§7. Study of base change in the covariant homological functors of modules

7.1. Functors of $A$-modules

7.1.1.

Given a ring $A$ (not necessarily commutative), we shall denote by $A{b}_A$ the category of left $A$-modules, and shall denote simply by Ab the category of $Z$-modules, identical with commutative groups. Let $T:A{b}_A\to Ab$ be a covariant additive functor, and let $M$ be an $(A,A)$-bimodule; $T(M)$ is then naturally equipped with a structure of right $A$-module. Indeed, for every $a\in A$, let us denote by $h_{a,M}$ (or simply $h_a$) the endomorphism $x\mapsto xa$ of the left $A$-module $M$. By hypothesis, $T(h_a)$ is an endomorphism of the $Z$-module $T(M)$; moreover, since $T$ is a covariant additive functor, we have, for $a\in A$, $b\in A$,

  T(h_{ab}) = T(h_b ∘ h_a) = T(h_b) ∘ T(h_a)   and   T(h_{a+b}) = T(h_a + h_b) = T(h_a) + T(h_b);

this proves that the map $(a,y)\mapsto T(h_a)(y)$ is an external composition law of right $A$-module on $T(M)$. In particular, $T(A_s)$ is a right $A$-module.

7.1.2.

When $A$ is a commutative ring, it follows from (7.1.1) that for every $A$-module $M$, $T(M)$ is naturally equipped with a structure of $A$-module; moreover, if $u:M\to N$ is a homomorphism of $A$-modules, we have, for every $a\in A$, $u\circ h_{a,M}=h_{a,N}\circ u$, whence $T(u)\circ T(h_{a,M})=T(h_{a,N})\circ T(u)$, which proves that $T(u):T(M)\to T(N)$ is a homomorphism of $A$-modules; we thus see that $T$ can be considered as a covariant additive functor from the category $A{b}_A$ into itself. More precisely, we have thereby defined a canonical equivalence between the category of covariant additive functors $A{b}_A\to Ab$ and the category of covariant $A$-linear functors $T:A{b}_A\to A{b}_A$, that is, those such that $T(h_{a,M})=h_{a,T(M)}$ for every $a\in A$. Since the inclusion functor $I:A{b}_A\to Ab$, sending each $A$-module to its underlying $Z$-module, is exact and faithful, the exactness properties of the two functors associated by the preceding equivalence are the same.

7.1.3.

The ring $A$ still being assumed commutative, let $B$ be an $A$-algebra (not necessarily commutative), and let $\rho :A\to B$ be the ring homomorphism corresponding to this algebra structure; this homomorphism defines a covariant

additive functor ${\rho }_{\ast }:M\mapsto M_{[\rho ]}$ from the category $A{b}_B$ of left $B$-modules into the category $A{b}_A$ of $A$-modules. By composition, we deduce a functor $T_{(B)}:A{b}_B{\to }^{{\rho }_{\ast }}A{b}_A{\to }^{T}Ab$, evidently covariant and additive, which we shall also denote by $T^{(B)}$ (for typographical reasons) or $T{\otimes }_{A}B$, and which we shall say is obtained from $T$ by extension of scalars from $A$ to $B$. Of course, if $B$ is commutative, one can consider $T_{(B)}$ as a functor from $A{b}_B$ into itself (7.1.2). When $B$ is commutative and $C$ a $B$-algebra, one sees at once that $T_{(C)}=(T_{(B)}{)}_{(C)}$; it is immediate that the extension of scalars is functorial and additive in $T$; moreover, when $T$ commutes with inductive limits or with direct sums (resp. is left exact, right exact, exact), the same is true of $T_{(B)}$: indeed, ${\rho }_{\ast }$ is exact and commutes with inductive limits and direct sums.

7.1.4.

Suppose still that $A$ is commutative, and let $T$ be an $A$-linear covariant additive functor $A{b}_A\to A{b}_A$, commuting with inductive limits. Then, for every multiplicative subset $S$ of $A$ and every $A$-module $M$, we have a canonical functorial isomorphism of $A$-modules

  T(S⁻¹ M) ⥲ S⁻¹ T(M).                                                       (7.1.4.1)

Suppose first that $S$ is the set of powers $f^n$ ($n\ge 0$) of an element $f\in A$. We know then that $M_f=\lim \to M_n$, where $(M_n,{\varphi }_{nm})$ is the inductive system of $A$-modules $M_n=M$, with ${\varphi }_{nm}:z\mapsto f^{n-m}z$ $(0_I,\mathrm{1.6.1})$; whence in this case the isomorphism (7.1.4.1) by virtue of the hypothesis on $T$. If next $S$ is arbitrary, $S^{-1}M$ is the inductive limit of the $M_f$ for $f\in S$ $(0_I,\mathrm{1.4.5})$, and one concludes in the same way. Moreover, the functoriality of the isomorphism (7.1.4.1) shows that it is an isomorphism of $S^{-1}A$-modules, and that one can therefore write, up to a canonical isomorphism,

  T_{(S⁻¹ A)}(S⁻¹ M) = S⁻¹ T(M) = T(S⁻¹ M).                                   (7.1.4.2)

When $S=A-\mathfrak{p}$ is the complement of a prime ideal $\mathfrak{p}$ of $A$, one writes $T_{\mathfrak{p}}$ instead of $T_{(A_{\mathfrak{p}})}$.

Proposition (7.1.5).

Under the hypotheses of (7.1.4), if $T_{\mathfrak{m}}$ is left exact (resp. right exact, exact) for every maximal ideal $\mathfrak{m}$ of $A$, then $T$ is left exact (resp. right exact, exact).

Proof. We know in fact that when two submodules $N$, $P$ of an $A$-module $M$ are such that $N_{\mathfrak{m}}=P_{\mathfrak{m}}$ for every maximal ideal $\mathfrak{m}$ of $A$, then $N=P$ (Bourbaki, Alg. comm., chap. II, § 3, n° 3, th. 1).

7.2. Characterizations of the tensor product functor

7.2.1.

Let $A$ be a ring (not necessarily commutative), $M$ (resp. $N$) a left (resp. right) $A$-module, $P$ a $Z$-module. Recall that giving a $Z$-homomorphism $v:N{\otimes }_{A}M\to P$ is equivalent to giving a $Z$-bilinear map $u:N\times M\to P$ such that $u(ta,x)=u(t,ax)$ for $a\in A$, $t\in N$, $x\in M$, the two maps being related by $v(t\otimes x)=u(t,x)$. On the other hand, giving $u$ is equivalent to giving a

$Z$-homomorphism $x\mapsto f_x$ of $M$ into ${\mathrm{Hom}}_Z(N,P)$ such that $f_{ax}(t)=f_x(ta)$ for $a\in A$, $t\in N$, $x\in M$, the two maps being related by $u(t,x)=f_x(t)$.

7.2.2.

