§ 4. Torsion product

In sections 4 to 8, we denote by $k$ a commutative ring, and by $A$ an associative unital $k$ -algebra. The role of $k$ is auxiliary; we mainly have in view the following three special cases:

a) we consider an arbitrary ring $A$ , we set $k=\mathbf{Z}$ and we endow $A$ with its natural structure of $\mathbf{Z}$ -algebra,

b) we consider an arbitrary ring $A$ , we take for $k$ the center of $A$ ,

c) we consider a commutative ring $A$ and we take $k=A$ .

1. Tensor product of two complexes

Let $(C,d)$ be a complex of right $A$ -modules and $(C',d')$ a complex of left $A$ -modules.

Let us endow the $k$ -module $C\otimes_A C'$ with the grading such that

(C\otimes_A C')_n=\sum_{p+q=n}(C_p\otimes C'_q)

and let $D$ be the unique $k$ -linear endomorphism of degree $(-1)$ of $C\otimes_A C'$ such that

\tag{1} D(x\otimes x')=dx\otimes x'+(-1)^p x\otimes d'x', \qquad x\in C_p,\ y\in C'_q,\ p,\ q\in\mathbf{Z}.

We have $D\circ D=0$ since, with the notations of (1)

D^2(x\otimes x')=ddx\otimes x'+(-1)^{p-1}dx\otimes d'x'+(-1)^pdx\otimes d'x'-x\otimes d'd'x'.

The complex of $k$ -modules $(C\otimes_A C',D)$ is called the tensor product complex of the complexes $(C,d)$ and $(C',d')$ .

Remark 1

When $C'$ is reduced to $C'_0=M$ , then $(C\otimes_A C')_n=C_n\otimes_A M$ and $D=d\otimes 1_M$ ; for example $C\otimes_A A_s$ is naturally identified with $C$ . Similarly, when $C$ is reduced to $C_0=P$ , then $(C\otimes_A C')_n=P\otimes_A C'_n$ and $D=1_P\otimes d$ .

Remark 2

For every integer $r$ , we have $(C\otimes_A C')(r)=C(r)\otimes_A C'$ , but $(C\otimes_A C')(r)$ and $C\otimes_A C'(r)$ do not in general have the same differential.

Let $p, q$ be two integers, $x \in Z_p(C), x' \in Z_q(C')$ ; then the element $x \otimes x'$ of $C_p \otimes C'_q$ belongs to $Z_{p+q}(C \otimes C')$ by (1); moreover, if $y \in C_{p+1}, y' \in C'_{q+1}$ , we have

(x + dy) \otimes (x' + d'y') = x \otimes x' + D(y \otimes x' + (-1)^p (x + dy) \otimes y');

by passing to the quotients, we deduce a

k

-linear mapping, called canonical

\gamma_{p,q}(C, C') : H_p(C) \otimes_A H_q(C') \to H_{p+q}(C \otimes_A C');

if we endow

H(C) \otimes_A H(C')

with the grading such that

(H(C) \otimes H(C'))_n = \sum_{p+q=n} H_p(C) \otimes_A H_q(C'),

the

\gamma_{p,q}

define a graded

k

-linear mapping of degree 0

\gamma(C, C') : H(C) \otimes_A H(C') \to H(C \otimes_A C').

Prop. 6 of II, p. 59, can be reformulated as follows:

Proposition 1

If the complexes $C$ and $C'$ are zero on the right, $C \otimes_A C'$ is zero on the right and the canonical $k$ -linear mapping*

\gamma_{0,0}(C, C') : H_0(C) \otimes_A H_0(C') \to H_0(C \otimes_A C')

is bijective.

Let $u : (C, d) \to (C_1, d_1)$ be a morphism of complexes of right $A$ -modules and $u' : (C', d') \to (C'_1, d'_1)$ a morphism of complexes of left $A$ -modules; then $u \otimes u' : C \otimes_A C' \to C_1 \otimes_A C'_1$ is a morphism of complexes of $k$ -modules; indeed, it is graded of degree 0, and if we denote by $D$ and $D_1$ the differentials of $C \otimes C'$ and $C_1 \otimes C'_1$ , we have for $p, q \in \mathbf{Z}, x \in C_p, x' \in C'_q$ ,

(u \otimes u') (D(x \otimes x')) = u(dx) \otimes u'(x') + (-1)^p u(x) \otimes u'(d'x') = = d_1 u(x) \otimes u'(x') + (-1)^p u(x) \otimes d'_1 u'(x') = D_1(u(x) \otimes u'(x')) .

Moreover the following diagram is commutative :

\begin{array}{ccc} H(C) \otimes_A H(C') & \xrightarrow{\gamma(C, C')} & H(C \otimes_A C') \\ H(u) \otimes H(u') \downarrow & & \downarrow H(u \otimes u') \\ H(C_1) \otimes_A H(C'_1) & \xrightarrow{\gamma(C_1, C'_1)} & H(C_1 \otimes_A C'_1) \end{array}

Let $A^\circ$ be the $k$ -algebra opposite to $A$ , $C^\circ$ (resp. ${C'}^\circ$ ) the complex $C$ (resp. $C'$ ) considered as a complex of left $A^\circ$ -modules (resp. right). Let us denote by

\sigma(C, C') : C \otimes_A C' \to {C'}^\circ \otimes_{A^\circ} C^\circ

the unique graded

k

-linear mapping of degree 0 such that, for

x \in C_p, x' \in C'_q, p, q \in \mathbf{Z}

, we have

\sigma(C, C') (x \otimes x') = (-1)^{pq} x' \otimes x .

Proposition 2

The mapping $\sigma(C, C') : C \otimes_A C' \to {C'}^\circ \otimes_A C^\circ$ is an isomorphism of complexes of $k$ -modules, whose reciprocal isomorphism is $\sigma({C'}^\circ, C^\circ)$ .

Since the mappings $\sigma(C, C')$ and $\sigma({C'}^\circ, C^\circ)$ are reciprocal to each other, it suffices to prove that $\sigma(C, C')$ is a morphism of complexes. Now, for

x \in C_p = C_p^\circ, \quad x' \in C'_q = C_{q'}^\circ, \quad p, q \in \mathbf{Z},

we have, denoting by

D'

the differential of

C' \otimes_{A^\circ} C

\begin{align*} \sigma(C, C') \circ D(x \otimes x') &= \sigma(C, C') (dx \otimes x' + (-1)^p x \otimes d'x') = \\ &= (-1)^{(p+1)q} x' \otimes dx + (-1)^{p+p(q+1)} d'x' \otimes x = (-1)^{pq} d'x' \otimes x + (-1)^{pq+q} x' \otimes dx \\ &= (-1)^{pq} D'(x' \otimes x) = D^\circ \circ \sigma(C, C') (x \otimes x'); \end{align*}

which gives

\sigma(C, C') \circ D = D^\circ \circ \sigma(C, C')

, whence the assertion sought.

The isomorphism $\sigma(C, C') : C \otimes_A C' \to {C'}^\circ \otimes_A C^\circ$ is called the commutation isomorphism of the tensor product of the complexes $C$ and $C'$ .

If $u : C \to C_1$ and $v : C' \to C'_1$ are two morphisms of complexes as above, we have a commutative diagram :

\begin{array}{ccc} C \otimes_A C' & \xrightarrow{\sigma(C, C')} & {C'}^\circ \otimes_A C^\circ \\ u \otimes u' \downarrow & & \downarrow u' \otimes u \\ C_1 \otimes_A C'_1 & \xrightarrow{\sigma(C_1, C'_1)} & C'_1{}^\circ \otimes_A C_1^\circ. \end{array}

Suppose for the end of this number that the ring $A$ is commutative (cf. No. 9 for the general case).

Let $C, C', C''$ be three complexes of $A$ -modules; the canonical homomorphism of $A$ -modules (III, p. 64)

\varphi : (C \otimes_A C') \otimes_A C'' \to C \otimes_A (C' \otimes_A C'')

is an isomorphism of complexes, as is immediately verified using the definitions.

