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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