Let $T:A{b}_A\to Ab$ be a covariant additive functor. We are going to define, for every left $A$-module $M$, a canonical homomorphism functorial in $M$, of $Z$-modules

  t_M : T(A_s) ⊗_A M → T(M).                                                 (7.2.2.1)

It will suffice for this, by virtue of (7.2.1), to define a $Z$-homomorphism $x\mapsto t_M^{\prime }(x)$ of $M$ into ${\mathrm{Hom}}_Z(T(A_s),T(M))$, such that $t_M^{\prime }(ax)(y)=t_M^{\prime }(x)(ya)$ for $a\in A$, $x\in M$ and $y\in T(A_s)$. Note for this that ${\mathrm{Hom}}_Z(T(A_s),T(M))$ is canonically equipped with a left $A$-module structure coming from the right $A$-module structure of $T(A_s)$, the external law being such that if $a\in A$, $v\in {\mathrm{Hom}}_Z(T(A_s),T(M))$, then $(a\cdot v)(y)=v(ya)$ for $y\in T(A_s)$. This being so, we define $t_M^{\prime }$ as the composite of the two canonical homomorphisms

  M ⥲ Hom_A(A_s, M) →^T Hom_Z(T(A_s), T(M)),

the second arrow being the map $u\mapsto T(u)$, the first the canonical isomorphism of $A$-modules $x\mapsto {\theta }_x$ such that ${\theta }_x(\xi )=\xi x$ for $\xi \in A$, $x\in M$. One has ${\theta }_{ax}={\theta }_x\circ h_a$, hence $T({\theta }_{ax})=T({\theta }_x\circ h_a)=T({\theta }_x)\circ T(h_a)$, and consequently, for $y\in T(A_s)$,

$$T({\theta }_{ax})(y)=T({\theta }_x)(T(h_a)(y))=T({\theta }_x)(ya)$$

by definition of the external law on $T(A_s)$, which proves the existence of $t_M$; it is immediate to verify that this homomorphism is functorial in $M$, that is, that for every homomorphism $w:M\to M^{\prime }$ of left $A$-modules, the diagram

                       t_M
  T(A_s) ⊗_A M  ─────────────→  T(M)
        │                        │
   1 ⊗ w│                        │T(w)                                       (7.2.2.2)
        ↓                        ↓
  T(A_s) ⊗_A M' ─────────────→  T(M')
                      t_{M'}

is commutative.

The functoriality of the homomorphism (7.2.2.1) shows that when $A$ is commutative, it is a homomorphism of $A$-modules (cf. (7.1.2)).

7.2.3.

When $A$ is commutative, one can more generally define a canonical homomorphism of $A$-modules

  T(N) ⊗_A M → T(N ⊗_A M)                                                    (7.2.3.1)

for every $A$-module $N$; it suffices in the construction of (7.2.2) to replace the homomorphism ${\theta }_x$ by the homomorphism of $A$-modules $N\to N{\otimes }_{A}M$ sending each $y\in N$ to $y\otimes x$. It is immediate that this homomorphism is functorial in $M$ and $N$.

In particular, if $B$ is an $A$-algebra (not necessarily commutative), one has a homomorphism functorial in $M$

  (T(M))_{(B)} = T(M) ⊗_A B → T(M ⊗_A B) = T_{(B)}(M_{(B)})                  (7.2.3.2)

which, by virtue of the functoriality of (7.2.3.1) in $M$, is a homomorphism of $B$-modules.

One has moreover the commutative diagram

                       t_M
   T(A) ⊗_A M  ───────────────→  T(M)
        │                         │
        │                         │                                          (7.2.3.3)
        ↓                         ↓
   T_{(B)}(B_s) ⊗_B M_{(B)} ───→ T_{(B)}(M_{(B)})
                       t_{M_{(B)}}

where the right vertical arrow is the composite homomorphism

  T(M) → T(M) ⊗_A B → T(M ⊗_A B) = T_{(B)}(M_{(B)})

of (7.2.3.2) and the canonical homomorphism; as for the left vertical arrow of (7.2.3.3), it is the homomorphism $T(A){\otimes }_{A}M\to T_{(B)}(B_s){\otimes }_B(B{\otimes }_{A}M)=T_{(B)}(B_s){\otimes }_{A}M$, where $T(A)\to T_{(B)}(B_s)=T(B)$ is $T(\rho )$, $\rho$ being considered as a homomorphism of $A$-modules $A\to B_{[\rho ]}$.

Lemma (7.2.4).

If $T$ is a covariant additive functor from $A{b}_A$ into Ab, commuting with direct sums, the canonical homomorphism $t_L$ (7.2.2.1) is an isomorphism for every free $A$-module $L$.

Proof. Indeed, one has $L={\oplus }_{\alpha \in I}{L}_{\alpha }$ where $L_{\alpha }$ is isomorphic to $A_s$ for every $\alpha \in I$; the definition of $t_M$ given in (7.2.2) shows that $t_L={\oplus }_{\alpha \in I}{t}_{L_{\alpha }}$, since

  T : Hom_A(A_s, L) → Hom_Z(T(A_s), T(L))

is the direct sum of the $Z$-linear maps $T_{\alpha }:{\mathrm{Hom}}_A(A_s,L_{\alpha })\to {\mathrm{Hom}}_Z(T(A_s),T(L_{\alpha }))$ by virtue of the hypothesis on $T$. We are thus reduced to proving the lemma for $L=A_s$; but $t_L$ is then none other than the canonical isomorphism $T(A_s){\otimes }_A{A}_s\stackrel{\sim }{\to }T(A_s)$ valid for every right $A$-module.

Proposition (7.2.5).

Let $T$ be a covariant additive functor from $A{b}_A$ into Ab, commuting with direct sums. The following conditions are equivalent:

• a) $T$ is right exact.
• b) The canonical homomorphism $t_M$ (7.2.2.1) is an isomorphism for every left $A$-module $M$.
• b') $T$ is semi-exact and the homomorphism $t_M$ is surjective for every left $A$-module $M$.
• c) $T$ is isomorphic to a functor in $M$ of the form $N{\otimes }_{A}M$, where $N$ is a right $A$-module.

Proof. It is clear that b) implies c) and that c) implies a); let us show that a) implies b).

Set $T^{\prime }(M)=T(A_s){\otimes }_{A}M$ for every left $A$-module $M$. There exists an exact sequence $L^{\prime }\to L\to M\to 0$, where $L$ and $L^{\prime }$ are two free left $A$-modules; since $T$ and $T^{\prime }$ are right exact, we have the commutative diagram

  T'(L') → T'(L) → T'(M) → 0
    │        │       │
    │t_{L'}  │t_L    │t_M
    ↓        ↓       ↓
  T(L')  →  T(L)  →  T(M)  → 0

where the two rows are exact; since $t_L$ and $t_{L^{\prime }}$ are isomorphisms by virtue of (7.2.4), the same is true of $t_M$ by the five lemma. Finally, it is clear that b) implies b'). To show that b') implies a), it suffices to prove

Lemma (7.2.5.1).

Let $\mathcal{K}$, ${\mathcal{K}}^{\prime }$ be two abelian categories, $F$, $G$ two covariant additive functors from $\mathcal{K}$ into ${\mathcal{K}}^{\prime }$, $f:F\to G$ a functorial morphism (T, I, 1.2) such that, for every object $E$ of the category $\mathcal{K}$, $f_E:F(E)\to G(E)$ is an epimorphism. Then, if $F$ is right exact and $G$ semi-exact, $G$ is right exact.

Proof. It all comes down to showing that for every epimorphism $v:E^{\prime }\to E$ in $\mathcal{K}$, $G(v):G(E^{\prime })\to G(E)$ is an epimorphism; one has the commutative diagram

            F(v)
  F(E') ─────────→  F(E)
   │                 │
   │f_{E'}           │f_E
   ↓                 ↓
  G(E') ─────────→  G(E)
            G(v)

in which $F(v)$, $f_{E^{\prime }}$ and $f_E$ are epimorphisms; hence so is $G(v)$.