More generally, let $(C^{(i)}, d^{(i)})_{i \in I}$ be a family of complexes of $A$ -modules, where the set $I$ is finite and totally ordered; we shall identify $I$ with the interval $[1, r]$ of $\mathbf{N}$ to simplify the notation. Let us endow the $A$ -module $C = \bigotimes_{i=1}^r C^{(i)}$ with the grading such that

C_n = \sum_{p_1 + p_2 + \cdots + p_r = n} (C^{(1)})_{p_1} \otimes (C^{(2)})_{p_2} \otimes \cdots \otimes (C^{(r)})_{p_r},

and define a graded

A

-endomorphism of degree

(-1)

C

D(x_1 \otimes \ldots \otimes x_r) = \sum_{j=1}^r (-1)^{p_1 + \cdots + p_{j-1}} x_1 \otimes \ldots \otimes x_{j-1} \otimes d_j x_j \otimes x_{j+1} \otimes \ldots \otimes x_r

where

x_i \in (C^{(i)})_{p_i}

for

i = 1, \ldots, n

. Then

(C, D)

is a complex of

A

-modules called the tensor product complex of the family

(C_i, d_i)

. For every strictly increasing sequence

r_0, ..., r_k

[0, r]

such that

r_0 = 0,\ r_k = r

, the canonical associativity isomorphism

\bigotimes_{j=0}^{k-1} \left( \bigotimes_{i=r_j+1}^{r_{j+1}} C^{(i)} \right) \to \bigotimes_{i=1}^r C^{(i)}

is an isomorphism of complexes.

We define as above a graded homomorphism of degree 0

\gamma((C^{(i)})) : \bigotimes_{i \in I} H(C^{(i)}) \to H \left( \bigotimes_{i \in I} C^{(i)} \right)

Remark 3

One can define the tensor product of a finite family of complexes without endowing the index set with a total order (X, p. 185, Exercise 3).

Remark 4

Suppose that each $C^{(i)}$ is endowed with a graded algebra structure compatible with its grading and such that the $d^{(i)}$ are antiderivations (III, p. 117). Let us then endow $\bigotimes_{i \in I} C^{(i)}$ with the left graded tensor product algebra structure of the given structures (III, p. 49). Then $D$ is an antiderivation. Indeed, using the associativity of the tensor product, we may assume that $I = \{1, 2\}$ ; let then $p_1,\ q_1,\ p_2,\ q_2 \in \mathbf{Z},\ x_1 \in (C^{(1)})_{p_1},\ y_1 \in (C^{(1)})_{q_1},\ x_2 \in (C^{(2)})_{p_2},\ y_2 \in (C^{(2)})_{q_2}$ ; we have

(D(x_1 \otimes x_2))(y_1 \otimes y_2) + (-1)^{p_1+p_2} (x_1 \otimes x_2)(D(y_1 \otimes y_2)) =

= (dx_1 \otimes x_2 + (-1)^{p_1} x_1 \otimes dx_2)(y_1 \otimes y_2) +

+ (-1)^{p_1+p_2} (x_1 \otimes x_2)(dy_1 \otimes y_2 + (-1)^{q_1} y_1 \otimes dy_2) =

= (-1)^{p_2 q_1} (dx_1) y_1 \otimes x_2 y_2 + (-1)^{p_1+(p_2-1)q_1} x_1 y_1 \otimes (dx_2) y_2 +

+ (-1)^{p_1+p_2+p_2(q_1-1)} x_1 dy_1 \otimes x_2 y_2 + (-1)^{p_1+p_2+q_1+p_2 q_1} x_1 y_1 \otimes x_2 dy_2

= (-1)^{p_2 q_1} [(dx_1) y_1 + (-1)^{p_1} x_1 dy_1] \otimes x_2 y_2 +

+ (-1)^{p_1+q_1+p_2 q_1} x_1 y_1 \otimes ((dx_2) y_2 + (-1)^{p_2} x_2 dy_2)

= (-1)^{p_2 q_1} [d(x_1 y_1) \otimes x_2 y_2 + (-1)^{p_1+q_1} x_1 y_1 \otimes d(x_2 y_2)]

= (-1)^{p_2 q_1} D(x_1 y_1 \otimes x_2 y_2) = D((x_1 \otimes x_2)(y_1 \otimes y_2)).

2. Tensor products and homotopy

Proposition 3

Let $C,\ C_1$ be two complexes of right $A$ -modules, $C',\ C'_1$ two complexes of left $A$ -modules, and $u : C \to C_1,\ v : C \to C_1,\ u' : C' \to C'_1,\ v' : C' \to C'_1$ be morphisms of complexes.

a) If $u$ and $u'$ are homotopic to $v$ and $v'$ respectively, then the two morphisms $u \otimes u'$ and $v \otimes v'$ from $C \otimes_A C'$ to $C_1 \otimes_A C'_1$ are homotopic.

b) If $u$ and $u'$ are homotopy equivalences, $u \otimes u'$ is a homotopy equivalence.

c) If $C$ or $C'$ is homotopic to zero, $C \otimes_A C'$ is homotopic to zero.

Let us denote by the same letter $d$ the differentials of the complexes $C,\ C_1,\ C',\ C'_1$ and by $D$ the differentials of the complexes $C \otimes_A C'$ and $C_1 \otimes_A C'_1$ .

If $u$ (resp. $u'$ ) is homotopic to $v$ (resp. $v'$ ), there exists a graded homomorphism of degree 1 $s:C\to C_1$ (resp. $s':C'\to C'_1$ ) such that

(2)\qquad u-v=ds+sd\qquad\text{(resp. }u'-v'=ds'+s'd\text{)}.

Let $S:C\otimes_A C'\to C_1\otimes_A C'_1$ be the unique graded homomorphism of degree 1 such that, for $x\in C_p,\ y\in C'_q,\ p,q\in\mathbf Z$ , we have

(3)\qquad S(x\otimes y)=s(x)\otimes u'(y)+(-1)^p v(x)\otimes s'(y).

Then, with the previous notations:

\begin{aligned} (DS+SD)(x\otimes y)&=D(sx\otimes u'y)+(-1)^pD(vx\otimes s'y)+S(dx\otimes y)\\ &\quad+(-1)^pS(x\otimes dy)=\\ &=dsx\otimes u'y+(-1)^{p+1}sx\otimes du'y+(-1)^pdvx\otimes s'y+vx\otimes ds'y\\ &\quad+sdx\otimes u'y+(-1)^{p-1}vdx\otimes s'y+(-1)^psx\otimes u'dy+vx\otimes s'dy\\ &=(ds+sd)(x)\otimes u'y+vx\otimes(ds'+s'd)(y)\\ &=(ux-vx)\otimes u'y+vx\otimes(u'y-v'y)=ux\otimes u'y-vx\otimes v'y. \end{aligned}

This gives $DS+SD=u\otimes u'-v\otimes v'$ , whence a).

Let us prove b). If $u$ and $u'$ are homotopy equivalences, there exist homomorphisms of complexes $\alpha:C_1\to C$ and $\alpha':C'_1\to C'$ such that $u\circ\alpha,\ \alpha\circ u',\ \alpha'\circ u'$ are homotopic respectively to $\mathrm{Id}_{C_1},\ \mathrm{Id}_C,\ \mathrm{Id}_{C'_1},\ \mathrm{Id}_{C'}$ . Then $(u\otimes u')\circ(\alpha\otimes\alpha')$ , which is equal to $(u\circ\alpha)\otimes(u'\circ\alpha')$ , is homotopic by a) to $\mathrm{Id}_{C_1}\otimes\mathrm{Id}_{C'_1}=\mathrm{Id}_{C_1\otimes C'_1}$ , while $(\alpha\otimes\alpha')\circ(u\otimes u')$ is homotopic to $\mathrm{Id}_{C\otimes C'}$ , whence b). Finally, c) follows from b) applied to the case where $C_1$ or $C'_1$ is zero.

Corollary 1

Let $C'$ be a split complex of left $A$ -modules such that $H(C')$ is flat. For every complex $C$ of right $A$ -modules, the canonical mapping

\gamma(C,C'):H(C)\otimes_AH(C')\longrightarrow H(C\otimes_A C')

is bijective.

By X, p. 35, def. 6, there exists a homotopy equivalence $u':C'\to H(C')$ . By prop. 3, $1_C\otimes u':C\otimes_A C'\to C\otimes_AH(C')$ is a homotopy equivalence; since

H(1_C\otimes u')\circ\gamma(C,C')=\gamma(C,H(C'))\circ(1_C\otimes H(u')),

and $H(1_C\otimes u')$ and $H(u')$ are bijective, it suffices to prove that $\gamma(C,H(C'))$ is bijective, and we are reduced to the case where $C'$ is flat and has zero differential. In this case the canonical exact sequences

\mathrm{(I)}\qquad 0\longrightarrow Z(C)\xrightarrow{i}C\xrightarrow{\partial}B(C)\longrightarrow0

\mathrm{(II)}\qquad 0\longrightarrow B(C)\xrightarrow{j}Z(C)\xrightarrow{\pi}H(C)\longrightarrow0

give exact sequences:

0 \longrightarrow Z(C) \otimes_A C' \xrightarrow{i\otimes 1} C \otimes_A C' \xrightarrow{\delta\otimes 1} B(C) \otimes_A C' \longrightarrow 0

0 \longrightarrow B(C) \otimes_A C' \xrightarrow{j\otimes 1} Z(C) \otimes_A C' \xrightarrow{\pi\otimes 1} H(C) \otimes_A C' \longrightarrow 0.

