§ 7. Composition product

We follow the general conventions of §4.

1. The homomorphism $\mathrm{Ext}_A(N, P) \otimes \mathrm{Ext}_A(M, N) \to \mathrm{Ext}_A(M, P)$

Let $M$ and $N$ be two left $A$ -modules. Consider the homomorphisms $p_M : L(M) \to M$ , $e_M : M \to I(M)$ , and $e_M \circ p_M : L(M) \to I(M)$ ; by Prop. 4 of X, p. 86, we deduce from this a bijective homomorphism

a_{M,N} = H(\mathrm{Homgr}_A(e_M \circ p_M, 1)) : H(\mathrm{Homgr}_A(I(M), I(N))) \to \mathrm{Ext}_A(M, N) .

Recall also (X, p. 82) that $H^n(\mathrm{Homgr}_A(I(M), I(N)))$ is the set of homotopy classes of degree-increasing morphisms from the complex $I(M)$ to the complex $I(N)$ . For example, if $f \in \mathrm{Hom}_A(M, N)$ , the homotopy class of $I(f)$ is sent by $a_{M,N}$ to $f$ .

If $P$ is a third left $A$ -module, we have by X, p. 99, a canonical $k$ -homomorphism

H(\mathrm{Homgr}_A(I(N), I(P))) \otimes_k H(\mathrm{Homgr}_A(I(M), I(N))) \to H(\mathrm{Homgr}_A(I(M), I(P)))

from which we deduce by transport via the isomorphisms $a_{N,P}$ , $a_{M,N}$ , $a_{M,P}$ a $k$ -homomorphism (called the composition homomorphism):

c_{M,N,P}: \mathrm{Ext}_A(N, P) \otimes_k \mathrm{Ext}_A(M, N) \to \mathrm{Ext}_A(M, P)

This corresponds to a bilinear mapping

(1)

\mathrm{Ext}_A(N, P) \times \mathrm{Ext}_A(M, N) \to \mathrm{Ext}_A(M, P)

which decomposes into homogeneous components

(2)

\mathrm{Ext}_A^i(N, P) \times \mathrm{Ext}_A^j(M, N) \to \mathrm{Ext}_A^{i+j}(M, P)

If $u \in \mathrm{Ext}_A(N, P)$ , $v \in \mathrm{Ext}_A(M, N)$ , the image of $(u, v)$ under (1) is called the composition product of $u$ and $v$ and is denoted $u \circ v$ . If $g$ (resp. $f$ ) is a degree-increasing morphism of degree $j$ (resp. $i$ ) from $I(M)$ to $I(N)$ (resp. from $I(N)$ to $I(P)$ ), whose homotopy class is $\bar{g}$ (resp. $\bar{f}$ ), then the composition product $a_{N,P}(\bar{f}) \circ a_{M,N}(\bar{g})$ is the image under $a_{M,P}$ of the homotopy class of the morphism $f \circ g$ from $I(M)$ to $I(P)$ .

Example 1

If $u \in \mathrm{Hom}_A(N, P)$ , $v \in \mathrm{Hom}_A(M, N)$ , then $u \circ v$ is the composite of $u$ and $v$ .

Example 2

If $u \in \mathrm{Hom}_A(N, P)$ , $v \in \mathrm{Ext}_A(M, N)$ , then

u \circ v = \mathrm{Ext}_A(l_M, u)(v) \in \mathrm{Ext}_A(M, P):

analogously, if $u \in \mathrm{Ext}_A(N, P)$ , $v \in \mathrm{Hom}_A(M, N)$ , then

u \circ v = \mathrm{Ext}_A(v, l_P)(u) \in \mathrm{Ext}_A(M, P).

This follows from the definitions and the remarks of X. p. 88.

If $Q, M, N, P$ are four left $A$ -modules, and if

u \in \mathrm{Ext}_A(N, P), \quad v \in \mathrm{Ext}_A(M, N), \quad w \in \mathrm{Ext}_A(Q, M),

then $(u \circ v) \circ w = u \circ (v \circ w)$ : the composition product is associative; we will therefore denote the composition of several elements without parentheses. In particular, by Example 2:

Example 3

Let $M, N, M', N'$ be four left $A$ -modules. If $u \in \mathrm{Ext}_A(M, N)$ , $f \in \mathrm{Hom}_A(M', M)$ , $g \in \mathrm{Hom}_A(N, N')$ , then

(3)

\mathrm{Ext}_A(f, g)(u) = g \circ u \circ f \in \mathrm{Ext}_A(M', N').