Remark (7.2.6).

For every right $A$-module $N$, set $T_N(M)=N{\otimes }_{A}M$ for every left $A$-module $M$, so that T_N is a covariant additive functor from $A{b}_A$ into Ab, right exact and commuting with direct sums. If one canonically identifies $T_N(A_s)$ with $N$, one verifies at once that the corresponding homomorphism (7.2.2.1) becomes the identity. One concludes that the right $A$-module $N$ in the statement of (7.2.5, c)) is determined up to unique isomorphism and is canonically isomorphic to $T(A_s)$. One can also say that the functorial morphisms $T⇝T(A_s)$ and $N⇝T_N$ constitute an equivalence (T, I, 1.2) of the category of right $A$-modules and the category of covariant additive functors $A{b}_A\to Ab$ that are right exact and commute with direct sums.

Proposition (7.2.7).

Let $A$ be a left artinian ring whose quotient by its radical $\mathfrak{m}$ is a field $k$. Let $T$ be a covariant additive functor from $A{b}_A$ into Ab, commuting with direct sums. The conditions of (7.2.5) are then also equivalent to

• d) $T$ is semi-exact and the homomorphism $T(ϵ):T(A_s)\to T(k)$ deduced from the canonical homomorphism $ϵ:A_s\to k$ is surjective.

Proof. It is clear that condition b') of (7.2.5) implies d); let us prove that d) implies b'). There exists an integer $n$ such that ${\mathfrak{m}}^n=0$; set, for every $A$-module $M$, $M_h={\mathfrak{m}}^{h}M$; we shall prove by descending induction on $h$ that $t_{M_h}$ is surjective. The proposition is evident for $h=n$; for $h<n$, one has an exact sequence

  0 → M_{h+1} → M_h → M_h / M_{h+1} → 0

and the induction hypothesis implies that $t_{M_{h+1}}$ is surjective. On the other hand, $M_h/M_{h+1}$ is annihilated by $\mathfrak{m}$ and is therefore an $(A/\mathfrak{m})$-module, in other words a direct sum of $A$-modules isomorphic to $k$. To prove that $t_{M_h/M_{h+1}}$ is surjective, it suffices therefore to prove that $t_k$ is, since $T$ commutes with direct sums. Now, by virtue of the commutativity of the diagram

                    t_{A_s}
  T(A_s) ⊗_A A_s ───────────→  T(A_s)
        │                        │
   1 ⊗ ε│                        │T(ε)
        ↓                        ↓
  T(A_s) ⊗_A k  ───────────→  T(k)
                     t_k

and of (7.2.4), hypothesis d) implies that $t_k$ is indeed surjective. To finish the proof, it will suffice to show that if one has an exact sequence $0\to M^{\prime }{\to }^{u}M{\to }^v{M}^{\prime \prime }\to 0$ of $A$-modules, such that $t_{M^{\prime }}$ and $t_{M^{\prime \prime }}$ are surjective, then $t_M$ is surjective. Now, one has a commutative diagram

  T'(M') ────→ T'(M)  ────→ T'(M'') ────→ 0
    │           │            │
    │t_{M'}     │t_M         │t_{M''}
    ↓           ↓            ↓
  T(M')  ────→ T(M)   ────→ T(M'')  ────→ Coker(T(v))
                      T(v)

in which the two rows are exact, by virtue of the hypothesis that $T$ is semi-exact. Since by the induction hypothesis $t_{M^{\prime }}$ and $t_{M^{\prime \prime }}$ are epimorphisms and the last vertical arrow is a monomorphism, the five lemma (M, I, 1.1) shows that $t_M$ is an epimorphism.

7.3. Exactness criteria for the homological functors of modules

Proposition (7.3.1).

Let $A$ be a ring (not necessarily commutative), $T_{\bullet }$ a covariant homological functor (T, II, 2.1) from the category $A{b}_A$ into the category Ab, commuting with direct sums. Let $p$ be an integer such that $T_p$ and $T_{p-1}$ are defined. The following conditions are equivalent:

• a) $T_p$ is right exact.
• b) $T_{p-1}$ is left exact.

• c) For every left $A$-module $M$, the canonical functorial homomorphism (7.2.2.1)

    T_p(A_s) ⊗_A M → T_p(M)                                                (7.3.1.1)
  

  is an isomorphism.
• d) For every left $A$-module $M$, the homomorphism (7.3.1.1) is an epimorphism.
• e) $T_p$ is isomorphic to a functor $M\mapsto N{\otimes }_{A}M$, where $N$ is a right $A$-module.

If moreover the conditions of (7.2.7) on $A$ and $\mathfrak{m}$ are satisfied, the preceding conditions are also equivalent to

• f) The canonical homomorphism $T_p(ϵ):T_p(A_s)\to T_p(k)$ is an epimorphism.

Proof. Since by definition of a homological functor, $T_i$ is semi-exact for every $i$ such that $T_i$ is defined, and since for every exact sequence $0\to M^{\prime }{\to }^{u}M{\to }^v{M}^{\prime \prime }\to 0$ one has $Ker(T_{i-1}(u))=Coker(T_i(v))$, it is clear that a) and b) are equivalent and the other assertions follow trivially from (7.2.5) and (7.2.7).

Corollary (7.3.2).

Let $A$ be a commutative ring. With the notations of (7.3.1), suppose $T_p$ is right exact. If $f\in A$ does not belong to the annihilator of any element $\ne 0$ of an $A$-module $M$, then $f$ does not belong to the annihilator of any element $\ne 0$ of $T_{p-1}(M)$. In particular, if $A$ is an integral domain, the $A$-module $T_{p-1}(A)$ is torsion-free.

Proof. Indeed, if $h_f$ denotes the homothety $x\mapsto fx$ of $M$, the hypothesis means that $h_f$ is injective; hence so is $T_{p-1}(h_f)$ by condition b) of (7.3.1).

Proposition (7.3.3).

Let $A$ be a ring, $T_{\bullet }$ a covariant homological functor from $A{b}_A$ into Ab, commuting with direct sums. Let $p$ be an integer such that $T_{p-1}$, $T_p$ and $T_{p+1}$ are defined. The following conditions are equivalent:

• a) $T_p$ is exact.
• b) $T_{p+1}$ and $T_p$ are right exact.
• c) $T_p$ and $T_{p-1}$ are left exact.
• d) $T_{p+1}$ is right exact and $T_{p-1}$ is left exact.
• e) For every $A$-module $M$, the canonical homomorphisms

    T_i(A_s) ⊗_A M → T_i(M)                                                (7.3.3.1)
  

  are isomorphisms for $i=p$ and $i=p+1$.
• e') For every $A$-module $M$, the canonical homomorphisms (7.3.3.1) are epimorphisms for $i=p$ and $i=p+1$.
• f) For every $A$-module $M$, the homomorphism (7.3.3.1) is an isomorphism for $i=p$ and $T_p(A_s)$ is a flat right $A$-module.
• f') For every $A$-module $M$, the homomorphism (7.3.3.1) is an epimorphism for $i=p$ and $T_p(A_s)$ is a flat right $A$-module.

Proof. The equivalence of conditions a), b), c), d) results from the equivalence of conditions a) and b) of (7.3.1). The equivalence of b), e) and e') results from the equivalence of a), c) and d) in (7.3.1). Finally, to say that $T_p(A_s)$ is flat means that the functor $M\mapsto T_p(A_s){\otimes }_{A}M$ is left exact; the equivalence of a), f) and f') again results from the equivalence of a), c), d) in (7.3.1).