Since $d=i\circ j\circ\delta$ , we have $D=d\otimes 1_{C'}=(i\otimes 1)\circ(j\otimes 1)\circ(\delta\otimes 1)$ , which shows that the canonical mappings $Z(C)\otimes_A C'\to Z(C\otimes_A C')$ and $B(C)\otimes_A C'\to B(C\otimes_A C')$ are bijective, hence also $\gamma(C,C')$ by passing to quotients.

Corollary 2

Let $\mathbf N$ be a flat left $A$ -module. For every complex $C$ of right $A$ -modules, the canonical homomorphisms

\gamma_n(C,\mathbf N): H_n(C)\otimes_A \mathbf N \longrightarrow H_n(C\otimes_A \mathbf N)

are bijective.

Corollary 3

Let $C'$ be a complex of left $A$ -modules such that $B(C')$ and $H(C')$ are projective. For every complex $C$ of right $A$ -modules, the mapping $\gamma(C,C')$ is bijective.

Indeed, $C'$ is split (X, p. 35, example 4) and $H(C')$ is projective; therefore we can apply corollary 1.

Remark 1

Using the commutation isomorphisms, one deduces from corollaries 1, 2 and 3 the analogous statements obtained by exchanging the roles of the two arguments of the tensor products.

Remark 2

We shall see below (X, p. 79, Cor. 4) that the conclusion of Cor. 1 is also true when one assumes $C'$ and $H(C')$ flat and $C'$ bounded on the right.

3. Tensor product by a flat complex bounded on the right

Lemma 1

Let C be a complex of right A-modules and E a complex of left A-modules. We assume that $H(C)=0$ and that E is flat and bounded on the right. Then $H(C\otimes_A E)=0$ .

For $k\in\mathbf Z$ , let $T^{(k)}$ be the subcomplex of $C\otimes_A E$ such that

T_n^{(k)}=\sum_{\substack{p+q=n\\q\leq k}} C_p\otimes_A E_q\ ;

then $T^{(k-1)}\subset T^{(k)}$ and we have an exact sequence of complexes

0\longrightarrow T^{(k-1)}\xrightarrow{i_k}T^{(k)}\xrightarrow{\pi}C\otimes_A E_k(-k)\longrightarrow 0

where $i_k$ is the canonical injection and where $\pi$ projects the preceding direct sum onto its factor $C_{n-k}\otimes_A E_k=(C\otimes_A E_k(-k))_n$ . By Cor. 2 above, we have $H(C\otimes_A E_k(-k))=0$ , hence $i_k$ is a homomorphism. We have $T^{(k)}=0$ for $k$ sufficiently small, since $E$ is bounded on the right, hence $H(T^{(k)})=0$ for all $k$ by induction on $k$ . Finally, the canonical morphism $\underset{\longrightarrow}{\lim}\,T^{(k)}\to C\otimes_A E$ is obviously an isomorphism, hence $H(C\otimes_A E)=0$ (X, p. 28, Prop. 1).

Lemma 2

If $u:C\to C'$ is a morphism of complexes of right A-modules and $E$ a complex of left A-modules, then the complexes $\operatorname{Con}(u)\otimes_A E$ and $\operatorname{Con}(u\otimes 1_E)$ are isomorphic.

By definition, $\operatorname{Con}(u)\otimes_A E$ is the graded module $(C'(-1)\oplus C)\otimes_A E$ endowed with the differential $D$ such that, for $x\in C_p$ , $y'\in C'(-1)_p=C'_{p-1}$ , $z\in E_q$ , we have

(4)

D(y',x)\otimes z=(-dy',dx-u(y'))\otimes z+(-1)^p(y',x)\otimes dz,

while $\operatorname{Con}(u\otimes 1_E)$ is the graded module $(C'\otimes_A E)(-1)\oplus(C\otimes_A E)$ endowed with the differential $D_1$ such that, for $x\in C_p$ , $y'\in C'_{p-1}$ , $z\in E_q$ , we have

D_1(y'\otimes z,x\otimes z)=(-dy'\otimes z-(-1)^{p-1}y'\otimes dz,dx\otimes z+(-1)^p x\otimes dz-u(y')\otimes z)

=(-dy'\otimes z,(dx-u(y'))\otimes z)+(-1)^p(y'\otimes dz,x\otimes dz),

whence the assertion.

Proposition 4

Let $u:C\to C'$ be a homology isomorphism of complexes of right A-modules and $E$ a complex of left A-modules, flat and bounded on the right. Then

u\otimes 1_E:C\otimes_A E\to C'\otimes_A E

is a homology isomorphism of complexes of $k$ -modules.

Indeed, by X, p. 38, Cor., $u$ (resp. $u\otimes 1_E$ ) is a homology isomorphism if and only if $H(\operatorname{Con}(u))=0$ (resp. $H(\operatorname{Con}(u\otimes 1_E))=0$ ). The conclusion then follows from Lemmas 1 and 2.

Remark

Using the commutation isomorphisms, one deduces from the preceding statements the analogous statements obtained by exchanging the roles of the two arguments of the tensor products.

4. Definition and first properties of the torsion product

For every A-module $E$ , we denote by $p_E:L(E)\to E$ the canonical free resolution of $E$ (X, p. 50).

Definition 1

Let $M$ be a right A-module and $N$ a left A-module. We call the torsion product of $M$ and $N$ the graded $k$ -module

(4)

\operatorname{Tor}^A(M,N)=H(L(M)\otimes_A L(N)).

The homogeneous components of $\operatorname{Tor}^A(M,N)$ are denoted

(5)

\operatorname{Tor}^A_n(M,N)=H_n(L(M)\otimes_A L(N)).

Since $L(M)$ and $L(N)$ are zero on the right, we have

(6)

\operatorname{Tor}_n^A(M, N) = 0 \quad \text{for } n < 0 .

Remark 1

We will see below (X, p. 107, Proposition 6) finiteness properties of the modules $\operatorname{Tor}^A(M, N)$ . For example, if $A$ is commutative noetherian and if $M$ and $N$ are $A$ -modules of finite type, every $A$ -module $\operatorname{Tor}_n^A(M, N)$ is of finite type.

Let $f : M \to M'$ be a homomorphism of right $A$ -modules and $g : N \to N'$ a homomorphism of left $A$ -modules. We define $\operatorname{Tor}^A(f, g) = H(L(f) \otimes_A L(g))$ ; this is a homomorphism of graded $k$ -modules

\operatorname{Tor}^A(f, g) : \operatorname{Tor}^A(M, N) \to \operatorname{Tor}^A(M', N')

whose homogeneous components are denoted

\operatorname{Tor}_n^A(f, g) : \operatorname{Tor}_n^A(M, N) \to \operatorname{Tor}_n^A(M', N') .

According to Proposition 1 of X, p. 62, the canonical homomorphism

\gamma_{0,0} : H_0(L(M)) \otimes_A H_0(L(N)) \to H_0(L(M) \otimes_A L(N))

is bijective; using the isomorphisms $M \to H_0(L(M))$ and $N \to H_0(L(N))$ , we deduce from it an isomorphism, called *canonical

(7)

\gamma_{M,N} : M \otimes_A N \to \operatorname{Tor}_0^A(M, N) .

We will always identify $\operatorname{Tor}_0^A(M, N)$ with $M \otimes_A N$ through this isomorphism. Hence the $k$ -linear map $\operatorname{Tor}_0^A(f, g)$ is identified with $f \otimes g$ .

Remark 2

The complex morphism $p_M \otimes p_N : L(M) \otimes_A L(N) \to M \otimes_A N$ induces on the degree 0 homology the isomorphism

\gamma_{M,N}^{-1} : \operatorname{Tor}_0^A(M, N) \to M \otimes_A N

inverse to $\gamma_{M,N}$ .

We have $L(1_M) = 1_{L(M)}$ , $L(1_N) = 1_{L(N)}$ , hence by passage to the homology:

(8)

\operatorname{Tor}^A(1_M, 1_N) = 1_{\operatorname{Tor}^A(M, N)} .