This provides a new proof of the $k$ -bilinearity of the mapping $(f, g) \to \mathrm{Ext}_A(f, g)$ (X, p. 88, Prop. 6).

2. Seven calculations of the composition product

==========

Let $M$ , $M'$ and $M''$ be three left $A$ -modules, $a : R \to M$ , $a' : R' \to M'$ and $a'' : R'' \to M''$ projective resolutions, $c : M \to E$ , $c' : M' \to E'$ and $c'' : M'' \to E''$ injective resolutions. It follows from X, p. 100, Th. 1 and p. 103, Prop. 2, that the diagram :

\begin{array}{cccccc} H(\mathrm{Homgr}_A(M, E')) & \longrightarrow & H(\mathrm{Homgr}_A(R, E')) & \longrightarrow & H(\mathrm{Homgr}_A(R, M')) \\ \uparrow & & \downarrow & & \uparrow \\ H(\mathrm{Homgr}_A(E, E')) & \xrightarrow{\varphi(E, E')} & \mathrm{Ext}_A(M, M') & \xleftarrow{\varphi(R, R')} & H(\mathrm{Homgr}_A(R, R')) \\ & \swarrow_{\varphi(M, E')} & & \searrow^{\varphi(R, M')} & \\ & & H(\mathrm{Homgr}_A(R, E')) & & \end{array}

where the unlabeled arrows are deduced canonically from $c, a, c', a'$ , is commutative, and that all the arrows are isomorphisms, which gives five descriptions of $\mathrm{Ext}_A(M, M')$ . Analogously, one obtains five descriptions of $\mathrm{Ext}_A(M', M'')$ , and as many of $\mathrm{Ext}_A(M, M'')$ .

Consider now the seven composition homomorphisms

H(\mathrm{Homgr}_A(C', C'')) \otimes_k H(\mathrm{Homgr}_A(C, C')) \to H(\mathrm{Homgr}_A(C, C''))

where one takes successively for $(C, C', C'')$ the seven triplets $(R, R', R''), (R, R', M''), (R, R', E''), (R, M', E''), (R, E', E''), (M, E', E''), (E, E', E'')$ .

!Diagram of the seven computations of the composition product

Fig. 1.

Identifying $H(\mathrm{Homgr}_A(C, C'))$ with $\mathrm{Ext}(M, M')$ by the isomorphism above, and analogously for $H(\mathrm{Homgr}_A(C', C''))$ and $H(\mathrm{Homgr}_A(C, C''))$ , one obtains seven homomorphisms

\mathrm{Ext}_A(M', M'') \otimes_k \mathrm{Ext}_A(M, M') \to \mathrm{Ext}_A(M, M'')

These seven homomorphisms coincide, and are independent of the choice of resolutions. In particular, they coincide with the homomorphism which was defined in No. 2, via the triplet $(I(M), I(M'), I(M''))$ . This follows indeed from the interpretation of the modules H(Homgr $(C, C')$ ) as module of homotopy classes of morphisms of complexes of C into C’, and from the fact that if in a diagram of complexes

\begin{array}{ccccc} C & \xrightarrow{\ f\ } & C' & \xrightarrow{\ g\ } & C''\\ \downarrow\scriptstyle{\alpha} & & \downarrow\scriptstyle{\alpha'} & & \downarrow\scriptstyle{\alpha''}\\ C_1 & \xrightarrow{\ f_1\ } & C'_1 & \xrightarrow{\ g_1\ } & C''_1 \end{array}

$\alpha'' \circ g$ is homotopic to $g_1 \circ \alpha'$ and $\alpha' \circ f$ homotopic to $f_1 \circ \alpha$ , then $\alpha'' \circ g \circ f$ is homotopic to $g_1 \circ f_1 \circ \alpha$ (X, p. 33, Prop. 4 and Cor.).