Corollary (7.3.4).

Suppose $A$ commutative, $T_p$ exact, and suppose moreover that $T_p(A)$ is an $A$-module of finite presentation. Then the function $x\mapsto ran{g}_{\kappa (x)}(T_p(\kappa (x)))$ is locally constant on $X=\mathrm{Spec}(A)$, hence constant if $\mathrm{Spec}(A)$ is connected.

Proof. Indeed, since $T_p(A)$ is a flat $A$-module by virtue of (7.3.3, f)), it is projective of finite type, and $(T_p(A){)}^{\sim }$ is therefore a locally free ${\mathcal{O}}_X$-module (Bourbaki, Alg. comm., chap. II, § 5, n° 2, th. 1); one has moreover $T_p(\kappa (x))=T_p(A){\otimes }_A\kappa (x)$ (7.3.3, e)) and we know that the rank at the point $x$ of the $A$-module $T_p(A)$ is locally constant (loc. cit.), whence the corollary.

Proposition (7.3.5).

Suppose that $A$ is a left artinian ring whose quotient by its radical $\mathfrak{m}$ is a field $k$. Then the conditions of (7.3.3) are also equivalent to each of the following:

• g) The canonical homomorphism $T_i(ϵ):T_i(A_s)\to T_i(k)$ is an epimorphism for $i=p$ and $i=p+1$.
• h) $T_p(ϵ)$ is an epimorphism and $T_p(A_s)$ is a flat right $A$-module (or, what amounts to the same (Bourbaki, Alg. comm., chap. II, § 3, n° 2, cor. 2 of prop. 5) a free $A$-module).

Suppose moreover that $A$ is commutative and the $A$-module $T_p(k)$ of finite length $d$. Then the preceding conditions are also equivalent to each of the following:

• i) For every $A$-module $M$ of finite length, one has

    long(T_p(M)) = d · long(M).                                            (7.3.5.1)
  

• j) One has

    long(T_p(A)) = d · long(A).                                            (7.3.5.2)
  

Proof. The equivalence of g) and h) with the conditions of (7.3.3) follows immediately from (7.2.7). To prove the other assertions, we shall use the following lemma:

Lemma (7.3.5.3).

Let $\mathcal{K}$, ${\mathcal{K}}^{\prime }$ be two abelian categories, $F:\mathcal{K}\to {\mathcal{K}}^{\prime }$ a covariant additive functor; suppose that $F$ is semi-exact, and that, for every simple object $S$ of $\mathcal{K}$, $F(S)$ is an object of finite length in ${\mathcal{K}}^{\prime }$. Then, for every object $E$ of finite length in $\mathcal{K}$, $F(E)$ is of finite length in ${\mathcal{K}}^{\prime }$. For every exact sequence $0\to E^{\prime }{\to }^{u}E{\to }^v{E}^{\prime \prime }\to 0$ of objects of finite length in $\mathcal{K}$, one has

  long F(E) ≤ long F(E') + long F(E'')                                       (7.3.5.4)

and for the two members of (7.3.5.4) to be equal, it is necessary and sufficient that the sequence

$$0\to F(E^{\prime })\to F(E)\to F(E^{\prime \prime })\to 0$$

be exact.

Proof. Indeed, the sequence $F(E^{\prime }){\to }^{F(u)}F(E){\to }^{F(v)}F(E^{\prime \prime })$ is exact by hypothesis; if one supposes $F(E^{\prime })$ and $F(E^{\prime \prime })$ of finite length, the same is true of $Im(F(u))$ and $Im(F(v))$, and since $Ker(F(v))=Im(F(u))$, $F(E)$ is of finite length and one has

  long F(E) = long Im(F(u)) + long Im(F(v)) ≤ long F(E') + long F(E'').      (7.3.5.5)

By induction on the length of $E$, this already proves the first assertion; moreover, the two members of (7.3.5.5) can be equal only if $longIm(F(u))=longF(E^{\prime })$ (which is equivalent to $longKer(F(u))=0$, or $Ker(F(u))=0$) and $longIm(F(v))=longF(E^{\prime \prime })$ (which is equivalent to $longCoker(F(v))=0$, or $Coker(F(v))=0$).

We now note that if $M$ is an $A$-module of finite length ($A$ being commutative), the quotients of a Jordan–Hölder sequence of $M$ are necessarily isomorphic to the $A$-module $k$; therefore, by (7.3.5.4) and induction on the length of $M$,

  long T_p(M) ≤ d · long(M).                                                 (7.3.5.6)

Moreover, it follows from (7.3.5.3) that if $T_p$ is exact, one has the equality (7.3.5.1); hence condition a) of (7.3.3) implies i); it is clear that i) implies j), and it remains to prove

Lemma (7.3.5.7).

The relation $long{T}_p(A)=d\cdot longA$ implies that $T_p(ϵ)$ is an epimorphism and that $T_p(A)$ is a flat $A$-module.

Proof. Indeed, starting from the exact sequence $0\to \mathfrak{m}\to A\to k\to 0$, it follows from (7.3.5.4) and (7.3.5.6) that one has

  long T_p(A) ≤ long T_p(𝔪) + long T_p(k) ≤ d(long 𝔪 + long k) = d · long A

and that equality can hold (7.3.5.3) only if the sequence

$$0\to T_p(\mathfrak{m})\to T_p(A)\to T_p(k)\to 0(\mathrm{7.3.5.8})$$

is exact. By virtue of (7.2.7) and (7.2.5), $T_p$ is isomorphic to a functor $M\mapsto N{\otimes }_{A}M$, and the exactness of the sequence (7.3.5.8) shows, by virtue of the exact sequence of Tor's, that one has $To{r}_1^A(N,k)=0$. One concludes that $N=T_p(A)$ is a flat $A$-module $(0_I,\mathrm{10.1.3})$.

Lemma (7.3.6).

Let $A$ be a ring, $T_{\bullet }$ a covariant homological functor from $A{b}_A$ into Ab, commuting with direct sums. Suppose $T_p$ and $T_{p+1}$ defined, and $T_p$ left exact. For $T_{p+1}$ to be exact, it is necessary and sufficient that $T_{p+1}(A_s)$ be a flat right $A$-module.

Proof. Indeed, one knows by (7.3.1) that the canonical homomorphism

  T_{p+1}(A_s) ⊗_A M → T_{p+1}(M)

is an isomorphism of functors; it suffices to apply the definition of a flat $A$-module.

Proposition (7.3.7).

Let $A$ be a ring, $T_{\bullet }$ a covariant homological functor from $A{b}_A$ into Ab, commuting with direct sums. Suppose there exists $i_0$ such that $T_i$ is exact for $i\le i_0$. Then, for every integer $p>i_0$, the following conditions are equivalent:

• a) $T_q$ is exact for $q\le p$;
• b) $T_q(A_s)$ is a flat right $A$-module for $q\le p$.
• c) For every $A$-module $M$, the canonical homomorphism $T_q(A_s){\otimes }_{A}M\to T_q(M)$ is surjective for $q\le p+1$.

Proof. The equivalence of a) and b) results from (7.3.6) by induction on $q$, since $T_{i_0}$ is exact by hypothesis; the equivalence of a) and c) results from the equivalence of conditions a) and e') in (7.3.3).

7.3.8.