If $f' : M' \to M''$ (resp. $g' : N' \to N''$ ) is a homomorphism of right $A$ -modules (resp. left $A$ -modules), we have $L(g' \circ g) = L(g') \circ L(g)$ and $L(f' \circ f) = L(f') \circ L(f)$ , hence

(9)

\operatorname{Tor}^A(f' \circ f, g' \circ g) = \operatorname{Tor}^A(f', g') \circ \operatorname{Tor}^A(f, g) .

Consider the morphisms of $k$ -complexes

L(M) \otimes_A N \xleftarrow{1 \otimes p_N} L(M) \otimes_A L(N) \xrightarrow{p_M \otimes 1} M \otimes_A L(N)

and the $k$ -homomorphisms they induce in homology:

H(L(M) \otimes_A N) \xleftarrow{\psi_M(N)} \mathrm{Tor}^A(M, N) \xrightarrow{\overline{\psi}_N(M)} H(M \otimes_A L(N)) ;

according to Proposition 4 of X, p. 67, $1 \otimes p_N$ and $p_M \otimes 1$ are homology morphisms. Therefore:

Proposition 5. — The $k$ -homomorphisms

\begin{aligned} & \psi_M(N) : \mathrm{Tor}^A(M, N) \to H(L(M) \otimes_A N) \\ & \overline{\psi}_N(M) : \mathrm{Tor}^A(M, N) \to H(M \otimes_A L(N)) \end{aligned}

are bijective.

Corollary. — If $M$ or $N$ is flat, $\mathrm{Tor}_i^A(M, N) = 0$ for $i \geqslant 0$ .

Suppose that $N$ (resp. $M$ ) is flat. Then $p_M \otimes 1 : L(M) \otimes_A N \to M \otimes_A N$ (resp. $1 \otimes p_N : M \otimes_A L(N) \to M \otimes_A N$ ) is a homology morphism (X, p. 67, Proposition 4), hence $H_i(L(M) \otimes_A N)$ (resp. $H_i(M \otimes_A L(N))$ ) vanishes for $i > 0$ .

Remark 3. — If $g : N \to N'$ is a homomorphism of left A-modules, then

(1_{L(M)} \otimes g) \circ (1_{L(M)} \otimes 1_N) = (1_{L(M)} \otimes 1_N) \circ (1_{L(M)} \otimes L(g)) ,

therefore the diagram

\begin{array}{ccc} \mathrm{Tor}^A(M, N) & \xrightarrow{\psi_{M(N)}} & H(L(M) \otimes_A N) \\ \mathrm{Tor}^A(1, g) \downarrow & & \downarrow H(1 \otimes g) \\ \mathrm{Tor}^A(M, N') & \xrightarrow{\psi_{M(N')}} & H(L(M) \otimes_A N') \end{array}

is commutative.

Analogously, if $f : M \to M'$ is a homomorphism of right A-modules, one has a commutative diagram :

\begin{array}{ccc} \mathrm{Tor}^A(M, N) & \xrightarrow{\overline{\psi}_{N(M)}} & H(M \otimes_A L(N)) \\ \mathrm{Tor}^A(f, 1) \downarrow & & \downarrow H(f \otimes 1) \\ \mathrm{Tor}^A(M', N) & \xrightarrow{\overline{\psi}_{N(M')}} & H(M' \otimes_A L(N)) . \end{array}

Proposition 6. — The mapping $(f, g) \mapsto \mathrm{Tor}^A(f, g) :$

\mathrm{Hom}_A(M, M') \times \mathrm{Hom}_A(N, N') \to \mathrm{Hom}_k(\mathrm{Tor}^A(M, N), \mathrm{Tor}^A(M', N'))

is $k$ *-bilinear.

Let $f \in \mathrm{Hom}_A(M, M')$ , $g_1, g_2 \in \mathrm{Hom}_A(N, N')$ , $\lambda_1, \lambda_2 \in k$ . Then the morphisms

\lambda_1(L(f) \otimes g_1) + \lambda_2(L(f) \otimes g_2) \quad \text{and} \quad L(f) \otimes (\lambda_1 g_1 + \lambda_2 g_2)

from $L(M) \otimes_A N$ into $L(M) \otimes_A N'$ coincide; by prop. 5 and Remark 3, one therefore has

(10)

\mathrm{Tor}^A(f, \lambda_1 g_1 + \lambda_2 g_2) = \lambda_1 \mathrm{Tor}^A(f, g_1) + \lambda_2 \mathrm{Tor}^A(f, g_2).

One reasons analogously for the mapping $f \mapsto \mathrm{Tor}^A(f, g)$ .

Corollary

Let $\lambda \in k$ . If $\lambda$ annihilates $M$ or $N$ , it annihilates $\mathrm{Tor}^A(M, N)$ .

Indeed, $\lambda . 1_{\mathrm{Tor}(M, N)} = \mathrm{Tor}(\lambda . 1_M, 1_N) = \mathrm{Tor}(1_M, \lambda . 1_N)$ .

Proposition 7

Let I and J be two sets, $(M_\alpha)_{\alpha \in I}$ a family of right A-modules, $(N_\beta)_{\beta \in J}$ a family of left A-modules. The homomorphism

\bigoplus_{\alpha \in I, \beta \in J} \mathrm{Tor}^A(M_\alpha, N_\beta) \to \mathrm{Tor}^A\left( \bigoplus_{\alpha \in I} M_\alpha, \bigoplus_{\beta \in J} N_\beta \right)

deduced from the canonical homomorphisms $M_\alpha \to \bigoplus M_\alpha$ and $N_\beta \to \bigoplus N_\beta$ is bijective.

It suffices to prove that for every right module $M$ (resp. every left module $N$ ), the canonical homomorphism

\bigoplus_{\beta \in J} \mathrm{Tor}^A(M, N_\beta) \to \mathrm{Tor}^A(M, \bigoplus_{\beta \in J} N_\beta)

(resp. $\bigoplus_{\alpha \in I} \mathrm{Tor}^A(M_\alpha, N) \to \mathrm{Tor}^A(\bigoplus_{\alpha \in I} M_\alpha, N)$ ) is bijective. Now this follows from the preceding, from Proposition 1 of X, p. 28, and from the canonical isomorphisms :

\bigoplus_{\beta} (L(M) \otimes_A N_\beta) \to L(M) \otimes_A (\bigoplus_{\beta} N_\beta),

\bigoplus_{\alpha} (M_\alpha \otimes_A L(N)) \to (\bigoplus_{\alpha} M_\alpha) \otimes_A L(N).

An analogous reasoning gives :

Proposition 8

Let I (resp. J) be a right filtered preordered set, (( $M_\alpha$ , $u_{\alpha', \alpha}$ ) (resp. (( $N_\beta$ , $v_{\beta', \beta}$ ))) an inductive system of right A-modules (resp. left) relative to I (resp. J). The homomorphism of graded k-modules

\lim_{\longrightarrow (\alpha, \beta) \in I \times J} \mathrm{Tor}^A(M_\alpha, N_\beta) \to \mathrm{Tor}^A\left( \lim_{\longrightarrow \alpha \in I} M_\alpha, \lim_{\longrightarrow \beta \in J} N_\beta \right),

deduced from the canonical A-homomorphisms $M_\alpha \to \lim_{\longrightarrow} M_\alpha$ and $N_\beta \to \lim_{\longrightarrow} N_\beta$ , is bijective.

In particular, taking $J = I$ and remarking that the $(\alpha, \alpha), \alpha \in I$ , form a cofinal subset of $I \times I$ , one obtains :

Corollary

Let I be a right filtered preordered set, $(M_i, u_{ji})$ (resp. $(N_i, v_{ji})$ ) an inductive system of right $A$ -modules (resp. left) relative to I. The homomorphism of graded $k$ -modules

\lim_{\longrightarrow i \in I} \operatorname{Tor}^A(M_i, N_i) \to \operatorname{Tor}^A\left(\lim_{\longrightarrow i \in I} M_i, \lim_{\longrightarrow i \in I} N_i\right),

deduced from the canonical $A$ -homomorphisms $M_i \to \lim_{\longrightarrow} M_i$ and $N_j \to \lim_{\longrightarrow} N_j$ is bijective.

5. Linking Homomorphisms and Exact Sequences

Let $M$ be a right $A$ -module, $N$ a left $A$ -module, $A^\circ$ the opposite ring of $A$ , $M^\circ$ the $A^\circ$ -left module underlying $M$ , $N^\circ$ the $A^\circ$ -right module underlying $N$ . We have $L(M^\circ ) = L(M)^\circ$ and $L(N^\circ ) = L(N)^\circ$ , whence a commutation isomorphism (X, p. 63, prop. 2)

\sigma(L(M), L(N)) : L(M) \otimes_A L(N) \to L(N^\circ) \otimes_{A^\circ} L(M^\circ).