In what follows, we shall use according to the case one or the other of the seven preceding constructions of the composition homomorphisms.

3. The class associated with an exact sequence

Proposition 1

Let $(C,d)$ and $(C',d')$ be two complexes of left $A$ -modules and $n,p,q$ three integers such that $p \geq q$ . For $p \geq i \geq q-1$ , let $f_i:C_i\longrightarrow C'_{i+n+1}$ be a homomorphism of $A$ -modules such that $f_p\circ d=0$ , $f_i\circ d=d'\circ f_{i+1}$ for $p>i\geq q-1$ , and $d'\circ f_{q-1}=0$ (see fig. 2).

\begin{array}{ccccccccccccc} C_{p+1}&\xrightarrow{\ d\ }&C_p&\xrightarrow{\ d\ }&C_{p-1}&\xrightarrow{\ d\ }&\cdots&\xrightarrow{\ d\ }&C_q&\xrightarrow{\ d\ }&C_{q-1}&\xrightarrow{\ d\ }&C_{q-2}\\ 0\downarrow&&\downarrow f_p&&\downarrow f_{p-1}&&&&\downarrow f_q&&\downarrow f_{q-1}&&\downarrow 0\\ C'_{p+n+2}&\xrightarrow{\ d'\ }&C'_{p+n+1}&\xrightarrow{\ d'\ }&C'_{p+n}&\xrightarrow{\ d'\ }&\cdots&\xrightarrow{\ d'\ }&C'_{q+n+1}&\xrightarrow{\ d'\ }&C'_{q+n}&\xrightarrow{\ d'\ }&C'_{q+n-1} \end{array}

Fig. 2.

Set $\alpha=f_{p-1}\circ d=d'\circ f_p$ , $\beta=f_{q-1}\circ d=d'\circ f_q$ , and let $a$ (resp. $b$ ) be the graded $A$ -homomorphism of degree $n$ from $C$ into $C'$ whose only non-zero bi-homogeneous component is $\alpha$ (resp. $\beta$ ). Then one has $a\in Z_n(\operatorname{Homgr}_A(C,C'))$ , $b\in Z_n(\operatorname{Homgr}_A(C,C'))$ and

a-(-1)^{(n+1)(p-q)}b\in B_n(\operatorname{Homgr}_A(C,C')).

One has $d'\circ\alpha=d'\circ f_{p-1}\circ d=0$ , $\alpha\circ d=f_{p-1}\circ d\circ d=0$ , so

a\in Z_n(\operatorname{Homgr}_A(C,C')) ;

Analogously $b\in Z_n(\operatorname{Homgr}_A(C,C'))$ . Set $\varepsilon=(-1)^{n+1}$ . We have, in the complex $\operatorname{Homgr}_A(C,C')$ the relations

Df_{p-1}=d'\circ f_{p-1}-\varepsilon f_{p-1}\circ d=f_{p-2}\circ d-\varepsilon\alpha

Df_i=d'\circ f_i-\varepsilon f_i\circ d=f_{i-1}\circ d-\varepsilon f_{i+1}\circ d\qquad (p-1>i>q)

Df_q=d'\circ f_q-\varepsilon f_q\circ d=\beta-\varepsilon f_q\circ d .

thence

\sum_{i=1}^{p-q} \varepsilon^i Df_{p-i} = \varepsilon^{p-q} \beta - \alpha,

which proves the lemma.

Consider two left A-modules M and N and an exact sequence of A-modules

(4)

0 \to N \to R_n \to R_{n-1} \to \ldots \to R_1 \to M \to 0.

By prop. 3 and 3 bis of X, p. 49, there exists a commutative diagram:

(5)

\begin{array}{ccccccccccccc} 0 & \longrightarrow & N & \xrightarrow{e_N} & I^0(N) & \longrightarrow & \ldots & \longrightarrow & I^{n-1}(N) & \xrightarrow{\delta^{n-1}} & I^n(N) & \xrightarrow{\delta^n} & I^{n+1}(N) \\ & & \uparrow & & \uparrow & & & & \uparrow & & \uparrow & & \uparrow \\ 0 & \longrightarrow & N & \longrightarrow & R_n & \longrightarrow & \ldots & \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 \\ & & \uparrow & & \uparrow & & & & \uparrow & & \uparrow & & \uparrow \\ L_{n+1}(M) & \xrightarrow{d_{n+1}} & L_n(M) & \xrightarrow{d_n} & L_{n-1}(M) & \longrightarrow & \ldots & \xrightarrow{d_1} & L_0(M) & \xrightarrow{p_M} & M & \longrightarrow & 0. \end{array}

