi-divisible for all k ∈ {1, . . . , t}. By Lemma 16 M is (Γ(x1 ,...,xt ) = Γb )-acyclic.
Corollary 19
Let M be an R-module such that Γb (M ) is quasi-divisible for every ideal b ⊆ R. Then M is
Γb -acyclic for every ideal b ⊆ R.
Corollary 20
Let I 0 be an injective R0 -module. Then I 0R is Γa -acyclic.
Proof. By Proposition 9 Γb (I 0R ) is quasi-divisible for every ideal b ⊆ R. Our claim follows now
directly by the above Corollary.
Proposition 21
Assume once more that f : R → R0 is flat. Let S ⊆ R with 0 ∈ S, and let M be an R-module such
that Γb (M ) is S-quasi-divisible for every S-ideal b ⊆ R. Then M ⊗R R0 is Γb0 -acyclic for every
f (S)-ideal b0 ⊆ R0 .
Proof. Let c ⊆ R be an S-ideal. Then by Lemma 12 ΓcR0 (M ⊗R R0 ) ∼
= Γc (M ) ⊗R R0 , and by
0
Lemma 13 Γc (M ) ⊗R R is f (S)-quasi-divisible. Since for every f (S)-ideal b0 there is an S-ideal b
such that bR0 = b0 , we see that Γb0 (M ⊗R R0 ) is f (S)-quasi-divisible for every f (S)-ideal b0 ⊆ R.
Therefore, by Proposition 18 we get our claim.
6
Corollary 22
Assume that f : R → R0 is flat. Let I be an injective R-module. Then I ⊗R R0 is ΓaR0 -acyclic.
Proof. By 3.14 Γa (I) is injective, and by Lemma 9 Γb (I) is quasi-divisible for every ideal b ⊆ R.
Thus, by the above Proposition, I ⊗R R0 is ΓbR0 -acyclic for every ideal b ⊆ R.
Remark and Definition 23
A) Let F be an additive functor from R-modules to R0 -modules, and let M be an R-module. A
right resolution ((E • , d• ); a) is said to be an F -acyclic resolution of M if E i is F -acyclic for all
i ∈ N0 .
B) By Corollary 20 injective modules are Γa -acyclic. So any R-module has a Γa -acyclic resolution.
Proposition 24
Let F be a left exact additive functor from R-modules to R0 -modules, let M be an R-module and
let ((E • , d• ); a) be an F -acyclic resolution of M . Then for all i ∈ N0
Ri F (M ) ∼
= H i (F (E • ), F (d• )).
Proof. We write d−1 := a : M → E 0 . First we proof by induction on j that for all n ∈ N and for
all j ∈ N0 there is an isomorphism Rn F (im(dj−1 )) ∼
= Rn+j F (M ). So, let j = 0, and fix n ∈ N.
−1
−1 ∼
As d = a is a monomorphism, we have im(d ) = M .
Now let j > 0. Consider the short exact sequence
0 → im(dj−2 ) → E j−1 → im(dj−1 ) → 0
and the associated long exact sequence
0 → R0 F (im(dj−2 )) → R0 F (E j−1 )
→ R0 F (im(dj−1 ))
1
j−2
→ R F (im(d )) → · · ·
→ Rn F (E j−1 )
n
j−1
n+1
j−2
→ R F (im(d )) → R
F (im(d )) → Rn+1 F (E j−1 )
→
··· .
Since E j−1 is F -acyclic, we have Rn F (E j−1 ) = 0 for all n > 0. This and the induction hypothesis
yield for all n > 0
Rn F (im(dj−1 )) ∼
= Rn+1 F (im(dj−2 )) ∼
= Rn+1+(j−1) F (M ) = Rn+j F (M ).
Because d−1 : M → E 0 is a monomorphism and F is left exact, we have by 3.3 C)
R0 F (M ) = F (M ) ∼
= F (im(d−1 )) ∼
= F (ker(d0 )) ∼
= ker(F (d0 )) = H 0 (F (E • ), F (d• )).