Passing to homology, $\sigma(L(M), L(N))$ induces a graded isomorphism of degree 0 $\sigma_{M,N} : \operatorname{Tor}^A(M, N) \to \operatorname{Tor}^{A^\circ}(N^\circ, M^\circ)$ called the commutation isomorphism of torsion products.

Let us note that $\sigma_{N^\circ, M^\circ} \circ \sigma_{M,N} = \mathrm{Id}_{\operatorname{Tor}(M,N)}$ and that $\sigma_{M,N}$ induces on the degree 0 terms the commutation homomorphism of the tensor product. On the other hand, if $f : M \to M'$ and $g : N \to N'$ are homomorphisms of $A$ -modules, then

\operatorname{Tor}^{A^\circ}(g, f) \circ \sigma_{M,N} = \sigma_{M', N'} \circ \operatorname{Tor}^A(f, g).

5. The linking homomorphisms and exact sequences

Let $M$ be a right $A$ -module. Let us recall that for every left $A$ -module $N$ , an isomorphism

\psi_M(N) : \operatorname{Tor}^A(M, N) \to H(L(M) \otimes_A N)

was defined in the previous section (X, p. 69, prop. 5). Let

(\mathscr{E})

0 \to N' \xrightarrow{u} N \xrightarrow{v} N'' \to 0

be an exact sequence of left $A$ -modules; then the sequence of $k$ -complexes

(\mathscr{ME})

0 \longrightarrow L(M) \otimes_A N' \xrightarrow{1 \otimes u} L(M) \otimes_A N \xrightarrow{1 \otimes v} L(M) \otimes_A N'' \longrightarrow 0

is then exact (X, p. 66, Lemma 1); denote by

\partial^{(\mathscr{ME})} : H(L(M) \otimes_A N'') \to H(L(M) \otimes_A N')

the corresponding linking homomorphism (X, p. 29).

Definition 2

The linking homomorphism of torsion products, relative to the module $M$ and the exact sequence $\mathscr{E}$ , is the composite homomorphism

\partial(M, \mathscr{E}) = \psi_M(N')^{-1} \circ \partial^{(M\mathscr{E})} \circ \psi_M(N'') : \mathrm{Tor}^A(M, N'') \to \mathrm{Tor}^A(M, N')

This is a graded $k$ -homomorphism of degree $-1$ , whose homogeneous components are denoted $\partial_n(M, \mathscr{E}) : \mathrm{Tor}_n^A(M, N'') \to \mathrm{Tor}_{n-1}^A(M, N')$ .

Theorem 1

The left-infinite sequence of $k$ -module homomorphisms

\cdots \longrightarrow \mathrm{Tor}_n^A(M, N') \xrightarrow{\mathrm{Tor}_n^A(1, u)} \mathrm{Tor}_n^A(M, N) \xrightarrow{\mathrm{Tor}_n^A(1, v)} \mathrm{Tor}_n^A(M, N'')

\xrightarrow{\partial_n(M, \mathscr{E})} \mathrm{Tor}_{n-1}^A(M, N') \xrightarrow{\mathrm{Tor}_{n-1}^A(1, u)} \cdots \xrightarrow{\mathrm{Tor}_1^A(1, v)} \mathrm{Tor}_1^A(M, N'')

\xrightarrow{\hat{c}_1(M, \mathscr{E})} M \otimes_A N' \xrightarrow{1 \otimes u} M \otimes_A N \xrightarrow{1 \otimes v} M \otimes_A N'' \longrightarrow 0

is exact.

To see this, consider the diagram

\begin{array}{ccccccccc} \mathrm{Tor}(M, N') & \xrightarrow{\mathrm{Tor}(1, u)} & \mathrm{Tor}(M, N) & \xrightarrow{\mathrm{Tor}(1, v)} & \mathrm{Tor}(M, N'') & \xrightarrow{\partial(M, \mathscr{E})} & \mathrm{Tor}(M, N') & \xrightarrow{\mathrm{Tor}(1, u)} & \mathrm{Tor}(M, N) \\ \psi_M(N') \downarrow & & \psi_M(N) \downarrow & & \psi_M(N'') \downarrow & & \psi_M(N') \downarrow & & \psi_M(N') \downarrow \\ H(L(M) \otimes N') & \xrightarrow{H(1 \otimes u)} & H(L(M) \otimes N) & \xrightarrow{H(1 \otimes v)} & H(L(M \otimes N'') & \xrightarrow{\partial^{(M\mathscr{ME})}} & H(L(M) \otimes N') & \xrightarrow{H(1 \otimes u)} & H(L(M) \otimes N) \end{array}

It commutes according to (X, p. 69, Remark 3) and Definition 2. Furthermore, the bottom row is exact (X, p. 30, Thm. 1), and the different $\psi_M$ are bijections (X, p. 69, prop. 5).

Corollary 1

If $\mathrm{Tor}_1^A(M, N'') = 0$ , the sequence

0 \longrightarrow M \otimes_A N' \xrightarrow{1 \otimes u} M \otimes_A N \xrightarrow{1 \otimes v} M \otimes_A N'' \longrightarrow 0

is exact.

Corollary 2

Let $0 \to C' \xrightarrow{u} C \xrightarrow{v} C'' \to 0$ be an exact sequence of complexes of left $A$ -modules, and let $E$ be a complex of right $A$ -modules. If $C''$ or $E$ is flat, the sequence

0 \longrightarrow E \otimes_A C' \xrightarrow{1 \otimes u} E \otimes_A C \xrightarrow{1 \otimes v} E \otimes_A C'' \longrightarrow 0

is exact.

Indeed, $\mathrm{Tor}_1^A(E, C'') = 0$ according to X, p. 69, Corollary to prop. 5.

Example

Let $a$ be an ideal of $A$ . The exact sequence

0 \to a \to A_s \to A/a \to 0

of left a-modules, gives rise to an exact sequence of torsion products, in which the terms Tor $_i^A$ (M, A) are zero for $i > 0$ . From this one deduces isomorphisms

\operatorname{Tor}_{i+1}^A(M, A/\alpha) \to \operatorname{Tor}_i^A(M, \alpha), \quad i > 0

and an exact sequence

0 \to \operatorname{Tor}_1^A(M, A\alpha) \to M \otimes_A \alpha \to M \otimes_A A \to M \otimes A \alpha \to 0 :

it follows that $\operatorname{Tor}_1^A(M, A/\alpha)$ is identified with the kernel of the canonical homomorphism $M \otimes_A \alpha \to M$ .

For example, taking for M a module of the form $A_d/b$ , where b is a right ideal of A, one obtains an isomorphism of $\operatorname{Tor}_1^A(A/b, A/\alpha)$ onto $(\alpha \cap b)/ba$ .

Proposition 9

Let f : M → M $_1$ be a homomorphism of right a-modules and

\begin{array}{ccccccccc} 0 & \longrightarrow & N' & \xrightarrow{u} & N & \xrightarrow{v} & N'' & \longrightarrow & 0 \\ & & g' \downarrow & & g \downarrow & & g'' \downarrow & & \\ 0 & \longrightarrow & N'_1 & \xrightarrow{u_1} & N_1 & \xrightarrow{v_1} & N''_1 & \longrightarrow & 0 \end{array}

(E $_1$ )

a commutative diagram with exact rows of homomorphisms of left a-modules. The diagram of k-modules

\begin{array}{ccc} \operatorname{Tor}^A(M, N'') & \xrightarrow{\partial(M, E)} & \operatorname{Tor}^A(M, N') \\ \operatorname{Tor}^A(f, g'') \downarrow & & \operatorname{Tor}^A(f, g') \downarrow \\ \operatorname{Tor}^A(M_1, N''_1) & \xrightarrow{\partial(M_1, E_1)} & \operatorname{Tor}^A(M_1, N'_1) \end{array}

is commutative.