Consider the two elements $b$ and $a$ of $\mathrm{Homgr}_A(L(M), I(N))$ whose only non-zero bihomogeneous components are

b^n = e_N \circ u_n : L_n(M) \to I^0(N) \quad \text{and} \quad a^n = v^n \circ p_M : L_0(M) \to I^n(N)

respectively.

Proposition 2

We have* $a, b \in Z^n(\mathrm{Homgr}_A(L(M), I(N)))$ . Moreover, the classes $\overline{a}$ and $\overline{b}$ of $a$ and $b$ in $H^n(\mathrm{Homgr}_A(L(M), I(N))) = \mathrm{Ext}_A^n(M, N)$ depend only on the exact sequence (4) and are equal.

By prop. 1, applied to the two extreme rows of (5) and to the composed vertical arrows, with $p = n, q = 0$ , we have $a, b \in Z^n(\mathrm{Homgr}_A(I(M), L(N)))$ and

a - b = a - (-1)^{(n+1)n} b \in B^n(\mathrm{Homgr}_A(L(M), I(N))) .

Since $a$ (resp. $b$ ) is independent of the choice of $u$ (resp. $v$ ), the element $\overline{a} = \overline{b}$ of $\mathrm{Ext}_A^n(M, N)$ is independent of the choice of the morphisms $u$ and $v$ , whence the proposition.

Definition 1

We call the class associated with the exact sequence* (4) the element $\theta$ of $\mathrm{Ext}_A^n(M, N)$ defined by $\theta = (-1)^{n(n+1)/2} \overline{a} = (-1)^{n(n+1)/2} \overline{b}$ .

Remark 1

Let (P, p) be a projective resolution of M. By X, p. 49. prop. 3, there exists a commutative diagram

\begin{array}{ccccccccccccc} 0 & \longrightarrow & N & \longrightarrow & R_n & \longrightarrow & \ldots & \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 \\ & & \uparrow & & \uparrow & & & & \uparrow & & \uparrow & & \uparrow \\ P_n & \longrightarrow & P_{n-1} & \longrightarrow & \ldots & \longrightarrow & P_0 & \xrightarrow{p} & M & \longrightarrow & 0. \end{array}

With the notations of § 6, θ is the image under φ(P, N) of the homotopy class of the morphism P → N defined by (−1)^{n(n+1)/2} \tilde{u}_n. Analogously, if (e, E) is an injective resolution of N, there exists a commutative diagram

\begin{array}{cccccccccccc} 0 & \longrightarrow & N & \longrightarrow & E^0 & \longrightarrow & \ldots \longrightarrow & E^{n-1} & \longrightarrow & E^n \\ & & ^{1_N}\uparrow & & ^{\tilde{v}^0}\uparrow & & & ^{\tilde{v}^{n-1}}\uparrow & & ^{\tilde{v}^n}\uparrow \\ 0 & \longrightarrow & N & \longrightarrow & R_n & \longrightarrow & \ldots \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 . \end{array}

and θ is the image under φ(M, E) of the homotopy class of the morphism M → E defined by (−1)^{n(n+1)/2} v^n. This follows from the construction of θ and the definitions of φ(P, N) and φ(M, E).

Remark 2

When $n = 0$ , the exact sequence (4) reads $0 \to N \xrightarrow{f} M \to 0$ , and the associated class is $f^{-1} \in \mathrm{Hom}_A(M, N) = \mathrm{Ext}^0_A(M, N)$ .