Now let i > 0. Then we have a short exact sequence
i−1
d
0 −→ im(di−2 ) −→ E i−1 −→ ker(di ) −→ 0,
i−1
where d
sequence
: E i−1 → ker(di ), e 7→ di−1 (e). This short exact sequence gives rise to the long exact
i−1
F (d
)
0 −→ F (im(di−2 )) −→ F (E i−1 ) −→ F (ker(di )) −→ R1 F (im(di−2 )) −→ R1 F (E i−1 ) = 0,
and therefore
H i (F (E • ), F (d• ))
∼ F (ker(di ))/ im(F (di−1 ))
= ker(F (di ))/ im(F (di−1 )) =
∼
∼
= R1 F (im(di−2 ))
= R1+(i−1) F (M )
7
= Ri F (M ).
Remark 25
By the previous Proposition, we can compute local cohomology using a Γa -acyclic resolution
instead of an injective resolution; i.e. if ((E • , d• ); a) is any Γa -cyclic resolution of an R-module
M , we have for all i ∈ N0
Hai (M ) ∼
= H i (Γa (E • ), Γa (d• )).
Theorem 26 (Independence Theorem)
Let M 0 be an R0 -module. Then for all i ∈ N0
i
0
0
∼ i
HaR
0 (M )R = Ha (M R ).
Proof. Let
d00
d0n
I 0• : 0 → I 00 → I 01 → · · · → I 0n → I 0n+1 → · · ·
be an injective resolving cocomplex of M 0 over R0 . Then by Corollary 20
d00R
d0nR
I 0•R : 0 → I 00R → I 01R → · · · → I 0nR → I 0n+1R → · · ·
is a Γa -acyclic resolving cocomplex of M 0 R . With Propositon 24 and Remark and Exercise 7 it
follows for all i ∈ N0
Hai (M 0R ) ∼
= ker(Γa (d0iR ))/ im(Γa (d0i−1R ))
= ker(ΓaR0 (d0i ))R / im(ΓaR0 (d0i−1 ))R
= ker(ΓaR0 (d0i )R )/ im(ΓaR0 (d0i−1 )R )
0
i
∼
= HaR
0 (M )R .
Theorem 27 (Flat Base Change)
Assume that f : R → R0 is flat, and let M be an R-module. Then, for each i ∈ N0 , there is an
isomorphism of R0 -modules
i
0
Hai (M ) ⊗R R0 ∼
0 (M ⊗R R ).
= HaR
Proof. Let i ∈ N0 , and let
d0
dn
I • : 0 → I 0 → I 1 → · · · → I n → I n+1 → · · ·
be an injective resolving cocomplex of M . As f is flat, there is a short exact sequence
0 → im(Γa (di−1 )) ⊗R R0 → ker(Γa (di )) ⊗R R0 → (ker(Γa (di ))/ im(Γa (di−1 ))) ⊗R R0 → 0,
and therefore
Hai (M ) ⊗R R0
∼
=
∼
=
∼
=
=
∼
=
H i (Γa (I • )) ⊗R R0
(ker(Γa (di )) ⊗R R0 )/(im(Γa (di−1 )) ⊗R R0 )
ker(Γa (di ) ⊗R R0 )/ im(Γa (di−1 ) ⊗R R0 )
H i (Γa (I • ) ⊗R R0 )
H i (Γa (I • ⊗R R0 )).
The last isomorphism in the above formula is due to Lemma 12. By Corollary 22 I • ⊗R R0 is a
ΓaR0 -acyclic resolution of M ⊗R R0 , and therefore by Proposition 24 we get H i (Γa (I • ⊗R R0 )) ∼
=
i
0
HaR
0 (M ⊗R R ).
Remark and Exercise 28
The isomorphisms in Theorem 26 and Theorem 27 are natural, i.e.
a) for all i ∈ N0 , there is a natural equivalence of functors from R0 -modules to R-modules
i
i
HaR
0 (•)R → Ha (•R );
0
b) if f : R → R is flat, then for all i ∈ N0 there is a natural equivalence of functors from R-modules
i
0
to R0 -modules Hai (•) ⊗R R0 → HaR
0 (• ⊗R R ).
8