This follows from X, p. 31, prop. 2, applied to the commutative diagram

\begin{array}{ccccccccc} 0 & \longrightarrow & L(M) \otimes_A N' & \xrightarrow{1 \otimes u} & L(M) \otimes_A N & \xrightarrow{1 \otimes v} & L(M) \otimes_A N'' & \longrightarrow & 0 \\ & & L(f) \otimes g' \downarrow & & L(f) \otimes g \downarrow & & L(f) \otimes g'' \downarrow & & \\ 0 & \longrightarrow & L(M_1) \otimes_A N'_1 & \xrightarrow{1 \otimes u_1} & L(M_1) \otimes_A N_1 & \xrightarrow{1 \otimes v_1} & L(M_1) \otimes_A N''_1 & \longrightarrow & 0 . \end{array}

In an analogous way, if N is a left a-module and

(F)

0 \to M' \xrightarrow{r} M \xrightarrow{s} M'' \to 0

an exact sequence of right a-modules, one defines linking homomorphisms

\partial(F, N) : \operatorname{Tor}^A(M'', N) \to \operatorname{Tor}^A(M', N)

\partial_n(F, N) : \operatorname{Tor}_n^A(M'', N) \to \operatorname{Tor}_{n-1}^A(M', N)

by $\partial(\mathscr F,\mathbf N)=\overline{\psi}_{\mathbf N}(M')^{-1}\circ\partial(\mathscr F^{\mathbf N})\circ\overline{\psi}_{\mathbf N}(M'')$ , where $\partial(\mathscr F^{\mathbf N})$ is the linking homomorphism of the exact sequence

(\mathscr F_{\mathbf N})\qquad 0\longrightarrow M'\otimes_A L(N)\longrightarrow M\otimes_A L(N)\longrightarrow M''\otimes_A L(N)\longrightarrow 0

deduced from $\mathscr F$ , and one has:

THEOREM 1 bis. — The left-unbounded sequence of homomorphisms of $k$ -modules

\longrightarrow \operatorname{Tor}_n^A(M',N)\xrightarrow{\operatorname{Tor}_n^A(r,1)}\operatorname{Tor}_n^A(M,N)\xrightarrow{\operatorname{Tor}_n^A(s,1)}\operatorname{Tor}_n^A(M'',N)\xrightarrow{\partial_n(\mathscr F,N)}\operatorname{Tor}_{n-1}^A(M',N)

\cdots\longrightarrow \operatorname{Tor}_1^A(M'',N)\xrightarrow{\partial_1(\mathscr F,N)}M'\otimes_A N\xrightarrow{r\otimes1}M\otimes_A N\xrightarrow{s\otimes1}M''\otimes_A N\longrightarrow0

is exact.

We leave to the reader the task of stating and proving the properties analogous to the corollaries of th. 1 and to prop. 9. Moreover:

Proposition 10

Let us denote by $(\mathscr F^\circ)$ the exact sequence of left $A$ -modules

0\longrightarrow M'\xrightarrow{r}M\xrightarrow{s}M''\longrightarrow0.

The diagram

\begin{array}{ccccc} \operatorname{Tor}^A(M'',N)&\xrightarrow{\partial(\mathscr F,N)}&\operatorname{Tor}^A(M',N)\\ \Big\downarrow{\sigma_{M'',N}}&&\Big\downarrow{\sigma_{M',N}}\\ \operatorname{Tor}^{A^\circ}(N^\circ,{M''}^\circ)&\xrightarrow{\partial(N^\circ,\mathscr F^\circ)}&\operatorname{Tor}^{A^\circ}(N^\circ,{M'}^\circ) \end{array}

is commutative.

Indeed, this follows from X, p. 31, prop. 2, applied to the commutative diagram

\begin{array}{ccccccccc} 0&\longrightarrow&M'\otimes_A L(N)&\xrightarrow{r\otimes1}&M\otimes_A L(N)&\xrightarrow{s\otimes1}&M''\otimes_A L(N)&\longrightarrow&0\\ &&\Big\downarrow{\sigma(M',L(N))}&&\Big\downarrow{\sigma(M,L(N))}&&\Big\downarrow{\sigma(M'',L(N))}\\ 0&\longrightarrow&L(N^\circ)\otimes_{A^\circ}{M'}^\circ&\xrightarrow{1\otimes r}&L(N^\circ)\otimes_{A^\circ}M^\circ&\xrightarrow{1\otimes s}&L(N^\circ)\otimes_{A^\circ}{M''}^\circ&\longrightarrow&0 . \end{array}

We shall see later other commutation relations (cf. X, p. 131, cor. 1*).*

6. Flat modules and torsion products

Theorem 2

Let E be a right $A$ -module. The following conditions are equi-* *valent :

(i) E is flat ;

(ii) for every left $A$ -module F, and every integer $n>0$ , one has

\operatorname{Tor}_n^A(E,F)=0\,;

(iii) For every cyclic left $A$ -module $F$ of finite presentation, we have

\operatorname{Tor}_1^A(E, F) = 0 ;

(iv) For every left ideal

a

A

of finite type, the canonical mapping

E \otimes_A a \to E

is injective; (v) For every exact sequence of right

A

-modules of the form

0 \to G \xrightarrow{\nu} H \xrightarrow{w} E \to 0,

and every left

A

-module

F

, the sequence

0 \longrightarrow G \otimes_A F \xrightarrow{\nu \otimes 1} H \otimes_A F \xrightarrow{w \otimes 1} E \otimes_A F \longrightarrow 0

is exact. (i)

\Rightarrow

(ii): This is the corollary to Prop. 5 of X, p. 69. (ii)

\Rightarrow

(iii): This is trivial. (iii)

\Leftrightarrow

(iv): Every cyclic left

A

-module of finite presentation is isomorphic to a quotient

A/a

, where

a

is a left ideal of finite type, so (iii) is equivalent to (iv) according to X, p. 72, Example. (iii)

\Rightarrow

(i): According to X, p. 8, Prop. 3, X, p. 72, Th. 1,

E

is flat as soon as

\operatorname{Tor}_1^A(E, F) = 0

for every left

A

-module

F

. If (iii) is satisfied, this is the case when

F

is cyclic and of finite presentation. According to X, p. 11, Prop. 7, every

A

-module (resp. every cyclic

A

-module) is a filtered inductive limit of modules of finite presentation (resp. of cyclic modules of finite presentation); hence, according to X, p. 70, Prop. 8, it suffices to prove that if

\operatorname{Tor}_1^A(E, F) = 0

when

F

is cyclic, then this is also the case when

F

is of finite type. We therefore proceed by induction on the cardinal of a system of generators

(f_1, \ldots, f_n)

F

; the exact sequence

0 \to Af_1 \to F \to F/Af_1 \to 0

gives rise to an exact sequence

\operatorname{Tor}_1^A(E, Af_1) \to \operatorname{Tor}_1^A(E, F) \to \operatorname{Tor}_1^A(E, F/Af_1),

so that

\operatorname{Tor}_1^A(E, F) = 0

since

\operatorname{Tor}_1^A(E, Af_1) = 0

and

\operatorname{Tor}_1^A(E, F/Af_1) = 0

by the induction hypothesis. (i)

\Rightarrow

(v): This is Corollary 2 to Th. 1 (X, p. 72). (v)

\Rightarrow

(iii): The exact sequence (X, p. 50)

0 \longrightarrow Z_0(E) \xrightarrow{i_E} L_0(E) \xrightarrow{p_E} E \longrightarrow 0

gives rise, for every left

A

-module

F

, to an exact sequence

0 \longrightarrow \operatorname{Tor}_1^A(E, F) \longrightarrow Z_0(E) \otimes_A F \xrightarrow{i_E \otimes 1} L_0(E) \otimes_A F \xrightarrow{p_E \otimes 1} E \otimes_A F \longrightarrow 0.

If (v) is satisfied, we have

\operatorname{Tor}_1^A(E, F) = 0

, whence (iii).

Corollary 1

Let $0 \to E' \to E \to E'' \to 0$ be an exact sequence of right $A$ -modules. Assume that $E''$ is flat. Then $E$ is flat if and only if $E'$ is flat.

Let $F$ be a left $A$ -module. Since $\operatorname{Tor}^A_i(E'', F)=0$ for $i=1,2$ (Th. 2, (i) $\Rightarrow$ (ii)), we have an exact sequence

0\longrightarrow \operatorname{Tor}^A_1(E',F)\longrightarrow \operatorname{Tor}^A_1(E,F)\longrightarrow 0

whence the assertion (Th. 2, (i)

\Leftrightarrow

(iii)).

Corollary 2. — Let

0\longrightarrow E_n\longrightarrow E_{n-1}\longrightarrow\cdots\longrightarrow E_1\longrightarrow0

be an exact sequence of right $A$ -modules. If $E_i$ is flat for $i=1,\ldots,n-1$ , then $E_n$ is flat.