4. Properties of the class associated with an exact sequence

Proposition 3

Let

(6)

0 \to P \to S_m \to S_{m-1} \to \ldots \to S_1 \xrightarrow{\lambda} N \to 0

(7)

0 \to N \twoheadrightarrow R_n \to R_{n-1} \to \ldots \to R_1 \to M \to 0

be two exact sequences of left A-modules with respective classes θ ∈ Ext^m_A(N, P) and θ' ∈ Ext^n_A(M, N). The class in Ext^{m+n}_A(M, P) associated with the exact sequence (8)

0 \to P \to S_m \to \ldots \to S_1 \xrightarrow{\mu \circ \lambda} R_n \to \ldots \to R_1 \to M \to 0

is the composition product θ ∘ θ'.

Choose commutative diagrams

\begin{array}{cccccccccccc} 0 & \longrightarrow & P & \longrightarrow & I^0(P) & \longrightarrow & \ldots \longrightarrow & I^{m-1}(P) & \xrightarrow{\delta_P^{m-1}} & I^m(P) \\ & & ^{l_P}\uparrow & & ^{w^0}\uparrow & & & ^{w^{m-1}}\uparrow & & ^{w^m}\uparrow \\ 0 & \longrightarrow & P & \longrightarrow & S_m & \longrightarrow & \ldots \longrightarrow & S_1 & \xrightarrow{\lambda} & N & \longrightarrow & 0 , \end{array}

\begin{array}{cccccccccccc} 0 & \longrightarrow & N & \xrightarrow{e_N} & I^0(N) & \longrightarrow & \ldots \longrightarrow & I^{n-1}(N) & \longrightarrow & I^n(N) \\ & & ^{1_N}\uparrow & & ^{v^0}\uparrow & & & ^{v^{n-1}}\uparrow & & ^{v^n}\uparrow \\ 0 & \longrightarrow & N & \xrightarrow{\mu} & R_n & \longrightarrow & \ldots \longrightarrow & R_1 & \longrightarrow & M & \longrightarrow & 0 . \end{array}

Since $I^m(P)$ is injective, there exists a homomorphism $h^0 \colon I^0(N) \rightarrow I^m(P)$ such that $w^m = h^0 \circ e_N$ ; according to X, p. 49, Proposition 3 bis, $h^0$ extends to a morphism of complexes $h \colon I(N) \rightarrow I(P)\, (-m)$ . Thus $w^m = h^0 \circ e_N = h^0 \circ v^0 \circ \mu$ , hence

\delta_I^{m-1} \circ w^{m-1} = w^m \circ \lambda = h^0 \circ v^0 \circ (\mu \circ \lambda),

and the following diagram is commutative:

\begin{array}{ccccccccccccccccccc} 0&\longrightarrow&P&\longrightarrow&I^0(P)&\longrightarrow&\cdots&\longrightarrow&I^{m-1}(P)&\longrightarrow&I^m(P)&\longrightarrow&I^{m+1}(P)&\longrightarrow&\cdots&\longrightarrow&I^{m+n}(P)\\ &&\uparrow^{i_P}&&\uparrow^{w^0}&&&&\uparrow^{w^{m-1}}&&\uparrow^{t^0}&&\uparrow^{t^1}&&&&\uparrow^{t^n}\\ 0&\longrightarrow&P&\longrightarrow&S_m&\longrightarrow&\cdots&\longrightarrow&S_1&\xrightarrow[\mu\circ\lambda]{}&R_n&\longrightarrow&R_{n-1}&\longrightarrow&\cdots&\longrightarrow&M&\longrightarrow&0 \end{array}

where $t^0=h^0\circ v^0$ , $t^1=(-1)^m h^1\circ v^1$ , $\ldots$ , $t^i=(-1)^{mi}h^i\circ v^i$ , $\ldots$ , $t^n=(-1)^{mn}h^n\circ v^n$ .

The class $\theta$ associated to (6) is that of $(-1)^{m(m+1)/2}w^m\in\operatorname{Homgr}_A^m(N,I(P))$ , corresponding by the isomorphism $\alpha_{N,P}$ to the class of $(-1)^{m(m+1)/2}h\in\operatorname{Homgr}_A^m(I(N),I(P))$ ; the class $\theta'$ associated to (7) is that of $(-1)^{n(n+1)/2}v^n\in\operatorname{Homgr}^n(M,I(N))$ , the class associated to (8) is that of $(-1)^{(m+n)(m+n+1)/2}t^n\in\operatorname{Homgr}^{m+n}(M,I(P))$ , hence the conclusion, according to the definition of the composition product (X, p. 114) and the formula

m(m+1)/2+n(n+1)/2=(m+n)(m+n+1)/2-mn.