If $A$ is a commutative ring, $B$ an $A$-algebra (not necessarily commutative), $T_{\bullet }$ a covariant homological functor from $A{b}_A$ into Ab, it follows from the definitions (7.1.3) that the functor from $A{b}_B$ into Ab obtained by extension of scalars from $A$ to $B$, and which we shall denote $T_{\bullet }^{(B)}=(T_{\bullet }{)}_{(B)}$, is again a homological functor.

Corollary (7.3.9).

Suppose that $T_{\bullet }$ satisfies the general conditions of (7.3.7) and commutes with inductive limits, and moreover that $A$ is an integral ring and all the $T_n(A)$ are $A$-modules of finite presentation. Then, for every integer $N$, there exists $f\in A-0$ such that the functor $T_{\bullet }^{(A_f)}:A{b}_{A_f}\to Ab$ is exact for $p\le N$.

Proof. By hypothesis, $T_i$ is exact for $i\le i_0$, hence $T_i(A)$ is flat for these values of $i$. By virtue of (7.3.7, b)), it suffices to take $f$ such that $T_p^{(A_f)}(A_f)=T_p(A_f)$ is a free $A_f$-module for $i_0<p\le N$. Now, one has $T_p(A_f)=(T_p(A){)}_f$ since $T_p$ commutes with inductive limits (7.1.4). If $x$ is the generic point of $\mathrm{Spec}(A)$, $(T_p(A){)}_x$ is a finite-dimensional vector space over the fraction field of $A$. Since each $T_p(A)$ is of finite presentation, there does indeed exist an $f$ with the desired property (Bourbaki, Alg. comm., chap. II, § 5, n° 1, cor. of prop. 2).

One will note that if there are only finitely many indices $i$ such that $T_i\ne 0$, there exists $f\in A-0$ such that all the $T_p^{(A_f)}$ are exact.

Corollary (7.3.10).

Suppose that $T_{\bullet }$ satisfies the general conditions of (7.3.7) and commutes with inductive limits, that $A$ is commutative and noetherian, and the $T_n(A)$ $A$-modules of finite type. Then, for every integer $N$, there exists a dense open set $U$ of $\mathrm{Spec}(A)$ such that, for every $p\le N$, the function $x\mapsto ran{g}_{\kappa (x)}(T_p(\kappa (x)))$ is constant on $U$.

Proof. Let $\mathfrak{p}$ be a minimal prime ideal of $A$; by hypothesis, the ring $B=A/\mathfrak{p}$ is integral and $\mathrm{Spec}(B)$ is identified with an irreducible component of the topological space $\mathrm{Spec}(A)$. We shall show by induction on $p\le N$ that there exists $f_p\in B-0$ such that, on setting $B^{\prime }=B_{f_p}$, $T_i^{(B^{\prime })}$ is exact and the $T_i(B^{\prime })$ are $B^{\prime }$-modules of finite type for $i\le p$. The proposition is true for $p\le i_0$ by virtue of the hypothesis, taking $f_p=1$ (hence $B^{\prime }=B=A/\mathfrak{p}$): for $T_p$ being then exact, $T_p(B)$ is isomorphic to $T_p(A)/T_p(\mathfrak{p})$, hence is an $A$-module (and a fortiori a $B$-module) of finite type. Let us argue by induction on $p$; $f_p$ is the canonical image in $B$ of an element $g_p\in A$, and if one sets $A^{\prime }=A_{g_p}$, one has $B^{\prime }=A^{\prime }/{\mathfrak{p}}^{\prime }$ where ${\mathfrak{p}}^{\prime }$ is a minimal prime ideal of $A^{\prime }$, equal to ${\mathfrak{p}}_{g_p}$. Since $T_i(A_{g_p})=(T_i(A){)}_{g_p}$, the $T_i(A^{\prime })$ are $A^{\prime }$-modules of finite type, so the functor $T_{\bullet }^{(A^{\prime })}$ satisfies the same hypotheses as $T_{\bullet }$, but replacing $i_0$ by $p$. One can therefore restrict to the case where $A^{\prime }=A$ and where $T_p$ is exact; the exact sequence $0\to \mathfrak{p}\to A\to A/\mathfrak{p}\to 0$ then gives the exact sequence $T_{p+1}(A)\to T_{p+1}(A/\mathfrak{p}){\to }^{\partial }{T}_p(\mathfrak{p})\to T_p(A)$, and since $T_p$ is exact, the rightmost arrow is injective, hence $T_{p+1}(A/\mathfrak{p})$ is a quotient of $T_{p+1}(A)$ and is consequently of finite type. We now note that the argument of (7.3.9) used the fact that the $T_p(A)$ are of finite type only for $p\le N$; one can therefore apply it to the integral ring $B$, the functor $T_{\bullet }^{(B)}$ and $N=p+1$, which completes the induction. This being so, there exists $f_N\in B-0$ such that $\mathrm{Spec}(B_{f_N})$ is an open set $V$ everywhere dense in $\mathrm{Spec}(B)$ meeting no other irreducible component of $\mathrm{Spec}(A)$. If the proposition

is proved for $B_{f_N}$, one will have an open set $W$ everywhere dense in $V$ in which the functions of the statement will be constant, since $A_x=(B_{f_N}{)}_x$ for every $x\in W$. Doing the same reasoning for every irreducible component of $\mathrm{Spec}(A)$, the corollary will be proved. One can therefore restrict to the case where $A$ is integral; the reasoning of (7.3.9) then proves the existence of $f\in A-0$ such that the $T_p(A_f)$ are free $A_f$-modules of finite type for $p\le N$, which entails the conclusion of (7.3.10) by virtue of (7.3.4).

Proposition (7.3.11).

Let $A$ be a commutative local ring, $k$ its residue field, $T_{\bullet }$ a covariant homological functor from $A{b}_A$ into Ab, commuting with direct sums. Suppose that there exists $i_0$ such that $T_i$ is exact for $i\le i_0$, and that all the $T_n(A)$ are $A$-modules of finite presentation. Then the equivalent conditions a), b), c) of (7.3.7) imply the two following ones, and are equivalent to them when the ring is moreover reduced:

• d) For every $x\in \mathrm{Spec}(A)$, one has $ran{g}_{\kappa (x)}{T}_q(\kappa (x))=ran{g}_k{T}_q(k)$ for $q\le p$.
• d') For every generic point $x_j$ of an irreducible component of $\mathrm{Spec}(A)$, one has

    rang_{κ(x_j)} T_q(κ(x_j)) = rang_k T_q(k)   for q ≤ p.
  

Proof. Since $T_q(A)$ is an $A$-module of finite presentation, condition b) of (7.3.7) is equivalent to saying that $T_q(A)$ is a free $A$-module for $q\le p$ (Bourbaki, Alg. comm., chap. II, § 3, n° 2, cor. 2 of prop. 5); condition c) implies that $T_q(\kappa (x))=T_q(A){\otimes }_A\kappa (x)$ for $q\le p$, hence the equivalent conditions of (7.3.7) imply d), and it is trivial that d) implies d'). It remains to prove that d') implies a) when $A$ is reduced. We argue by induction on $q\le p$, since $T_q$ is exact for $q\le i_0$. Suppose then that $T_k$ is exact for $k\le q<p$ and let us show that $T_{q+1}(A)$ is a free $A$-module. By virtue of the induction hypothesis, $T_{q+1}(A){\otimes }_{A}M$ is isomorphic to $T_{q+1}(M)$ for every $A$-module $M$, by condition c) of (7.3.7) and (7.3.3); applying this property to $M=\kappa (x_j)$ and $M=k$, one finds, by virtue of hypothesis d'), that

  rang_{κ(x_j)} (T_{q+1}(A) ⊗_A κ(x_j)) = rang_k T_{q+1}(k)

for every $j$; but this implies that $T_{q+1}(A)$ is free (Bourbaki, Alg. comm., chap. II, § 3, n° 2, prop. 7), which completes the proof.