7. Künneth Formula

In this section, we consider a complex $(C,d)$ of right $A$ -modules and a complex $(C',d')$ of left $A$ -modules. Let (I)

0\longrightarrow Z(C)\xrightarrow{i}C\xrightarrow{\delta}B(C)(-1)\longrightarrow0,

(II)

0\longrightarrow B(C)\xrightarrow{i}Z(C)\xrightarrow{p}H(C)\longrightarrow0;

be the canonical exact sequences.

From $\delta$ we deduce a $k$ -homomorphism

H(\delta\otimes 1):H(C\otimes_A C')\longrightarrow H(B(C)\otimes_A C')(-1);

from (II) we deduce a connecting homomorphism

\partial(\mathrm{II},H(C')): \operatorname{Tor}^A_1(H(C),H(C'))\longrightarrow B(C)\otimes_A H(C');

if we equip $\operatorname{Tor}^A_1(H(C),H(C'))$ with the grading whose homogeneous component of degree $n$ is

\bigoplus_{p+q=n}\operatorname{Tor}^A_1(H_p(C),H_q(C')),

this connecting homomorphism is graded of degree 0. We also have a canonical homomorphism (X, p. 62)

\gamma(B(C),C'): B(C)\otimes_A H(C')\longrightarrow H(B(C)\otimes_A C').

With these notations:

Theorem 3. — *Assume that the $A$ -modules $B(C)$ and $Z(C)$ are flat. There exists a unique homomorphism of graded $k$ -modules, of degree $-1$ ,

\alpha:H(C\otimes_A C')\longrightarrow\operatorname{Tor}^A_1(H(C),H(C'))

making the following diagram commute*

\begin{array}{ccc} H(C\otimes_A C') & \xrightarrow{\alpha} & \operatorname{Tor}^A_1(H(C),H(C'))(-1)\\ {\scriptstyle H(\delta\otimes1)}\downarrow & & \downarrow{\scriptstyle\partial(\mathrm{II},H(C'))}\\ H(B(C)\otimes_A C')(-1) & \xleftarrow{\gamma(B(C),C')} & (B(C)\otimes_A H(C'))(-1). \end{array}

The sequence of graded $k$ -modules

0 \longrightarrow \mathrm{H}(C) \otimes_A \mathrm{H}(C') \xrightarrow{\gamma(C,C')} \mathrm{H}(C \otimes_A C') \xrightarrow{\alpha} \mathrm{Tor}_1^A(\mathrm{H}(C), \mathrm{H}(C'))(-1) \longrightarrow 0

is exact.

Thus, for each $n$ , we have an exact sequence

(11)

\begin{array}{cccccc} 0 & \longrightarrow & \bigoplus_{p+q=n} \mathrm{H}_p(C) \otimes_A \mathrm{H}_q(C') & \xrightarrow{\gamma_n(C,C')} & \mathrm{H}_n(C \otimes_A C') \\ & & & \xrightarrow{\alpha_n} & \bigoplus_{p+q=n-1} \mathrm{Tor}_1^A(\mathrm{H}_p(C), \mathrm{H}_q(C')) & \longrightarrow 0 . \end{array}

For simplicity, set $B = B(C)$ , $Z = Z(C)$ , $H = H(C)$ , and $H' = H(C')$ .

Since $B$ is flat, from (I) we deduce an exact sequence (X, p. 72, Corollary 2)

(12)

0 \longrightarrow Z \otimes_A C' \xrightarrow{j \otimes 1} C \otimes_A C' \xrightarrow{\delta \otimes 1} (B \otimes_A C')(-1) \longrightarrow 0 .

Lemma 3

The connecting homomorphism $\mathrm{H}(B \otimes_A C') \to \mathrm{H}(Z \otimes_A C')$ associated with the exact sequence (12) is equal to $\mathrm{H}(i \otimes 1)$ .

Indeed, let $a \in \mathrm{H}(B \otimes_A C')$ ; since $B$ is flat, $a$ belongs to the image of $B \otimes_A Z(C')$ , hence can be written as $\sum_\lambda da_\lambda \otimes b_\lambda$ , where $a_\lambda \in C$ , $b_\lambda \in C'$ , and $db_\lambda = 0$ . The image of the class of $a$ under the connecting homomorphism is, by definition, the class of $D(\sum a_\lambda \otimes b_\lambda) = \sum da_\lambda \otimes b_\lambda = (i \otimes 1)(a)$ , which proves the lemma.

The associated homology exact sequence to (12) is therefore

\mathrm{H}(B \otimes_A C') \xrightarrow{\mathrm{H}(i \otimes 1)} \mathrm{H}(Z \otimes_A C') \xrightarrow{\mathrm{H}(j \otimes 1)} \mathrm{H}(C \otimes_A C')

\xrightarrow{\mathrm{H}(\delta \otimes 1)} \mathrm{H}(B \otimes_A C')(-1) \xrightarrow{\mathrm{H}(i \otimes 1)} \mathrm{H}(Z \otimes_A C')(-1) .

On the other hand, since $Z$ is flat, from (II) we obtain a sequence of graded $k$ -modules

0 \longrightarrow \operatorname{Tor}_1^A(H, H') \xrightarrow{\partial(\mathrm{II}, H')} B \otimes_A H' \xrightarrow{i \otimes 1} Z \otimes_A H' \xrightarrow{p \otimes 1} H \otimes_A H' \longrightarrow 0 ;

finally, we have the canonical homomorphisms of No. 1

\gamma_B = \gamma(B, C') : B \otimes_A H' \to \mathrm{H}(B \otimes_A C')

\gamma_Z = \gamma(Z, C') : Z \otimes_A H' \to \mathrm{H}(Z \otimes_A C')

\gamma_C = \gamma(C, C') : H \otimes_A H' \to \mathrm{H}(C \otimes_A C') ,

whence a diagram of graded $k$ -modules, with exact rows:

\begin{array}{ccccccccc} B \otimes H' & \xrightarrow{i \otimes 1} & Z \otimes H' & \xrightarrow{p \otimes 1} & H \otimes H' & \longrightarrow & 0 \\ \downarrow \gamma_B & & \downarrow \gamma_Z & & \downarrow \gamma_C & & \\ H(B \otimes C') & \xrightarrow{H(i \otimes 1)} & H(Z \otimes C') & \xrightarrow{H(j \otimes 1)} & H(C \otimes_A C') & \xrightarrow{H(\delta \otimes 1)} & H(B \otimes C')(-1) & \xrightarrow{H(i \otimes 1)} & H(Z \otimes C')(-1) \end{array}

0 \longrightarrow \operatorname{Tor}_1^A(H, H')(-1) \xrightarrow{\partial(\mathrm{II}, H')} (B \otimes H')(-1) \xrightarrow{i \otimes 1} (Z \otimes H')(-1),

which is commutative by definition of the homomorphisms $\gamma$ . But since the complexes $B$ and $Z$ are split and flat, $\gamma_B$ and $\gamma_Z$ are bijective (X, p. 65, Corollary 1). From this, on the one hand, we deduce that $\gamma_C$ is injective and has image equal to $\operatorname{Ker} H(\delta \otimes 1)$ , and on the other hand, that $\gamma_B \circ \partial(\mathrm{II}, H')$ is injective with image equal to $\operatorname{Im} H(\delta \otimes 1)$ . The theorem follows immediately from this.

Corollary 1

If $B(C)$ and $Z(C)$ are flat, then for every left $A$ -module $N$ and every integer $n$ , there is an exact sequence

(13)

0 \longrightarrow H_n(C) \otimes_A N \xrightarrow{\gamma_n} H_n(C \otimes_A N) \xrightarrow{\alpha_n} \operatorname{Tor}_1^A(H_{n-1}(C), N) \longrightarrow 0 .

Corollary 2

Suppose $B(C)$ and $B(C')$ are projective and $Z(C)$ is flat. Then the sequences of $k$ -modules (11) and (13) are exact and split.* This follows from the theorem and the following lemma:

Lemma 4

If $B(C)$ and $B(C')$ are projective, then the canonical homomorphism*

\gamma(C, C') : H(C) \otimes_A H(C') \to H(C \otimes_A C')

has a $k$ -linear retraction. Indeed by X, p. 65, remark b), there exist homologisms

\varphi : C \to H(C)

and

\varphi' : C' \to H(C')

such that

H(\varphi) = 1_{H(C)}

and

H(\varphi') = 1_{H(C')}

. In the commutative diagram

\begin{array}{ccc} H(C) \otimes_A H(C') & \xrightarrow{\gamma(C, C')} & H(C \otimes_A C') \\ \downarrow H(\varphi) \otimes H(\varphi') & & \downarrow H(\varphi \otimes \varphi') \\ H(C) \otimes_A H(C') & \xrightarrow{\gamma(H(C), H(C'))} & H(C) \otimes_A H(C') \end{array}

$H(\varphi) \otimes H(\varphi')$ and $\gamma(H(C), H(C'))$ are the identity, whence the assertion.