4. — Considérons un diagramme commutatif de $A$ -modules à lignes exactes

\begin{array}{ccccccccccccccc} 0&\longrightarrow&N&\longrightarrow&R_n&\longrightarrow&R_{n-1}&\longrightarrow&\cdots&\longrightarrow&R_1&\longrightarrow&M&\longrightarrow&0\\ &&\downarrow^{g}&&\downarrow&&\downarrow&&&&\downarrow&&\downarrow^{f}&&\\ 0&\longrightarrow&N'&\longrightarrow&R'_n&\longrightarrow&R'_{n-1}&\longrightarrow&\cdots&\longrightarrow&R'_1&\longrightarrow&M'&\longrightarrow&0. \end{array}

Let $\theta$ (resp. $\theta'$ ) be the class of the first (resp. second) row in

\operatorname{Ext}_A^n(M,N)\quad\text{(resp. }\operatorname{Ext}_A^n(M',N')\text{)}.

In $\operatorname{Ext}_A^n(M,N')$ , we have $\theta'\circ f=g\circ\theta$ .

Indeed, consider a commutative diagram

\begin{array}{ccccccccccccc} L_n(M)&\xrightarrow{d_n}&L_{n-1}(M)&\longrightarrow&\cdots&\longrightarrow&L_0(M)&\xrightarrow{p_M}&M&\longrightarrow&0\\ \downarrow^{u_n}&&\downarrow&&&&\downarrow&&\downarrow^{1}\\ 0&\longrightarrow&N&\longrightarrow&R_n&\longrightarrow&\cdots&\longrightarrow&R_1&\longrightarrow&M&\longrightarrow&0\\ &&\downarrow^{g}&&\downarrow&&&&\downarrow&&\downarrow^{f}\\ 0&\longrightarrow&N'&\longrightarrow&R'_n&\longrightarrow&\cdots&\longrightarrow&R'_1&\longrightarrow&M'&\longrightarrow&0\\ &&\downarrow&&\downarrow&&&&\downarrow&&\downarrow^{v^n}\\ 0&\longrightarrow&N'&\xrightarrow{\epsilon_{N'}}&I^0(N')&\xrightarrow{\delta^0}&\cdots&\longrightarrow&I^{n-1}(N')&\xrightarrow{\delta^{n-1}}&I^n(N') \end{array}

By definition, $\theta'\circ f$ is the class of $(-1)^{n(n+1)/2}v^n\circ f\circ p_M\in\operatorname{Homgr}^n(L(M),I(N'))$ , while $g\circ\theta$ is the class of $(-1)^{n(n+1)/2}\epsilon_{N'}\circ g\circ u_n$ . According to Lemma 1, applied to the two extreme lines of the diagram, these two classes are equal.

Corollary 1

Consider a commutative diagram with exact rows

\begin{array}{cccccccccccc} 0 & \rightarrow & N & \rightarrow & R_n & \rightarrow & \cdots \rightarrow & R_1 & \rightarrow & M & \rightarrow & 0 \\ \downarrow & & \downarrow & & \downarrow & & & \downarrow & & \downarrow & & \downarrow \\ 0 & \rightarrow & N & \rightarrow & R'_n & \rightarrow & \cdots \rightarrow & R'_1 & \rightarrow & M & \rightarrow & 0 ; \end{array}

the two rows of the diagram have the same associated class in $\mathrm{Ext}_A^n(M, N)$ .

Corollary 2

Let

0 \to N \xrightarrow{f_{n-1}} R_n \xrightarrow{f_n} R_{n-1} \to \cdots \xrightarrow{f_2} R_1 \xrightarrow{f_1} M \to 0

be an exact sequence, $\theta \in \mathrm{Ext}_A^n(M, N)$ its associated class, $a_1, \ldots, a_{n+1}$ invisible elements of $k$ . The associated class of the exact sequence

0 \to N \xrightarrow{a_{n+1} f_{n+1}} R_n \xrightarrow{a_n f_n} R_{n-1} \to \cdots \xrightarrow{a_2 f_2} R_1 \xrightarrow{a_1 f_1} M \to 0

is $(a_1^{-1}\, a_2^{-1}\, \ldots\, a_{n+1}^{-1})\ \theta$ .