The preceding results will be considerably improved for the homological functors of particular type that we shall study in (7.4); we shall obtain in fact exactness criteria involving only one of the $T_p$.

7.4. Exactness criteria for the functors $H_{\bullet }(P_{\bullet }{\otimes }_{A}M)$

7.4.1.

Let $A$ be a ring (not necessarily commutative), $P_{\bullet }$ a complex of flat right $A$-modules. Since the functor $M\mapsto P_k{\otimes }_{A}M$ is then exact on $A{b}_A$ for every $k$, the $\partial$-functor

  T_•(M) = H_•(P_• ⊗_A M)                                                    (7.4.1.1)

is a homological functor from $A{b}_A$ into Ab, evidently $A$-linear when $A$ is commutative (7.1.2), and commuting with inductive limits.

If $A$ is commutative, then, for every $A$-algebra $B$, the homological functor $T_{\bullet }^{(B)}$ (7.3.8) is given by definition by

  T_•^{(B)}(N) = H_•(P_• ⊗_A N_{[ρ]})                                        (7.4.1.2)

where $\rho :A\to B$ is the homomorphism defining the algebra structure of $B$; since one can also write $P_{\bullet }{\otimes }_A{N}_{[\rho ]}=P_{\bullet }{\otimes }_A(B{\otimes }_{B}N{)}_{[\rho ]}=(P_{\bullet }{\otimes }_{A}B){\otimes }_{B}N$, one sees that one has

  T_•^{(B)}(N) = H_•(P'_• ⊗_B N)                                             (7.4.1.3)

for every $B$-module $N$, $P_{\bullet }^{\prime }$ being the complex $P_{\bullet }{\otimes }_{A}B$ of flat $B$-modules $(0_I,\mathrm{6.2.1})$.

Proposition (7.4.2).

Under the general conditions of (7.4.1), and for a given integer $p\in Z$, the following properties are equivalent:

• a) $T_p$ is left exact (or, what amounts to the same, $T_{p+1}$ is right exact).
• b) $Z_p^{\prime }(P_{\bullet })=Coker(P_{p+1}\to P_p)$ is a flat right $A$-module.
• c) There exists a complex $P_{\bullet }^{\prime }$ of flat right $A$-modules such that the differential

    d'_{p+1} : P'_{p+1} → P'_p
  

  is zero, and an isomorphism of homological functors from $H_{\bullet }(P_{\bullet }{\otimes }_{A}M)$ onto $H_{\bullet }(P_{\bullet }^{\prime }{\otimes }_{A}M)$.

Proof. By definition, one has an exact sequence functorial in $M$

  0 → T_p(M) → Z'_p(P_• ⊗ M) → P_{p−1} ⊗ M

where $Z_p^{\prime }(P_{\bullet }\otimes M)=Coker(P_{p+1}\otimes M\to P_p\otimes M)=Z_p^{\prime }(P_{\bullet })\otimes M$ by virtue of the right exactness of the tensor product. For every homomorphism $f:M\to N$, one therefore has a commutative diagram

  0 → T_p(M) ──── Z'_p(P_•) ⊗ M ──── P_{p−1} ⊗ M
        │              │                │
        │u             │v               │w                                   (7.4.2.1)
        ↓              ↓                ↓
  0 → T_p(N) ──── Z'_p(P_•) ⊗ N ──── P_{p−1} ⊗ N

whose rows are exact. If $f$ is a monomorphism, so is $w$ since $P_{p-1}$ is flat; if $T_p$ is left exact, $u$ is also a monomorphism; one concludes that $v$ is a monomorphism, which implies that $Z_p^{\prime }(P_{\bullet })$ is flat. Conversely, if it is so, $v$ is a monomorphism for every monomorphism $f:M\to N$, hence the diagram (7.4.2.1) shows that $u$ is a monomorphism, and consequently $T_p$ (which is already semi-exact) is left exact. Thus a) and b) are equivalent. It is immediate that c) implies a), for if $d_{p+1}^{\prime }:P_{p+1}^{\prime }\to P_p^{\prime }$ is zero, and $0\to M^{\prime }\to M\to M^{\prime \prime }\to 0$ is an exact sequence of $A$-modules, the boundary operator $\partial$ in the exact sequence

  H_{p+1}(P'_• ⊗ M'') →^∂ H_p(P'_• ⊗ M') → H_p(P'_• ⊗ M)

is zero by definition (M, IV, 1), hence $T_p$ is left exact. Let us show conversely that b) implies c). If $Z_{p+1}(P_{\bullet })=Ker(P_{p+1}\to P_p)$, one has an exact sequence

$$0\to Z_{p+1}(P_{\bullet })\to P_{p+1}\to Z_p^{\prime }(P_{\bullet })\to 0$$

in which $P_{p+1}$ and $Z_p^{\prime }(P_{\bullet })$ are flat, hence $Z_{p+1}(P_{\bullet })$ is flat $(0_I,\mathrm{6.1.2})$. We shall take

  P'_i = P_i  for  i ≠ p  and  i ≠ p+1,   P'_p = Z'_p(P_•)  and  P'_{p+1} = Z_{p+1}(P_•);

for the differential $d_i^{\prime }:P_i^{\prime }\to P_{i-1}^{\prime }$, we shall take that of the complex $P_{\bullet }$ for $i\ne p$ and $i\ne p+1$, 0 for $i=p+1$ and for $i=p$ the homomorphism $Z_p^{\prime }(P_{\bullet })\to P_{p-1}$ deduced from $d_p$ by passage to the quotient. Since the $P_i$ are flat, one has

  Z'_i(P_• ⊗ M) = Z'_i(P_•) ⊗ M,  Z_i(P_• ⊗ M) = Z_i(P_•) ⊗ M  and  B_i(P_• ⊗ M) = B_i(P_•) ⊗ M

(setting $B_i(P_{\bullet })=Im(P_{i+1}\to P_i)$); one concludes at once for every $M$ the functorial isomorphisms $H_i(P_{\bullet }\otimes M)\stackrel{\sim }{\to }{H}_i(P_{\bullet }^{\prime }\otimes M)$ for every $i$, and the verification of the fact that this is an isomorphism of $\partial$-functors follows without difficulty from the definition of $\partial$ (M, IV, 1).

One notes that the conditions of (7.4.2) also imply that $B_p(P_{\bullet })$ is flat, for one has an exact sequence $0\to B_p(P_{\bullet })\to P_p\to Z_p^{\prime }(P_{\bullet })\to 0$, in which $P_p$ and $Z_p^{\prime }(P_{\bullet })$ are flat $(0_I,\mathrm{6.1.2})$.

Corollary (7.4.3).

Suppose that $A$ is a noetherian regular ring of dimension 1 (in other words a product of Dedekind rings $(0_{IV},\mathrm{17.1.3}and\mathrm{17.3.7})$, for example a principal ring). Then, for $T_p$ to be left exact, it is necessary and sufficient that $T_p(A)$ be a flat $A$-module. For $T_p$ to be exact, it is necessary and sufficient that $T_p(A)$ and $T_{p-1}(A)$ be flat $A$-modules.