Corollary 3 (universal coefficient formula). — Suppose the ring $A$ is principal. If the complexes $C$ and $C'$ are free, the sequences of $A$ -modules (11) are exact and split; if the complex $C$ is free, the sequences of $A$ -modules (13) are exact and split for every $A$ -module $N$ . Indeed, $B(C), Z(C)$ and $B(C')$ are submodules of the free modules $C, C, C'$ , so are free (VII, § 3, cor. 2 to th. 1), and we apply cor. 2.

Corollary 4 (« Künneth formula »)

Suppose C bounded on the right, C and H(C) flat; then the canonical homomorphism

\gamma(C, C') : H(C) \otimes_A H(C') \to H(C \otimes_A C')

is bijective.

By the theorem, it suffices to proving that B(C) and Z(C) are flat. Now one has exact sequences

\begin{align*} 0 &\to B_n(C) \to Z_n(C) \to H_n(C) \to 0 \\ 0 &\to Z_n(C) \to C_n \to B_{n-1}(C) \to 0, \end{align*}

whence, by X, p. 75, cor. 1, implications (B_{n-1}(C) is flat) $\Rightarrow$ (Z_n(C) is flat) $\Rightarrow$ (B_n(C) is flat) ; we conclude by remarking that B_n(C) = 0 for n small enough.

Corollary 5

Let u : C \to C' be a homologism of complexes of right a-modules, flat and bounded on the right. For every complex E of left a-modules, the morphism $u \otimes 1_E : C \otimes_A E \to C' \otimes_A E$ is a homologism.

Indeed, Con (u) is a flat complex, bounded on the right and of zero homology; then one has H(Con (u) \otimes_A E) = 0 by cor. 4, hence H(Con (u \otimes 1_E)) = 0 (X, p. 67, lemma 2), and $u \otimes 1_E$ is a homologism.

8. Bounded and flat complexes over a noetherian ring

Proposition 11

Suppose A left noetherian, and let C be a bounded and flat complex of left a-modules such that H(C) is a finitely generated a-module. Let a and b be two integers such that a \leq b and H_n(C) = 0 for n < a, C_n = 0 for n > b. There exists a complex P of left a-modules such that P_n is projective and finitely generated for each n, and P_n = 0 for n \notin [a, b], and a homologism u : P \to C. Moreover, for every complex E of right a-modules, the homomorphism

H(1_E \otimes u) : H(E \otimes_A P) \to H(E \otimes_A C) \quad \text{is bijective}.

By X, p. 53, prop. 7, there exists a complex (L, d) such that L_n is free and finitely generated for each n, and zero when n < a, and a homologism f : L \to C. Let P be the quotient complex L/L', where L'_n = 0 for n < b, L'_n = L_n for n > b, L'_b = B_b(L). Since C_n = 0 for n > b, f(L') = 0, so f factors through a morphism of complexes u : P \to C.

\begin{array}{ccccccccc} \ldots & \longrightarrow & L_{b+1} & \xrightarrow{d_{b+1}} & L_b & \xrightarrow{d_b} & L_{b-1} & \longrightarrow & \ldots \\ & & \downarrow & & \downarrow & & \downarrow & & \\ & & 0 & \longrightarrow & P_b & \longrightarrow & P_{b-1} & \longrightarrow & \ldots \\ & & \downarrow & & \downarrow^{u_b} & & \downarrow^{u_{b-1}} & & \\ & & 0 & \longrightarrow & C_b & \longrightarrow & C_{b-1} & \longrightarrow & \ldots \end{array}

Since $f$ is a homologism, one has $H(\operatorname{Con}(f))=0$ , whence an exact sequence

\cdots \longrightarrow L_{b+1} \xrightarrow{d_{b+1}} L_b \longrightarrow L_{b-1}\oplus C_b \longrightarrow L_{b-2}\oplus C_{b-1} \longrightarrow \cdots

One then has an exact sequence

0\longrightarrow P_b\longrightarrow L_{b-1}\oplus C_b\longrightarrow L_{b-2}\oplus C_{b-1}\longrightarrow\cdots.

This shows on the one hand that the cone of $u$ is of zero homology, so that $u$ is a homologism, on the other hand that the module $P_b$ is flat (X, p. 76, cor. 2) ; since $P_b$ is finitely generated as a quotient of $L_{b+1}$ , it is projective (X, p. 13, cor.). The pair $(P,u)$ then satisfies the required condition. The last assertion follows from X, p. 79, cor. 5.

\* Example. — Let A be a commutative noetherian ring, X a proper and flat A-scheme, $\mathscr F$ a coherent $\mathscr C_X$ -module, flat over A. There exists a bounded complex P consisting of finitely generated projective A-modules such that for every A-module M, $H(X,\mathscr F\otimes_A M)$ identifies naturally with $H(P\otimes_A M)$ . Indeed, let $\mathscr U$ be a covering of X by a finite number of affine open sets, $\mathscr C(\mathscr U,\mathscr F)$ the associated Čech complex. One shows that $H^i(\mathscr C(\mathscr U,\mathscr F))$ is isomorphic to the A-module $H^i(X,\mathscr F)$ , and that the latter is finitely generated; moreover, for every A-module M, the complex $\mathscr C(\mathscr U,\mathscr F)\otimes_A M$ is isomorphic to $\mathscr C(\mathscr U,\mathscr F\otimes_A M)$ . Applying Prop. 11 to the complex $\mathscr C(\mathscr U,\mathscr F)$ (which is bounded), one obtains a complex P that answers the question.

For every point $y$ of Spec (A), let $\kappa(y)$ denote the residue field of A at $y$ , $X_y=X\otimes_A\kappa(y)$ the fibre of X over $y$ , $\mathscr F_y=\mathscr F\otimes_A\kappa(y)$ , and set $h^p(y)=\dim_{\kappa(y)}H^p(X_y,\mathscr F_y)$ for $p\geq 0$ .

One easily deduces the following results from the existence of the complex P :

(i) the function $h^p$ is upper semi-continuous on Spec (A) ;

(ii) the function $\displaystyle\sum_{p\geq 0}(-1)^p h^p$ is locally constant on Spec (A). \*

9. Generalization to multimodule complexes

Let B and B′ be two rings, C a complex of (B, A)-bimodules, C′ a complex of (A, B′)-bimodules (X, p. 43); then $(C\otimes_A C',D)$ (X, p. 61) is a complex of (B, B′)-bimodules and the canonical homomorphism

\gamma:H(C)\otimes_A H(C')\longrightarrow H(C\otimes_A C')

is compatible with the (B, B′)-bimodule structures of the two members.

If B″ is a third ring, and C″ a complex of (B′, B″)-bimodules, the canonical homomorphism (II, p. 64, Prop. 8)

(C\otimes_A C')\otimes_{B'}C''\longrightarrow C\otimes_A(C'\otimes_{B'}C'')

is an isomorphism of complexes of (B, B″)-bimodules.

More generally, we leave it to the reader to develop the theory of tensor products of finite, totally ordered families of multimodule complexes on the model of No. 1 (X, p. 63) and of II, pp. 65–72 (associativity, commutativity, … isomorphisms).

Let B and B' be two rings, M a (B, A)-bimodule, N an (A, B')-bimodule : then $L(M)\otimes_A L(N)$ is a complex of (B, B')-bimodules, so that $\operatorname{Tor}^A(M,N)$ is endowed with a natural structure of graded (B, B')-bimodule; on the term of degree 0, this structure coincides with that of $M\otimes_A N$ .

If $\lambda\in B$ , $\lambda'\in B'$ , and if we denote by $\lambda_M,\lambda_N,\lambda_T,\lambda'_T$ the homotheties $x\mapsto\lambda x$ , $y\mapsto y\lambda'$ , $z\mapsto\lambda z$ , $z\mapsto z\lambda'$ of $M$ , $N$ , $\operatorname{Tor}^A(M,N)$ , $\operatorname{Tor}^A(M,N)$ respectively, then

\lambda_T=\operatorname{Tor}^A(\lambda_M,1_N),\qquad \lambda'_T=\operatorname{Tor}^A(1_M,\lambda_N'),

which gives another description of the bimodule structure of $\operatorname{Tor}^A(M,N)$ . We leave it to the reader to generalize Nos. 5 and 7 to the case of multimodule complexes.

Exercises

See the exercises for § 4.