Indeed, there is a commutative diagram

\begin{array}{ccccccccccccccc} 0 & \rightarrow & N & \xrightarrow{a_{n+1} f_{n+1}} & R_n & \rightarrow & \cdots & \rightarrow & R_2 & \xrightarrow{a_2 f_2} & R_1 & \xrightarrow{a_1 f_1} & M & \rightarrow & 0 \\ & & \downarrow a_1 \ldots a_{n+1} & & \downarrow a_1 \ldots a_n & & & \downarrow a_1 a_2 & & \downarrow a_1 & & \downarrow 1 \\ 0 & \rightarrow & N & \xrightarrow{f_{n+1}} & R_n & \rightarrow & \cdots & \rightarrow & R_2 & \xrightarrow{f_2} & R_1 & \xrightarrow{f_1} & M & \rightarrow & 0 . \end{array}

and we apply the proposition.

Corollary 3

Let $0 \to N \xrightarrow{f_{n-1}} R_n \xrightarrow{f_n} \cdots \to R_1 \xrightarrow{f_1} M \to 0$ be an exact sequence, $\theta$ its class in $\mathrm{Ext}_A^n(M, N)$ , $u \colon M' \to M$ and $v \colon N \to N'$ homomorphisms of $A$ -modules.

a) The element $v \circ \theta$ of $\mathrm{Ext}_A^n(M, N')$ is equal to the class of the exact sequence

0 \to N' \xrightarrow{f_{n'+1}} R'_n \xrightarrow{f_{n'}} R_{n-1} \xrightarrow{f_{n-1}} \cdots \to R_1 \to M \to 0 ,

where $R'_n$ is the $A$ -quotient module of $R_n \oplus N'$ by the submodule formed by the pairs $(f_{n+1}(x), -v(x))$ for $x \in N$ , and where $f_{n'+1}'$ (resp. $f_{n'}'$ ) is deduced from the canonical injection (resp. from $(f_n, 0)$ ) by passing to quotients.

b) The element $\theta \circ u$ of $\mathrm{Ext}_A^n(M', N)$ is the class of the exact sequence

0 \to N \to R_n \to \cdots \to R_2 \xrightarrow{f_2''} R'_1 \xrightarrow{f_1''} M' \to 0 ,

where $R'_1$ is the fibre product $R_1 \times_M M'$ , that is to say (I, p. 44) the submodule of $R_1 \times M'$ formed by the pairs $(x, y)$ such that $f_1(x) = u(y)$ , and where $f_2''$ (resp. $f_1''$ ) is deduced from $(f_2, 0)$ (resp. from the second projection).

Let us prove for example $a$ . Let $z$ be an element of $R'_n$ such that $f'_n(z) = 0$ ; if $z$ is the class of a pair $(x, y)$ , with $x \in R_n$ , $y \in N'$ , one has $f_n(x) = 0$ , so that there exists an element $t \in N$ such that $x = f_{n+1}(t)$ . Then one has $z = f'_{n+1}(y + v(t))$ , which proves that $\mathrm{Ker}\ f'_n = \mathrm{Im}\ f'_{n+1}$ . The injectivity of $f'_{n+1}$ follows from that of $f_{n+1}$ .

Let $j : R_n \to R'_n$ be the homomorphism deduced from the canonical injection; one has a commutative diagram of exact sequences:

\begin{array}{ccccccccccccc} 0 & \longrightarrow & N & \xrightarrow{f_{n+1}} & R_n & \xrightarrow{f_n} & R_{n-1} & \longrightarrow & \ldots & \longrightarrow & M & \longrightarrow & 0 \\ & & \downarrow v & & \downarrow j & & \downarrow 1 & & & & \downarrow 1 & & \\ 0 & \longrightarrow & N' & \xrightarrow{f'_{n+1}} & R'_n & \xrightarrow{f'_n} & R'_{n-1} & \longrightarrow & \ldots & \longrightarrow & M & \longrightarrow & 0 ; \end{array}

assertion $a$ then follows from the proposition.