Proof. Recall that for a module $M$ over a Dedekind ring, it amounts to the same to say that $M$ is flat or that it is torsion-free $(0_I,\mathrm{6.3.3}and\mathrm{6.3.4})$; under the hypotheses of (7.4.3), every submodule of a flat $A$-module is therefore flat.

The second assertion of (7.4.3) results from the first, since to say that $T_p$ is exact means that $T_p$ and $T_{p-1}$ are left exact. To prove the first assertion, note that one has an exact sequence

$$0\to H_p(P_{\bullet })\to Z_p^{\prime }(P_{\bullet })\to B_{p-1}(P_{\bullet })\to 0$$

in which $B_{p-1}(P_{\bullet })$ is a flat $A$-module, as a submodule of the flat $A$-module $P_{p-1}$. It is therefore equivalent to say that $H_p(P_{\bullet })$ is flat or that $Z_p^{\prime }(P_{\bullet })$ is flat $(0_I,\mathrm{6.1.2})$.

The most important applications of (7.4.2) are the following:

Proposition (7.4.4).

Let $A$ be a noetherian ring, $P_{\bullet }$ a complex of flat $A$-modules: suppose, either that the $P_i$ are of finite type, or that the $H_i(P_{\bullet })$ are $A$-modules of finite type and that there exists $i_0$ such that $H_i(P_{\bullet })=0$ for $i<i_0$. Let $T_{\bullet }$ be the homological functor defined by (7.4.1.1). Then the set $U$ of $y\in \mathrm{Spec}(A)$ such that $(T_p{)}_y$ (7.1.4) is right exact (resp. left exact, exact) is open in $\mathrm{Spec}(A)$.

Proof. In the second hypothesis on $P_{\bullet }$, one can replace $P_{\bullet }$ by a complex $P_{\bullet }^{\prime }$ of

free $A$-modules of finite type for which the functor $H_{\bullet }(P_{\bullet }^{\prime }{\otimes }_{A}M)$ is isomorphic (as a $\partial$-functor) to $T_{\bullet }(M)$ (0, 11.9.3). One can therefore always reduce to the first hypothesis, and in this case the $Z_p^{\prime }(P_{\bullet })$ are of finite type; moreover, one can restrict to proving the assertions relative to left exactness (cf. (7.4.2, a))). This being so, let $x\in U$; since the functor $M\mapsto M_x$ is exact, one has $(Z_p^{\prime }(P_{\bullet }){)}_x=Z_p^{\prime }((P_{\bullet }{)}_x)$, and (taking (7.4.1.3) into account) the hypothesis implies, by virtue of (7.4.2, b)), that $(Z_p^{\prime }(P_{\bullet }){)}_x$ is a flat $A_x$-module, hence free since it is of finite type and $A_x$ is a noetherian local ring (Bourbaki, Alg. comm., chap. II, § 3, n° 2, cor. 2 of prop. 5). One concludes that there is $f\in A$ such that $(Z_p^{\prime }(P_{\bullet }){)}_f$ is free over $A_f$ (Bourbaki, Alg. comm., chap. II, § 5, n° 1, cor. of prop. 2), and a fortiori $(Z_p^{\prime }(P_{\bullet }){)}_y$ is free over $A_y$ for every $y\in D(f)$, which completes the proof by virtue of (7.4.2, b)).

Corollary (7.4.5).

Under the hypotheses of (7.4.4), suppose moreover that $A$ is integral. Then the set $U$ of $x\in \mathrm{Spec}(A)$ such that $(T_p{)}_x$ is exact is open and non-empty.

Proof. It suffices to prove that $(T_p{)}_x$ is exact for the generic point $x$ of $\mathrm{Spec}(A)$, which is immediate since $A_x$ is a field, hence every additive functor on $A{b}_{A_x}$ is exact.

Proposition (7.4.6).

Under the general hypotheses of (7.4.4), conditions a), b) and c) of (7.4.2) are also equivalent to:

• d) There exists an $A$-module $Q$ and a functorial isomorphism

    T_p(M) ⥲ Hom_A(Q, M).                                                  (7.4.6.1)
  

Moreover, the $A$-module $Q$ is determined up to unique isomorphism by this property, and it is of finite type.

Proof. The uniqueness of $Q$ is a particular case of the uniqueness of a representative object of a representable functor (0, 8.1.5). It is clear that the second member of (7.4.6.1) is left exact. Conversely, to prove the existence of $Q$, when $T_p$ is left exact, one can first, as in (7.4.4), reduce to the case where the $P_i$ are flat and of finite type, hence (since $A$ is noetherian) projective of finite type (Bourbaki, Alg. comm., chap. II, § 5, n° 2, cor. of th. 1). The dual ${\check{P}}_i$ of $P_i$ is then also a projective $A$-module of finite type, $P_i$ is canonically isomorphic to the dual of ${\check{P}}_i$, and the canonical homomorphism $P_i{\otimes }_{A}M\to {\mathrm{Hom}}_A({\check{P}}_i,M)$ is bijective (Bourbaki, Alg., chap. II, 3rd ed., § 4, n° 2, prop. 2). One knows on the other hand (7.4.2, c)) that one can suppose $d_{p+1}:P_{p+1}\to P_p$ is zero, hence one has an exact sequence

  0 → T_p(M) → P_p ⊗ M →^v P_{p−1} ⊗ M

where $v=d_p\otimes 1$. Set then $Q^{\prime }=Ker(d_p)$, so that one has the exact sequence $0\to Q^{\prime }{\to }^w{P}_p{\to }^{d_p}{P}_{p-1}$, whence by transposition the exact sequence ${\check{P}}_{p-1}{\to }^{{}^t{d}_p}{\check{P}}_p{\to }^{{}^{t}w}{\check{Q}}^{\prime }\to 0$. We shall see that $Q={\check{Q}}^{\prime }=Coker({}^t{d}_p)$ answers the question. Indeed, one has the exact sequence

  0 → Hom(Q, M) → Hom(P̌_p, M) →^{v'} Hom(P̌_{p−1}, M)

where $v^{\prime }=\mathrm{Hom}({}^t{d}_p,1)$; when one canonically identifies $P_i\otimes M$ to $\mathrm{Hom}({\check{P}}_i,M)$, $v^{\prime }$ is therefore identified with $v=d_p\otimes 1$, and one has consequently the functorial isomorphism $T_p(M)\stackrel{\sim }{\to }{\mathrm{Hom}}_A(Q,M)$ sought. Moreover, $Q$, being a quotient of ${\check{P}}_p$, is of finite type.

Proposition (7.4.7).

Suppose the general conditions of (7.4.4) satisfied. Then, for every $A$-module $M$ of finite type:

• (i) The $T_i(M)$ are $A$-modules of finite type.
• (ii) For every ideal $\mathfrak{m}$ of $A$, the canonical homomorphism

    (T_i(M))^∧ → lim←_n T_i(M ⊗_A (A/𝔪^{n+1}))                             (7.4.7.1)
  

  (where the left member is the Hausdorff completion of $T_i(M)$ for the $\mathfrak{m}$-preadic topology) is bijective.

Proof. As in (7.4.4), one reduces first to the case where the $P_i$ are of finite type; $A$ being noetherian, the submodules of $P_i{\otimes }_{A}M$ are of finite type, whence trivially assertion (i). As for assertion (ii), it follows more generally from the following lemma:

Lemma (7.4.7.2).

Let $A$ be a noetherian ring, $u:E\to F$ a homomorphism of $A$-modules of finite type. For every $A$-module of finite type, set $K(M)=Ker(u\otimes 1_M)$, $C(M)=Coker(u\otimes 1_M)$; then the canonical homomorphisms

  (K(M))^∧ → lim←_n K(M_n),    (C(M))^∧ → lim←_n C(M_n)                      (7.4.7.3)

(where one has set $M_n=M{\otimes }_A(A/{\mathfrak{m}}^{n+1})=M/{\mathfrak{m}}^{n+1}M$) are bijective for every ideal $\mathfrak{m}$ of $A$.

Proof. Since $E\otimes M$ and $F\otimes M$ are of finite type, and the functor $M\mapsto \hat{M}$ is exact in the category of $A$-modules of finite type $(0_I,\mathrm{7.3.3})$, $(K(M){)}^{\wedge }$ and $(C(M){)}^{\wedge }$ are respectively the kernel and the cokernel of $(u\otimes 1{)}^{\wedge }:(E\otimes M{)}^{\wedge }\to (F\otimes M{)}^{\wedge }$. The left exactness of the functor $\lim \leftarrow$ shows therefore that $(K(M){)}^{\wedge }=\lim {\leftarrow }_{n}K(M_n)$; on the other hand, the right exactness of the tensor product proves that $C(M_n)=C(M){\otimes }_A(A/{\mathfrak{m}}^{n+1})$, hence $(C(M){)}^{\wedge }=\lim {\leftarrow }_{n}C(M_n)$ by definition.

Remark (7.4.8).

Taking (6.10.5) and (6.10.6) into account, one sees that, given an additional flatness hypothesis, (7.4.7) gives back the fact that (4.1.7.1) is an isomorphism, that is, the essence of the "first comparison theorem" for proper morphisms; moreover, the statement applies not only to a coherent ${\mathcal{O}}_X$-module, but to a complex of such modules. It would be interesting to obtain a statement comprising at the same time (7.4.7) and (4.1.7.1) as particular cases. One will note that when the $P_i$ are of finite type, the proof of (7.4.7) does not use the fact that they are flat modules; it would be worthwhile examining whether the conclusion of (7.4.7) is still valid when the $P_i$ are not supposed flat nor of finite type, but the $H_i(P_{\bullet })$ are supposed of finite type for every $i$ and zero for $i<i_0$. Is it then possible to replace $P_{\bullet }$ by a complex $P_{\bullet }^{\prime }$ of $A$-modules of finite type such that the functors $H_{\bullet }(P_{\bullet }\otimes M)$ and $H_{\bullet }(P_{\bullet }^{\prime }\otimes M)$ (which are no longer homological) are still isomorphic?

7.5. The case of noetherian local rings

7.5.1.

Let $A$ be a noetherian local ring, $\mathfrak{m}$ its maximal ideal, and for every $A$-module $M$, denote by $\hat{M}$ its Hausdorff completion for the $\mathfrak{m}$-preadic topology, isomorphic to $\lim \leftarrow (M{\otimes }_A(A/{\mathfrak{m}}^{n+1}))=\lim \leftarrow (M/{\mathfrak{m}}^{n+1}M)$. Let $T$ be a covariant additive functor from $A{b}_A$ into Ab; the canonical homomorphisms (7.2.3.1)

  T(M) ⊗_A (A/𝔪^{n+1}) → T(M ⊗_A (A/𝔪^{n+1}))

evidently form a projective system of $A$-homomorphisms, which thus give in the limit an Â-homomorphism functorial in $M$

$$(T(M){)}^{\wedge }\to \lim {\leftarrow }_{n}T(M_n)(\mathrm{7.5.1.1})$$

where one has set $M_n=M{\otimes }_A(A/{\mathfrak{m}}^{n+1})$, $A_n=A/{\mathfrak{m}}^{n+1}$.

Proposition (7.5.2).

Let $A$ be a noetherian local ring with maximal ideal $\mathfrak{m}$, $k=A/\mathfrak{m}$ its residue field, $T$ a covariant additive functor from $A{b}_A$ into Ab, semi-exact and commuting with inductive limits. Suppose moreover that for every $A$-module of finite type $M$, $T(M)$ is an $A$-module of finite type and that the canonical homomorphism (7.5.1.1) is an isomorphism. Under these conditions, the following properties are equivalent:

• a) $T$ is right exact.
• b) For every $n$, the functor $N\mapsto T(N)$ is right exact in the category of $A_n$-modules of finite type (which amounts to saying that $T$ is right exact in the category of $A$-modules of finite length).
• c) The canonical homomorphism $T(ϵ):T(A)\to T(k)$ is surjective.
• d) For every $n$ sufficiently large, the canonical homomorphism $T(A_n)\to T(k)$ is surjective.

Proof. It is clear that a) implies b). Let us show that b) implies a), that is, that if $u:M\to N$ is an epimorphism of $A$-modules, $T(u)$ is an epimorphism. Since $T$ commutes with inductive limits and the functor $\lim \to$ is exact in the category of modules (for filtered index sets), one can restrict to the case where $M$ and $N$ are of finite type. Since $T(M)$ and $T(N)$ are then of finite type, and $A$ is a noetherian local ring, it suffices to show that $(T(u){)}^{\wedge }:(T(M){)}^{\wedge }\to (T(N){)}^{\wedge }$ is surjective $(0_I,\mathrm{7.3.5}and{0}_I,\mathrm{6.4.1})$. By hypothesis, $(T(M){)}^{\wedge }$ and $(T(N){)}^{\wedge }$ are respectively $\lim {\leftarrow }_{n}T(M_n)$ and $\lim {\leftarrow }_{n}T(N_n)$, hence $(T(u){)}^{\wedge }$ is the limit of the projective system of homomorphisms $T(u\otimes 1_{A_n}):T(M_n)\to T(N_n)$. Now, b) means that these homomorphisms are surjective; moreover, $T(M_n)$ is an $A_n$-module of finite type, and $A_n$ is an artinian ring by hypothesis; one concludes that $(T(u){)}^{\wedge }$ is surjective (0, 13.1.2 and 13.2.2). It is clear that a) implies c), and since $T(ϵ)$ factors as $T(A)\to T(A_n)\to T(k)$, c) implies d); finally, it follows from (7.2.7) that b) and d) are equivalent since $T$ is semi-exact in $A{b}_{A_n}$, which completes the proof.

Corollary (7.5.3).

Under the general hypotheses of (7.5.2), if $T(k)=0$, then $T(M)=0$ for every $A$-module $M$.

Proof. Since $k$ is the only simple $A$-module, one deduces from (7.3.5.4) that $T(E)=0$ for every $A$-module of finite length $E$. If now $M$ is of finite type, $(T(M){)}^{\wedge }$ is isomorphic to $\lim {\leftarrow }_{n}T(M_n)$, and since the $M_n$ are of finite length, one has $(T(M){)}^{\wedge }=0$; since $T(M)$ is of finite type by hypothesis, it is isomorphic to a submodule of $(T(M){)}^{\wedge }$ $(0_I,\mathrm{7.3.5})$, hence one has $T(M)=0$. Finally, for an arbitrary $A$-module $M$, $T(M)$ is the inductive limit of the $T(N_{\alpha })$ for the submodules of finite type $N_{\alpha }$ of $M$, which completes the proof.