The proof of $b$ is analogous.

Remark

Let $\theta \in \mathrm{Ext}^n_A(M, N)$ , resp. $\theta' \in \mathrm{Ext}^n_A(M', N')$ , be the class of an exact sequence

0 \to N \xrightarrow{f_{n+1}} R_n \to \ldots \to R_1 \xrightarrow{f_1} M \to 0,

resp.

0 \to N' \xrightarrow{f'_{n+1}} R'_n \to \ldots \to R'_1 \xrightarrow{f'_1} M' \to 0.

Let $i_N,\ i_{N'}$ , be the canonical injections of $N$ and $N'$ into $N \oplus N'$ , $q_M,\ q_{M'}$ , the projections of $M \oplus M'$ onto $M$ and $M'$ . Consider the homomorphism

m = \mathrm{Ext}\,(q_M,\ i_N) \oplus \mathrm{Ext}\,(q_{M'},\ i_{N'})

from $\mathrm{Ext}_A(M, N) \oplus \mathrm{Ext}_A(M', N')$ into $\mathrm{Ext}_A(M \oplus M', N \oplus N')$ . *The element

m(\theta, \theta') = i_N \circ \theta \circ q_M + i_{N'} \circ \theta' \circ q_{M'}

is the class of the exact sequence

0 \to N \oplus N' \xrightarrow{f_{n-1} \oplus f'_{n-1}} R_n \oplus R'_n \to \ldots \to R_1 \oplus R'_1 \xrightarrow{f_1 \oplus f'_1} M \oplus M' \to 0 .

Indeed, if one denotes this class by $\theta''$ , it follows from prop. 4 that one has

\theta'' \circ i_M = i_N \circ \theta = m(\theta, \theta') \circ i_M \quad \text{and} \quad \theta'' \circ i_{M'} = i_{N'} \circ \theta = m(\theta, \theta') \circ i_{M'} ;

by X, p. 89, prop. 7, this entails $\theta'' = m(\theta, \theta')$ .

5. Relation between exact sequences and elements of $\mathrm{Ext}_A(M, N)$

Theorem 1

Let $n$ be an integer $\geqslant 1$ , $M$ and $N$ two $A$ -modules.

a) Every element of $\mathrm{Ext}^n_A(M, N)$ is the class of an exact sequence (X, p. 117, def. 1).

b) Let $0 \to N \xrightarrow{f_{n+1}} R_n \xrightarrow{f_n} \ldots \to R_1 \xrightarrow{f_1} M \to 0$ and $0 \to N \xrightarrow{f_{n+1}} R'_n \xrightarrow{f_n} \ldots \to R'_1 \xrightarrow{f_1} M \to 0$

be exact sequences, $\theta$ and $\theta'$ the associated classes. The following conditions are equivalent:

(i) $\theta = \theta'$ ;

(ii) there exists a commutative diagram with exact rows:

\begin{array}{ccccccccccccc} 0 & \longrightarrow & N & \xrightarrow{f_{n+1}} & R_n & \longrightarrow & \ldots & \longrightarrow & R_1 & \xrightarrow{f_1} & M & \longrightarrow & 0 \\ & & 1_N \uparrow & & \uparrow & & & & \uparrow & & 1_M \uparrow \\ 0 & \longrightarrow & N & \longrightarrow & R''_n & \longrightarrow & \ldots & \longrightarrow & R''_1 & \longrightarrow & M & \longrightarrow & 0 \\ & & 1_N \downarrow & & \downarrow & & & & \downarrow & & 1_M \downarrow \\ 0 & \longrightarrow & N & \xrightarrow{f'_{n+1}} & R'_n & \longrightarrow & \ldots & \longrightarrow & R'_1 & \xrightarrow{f'_1} & M & \longrightarrow & 0 ; \end{array}

(iii) there exists a commutative diagram with exact rows:

\begin{array}{ccccccccccccc} 0 & \longrightarrow & N & \xrightarrow{f_{n+1}} & R_n & \longrightarrow & \ldots & \longrightarrow & R_1 & \xrightarrow{f_1} & M & \longrightarrow & 0 \\ & & 1_N \downarrow & & \downarrow & & & & \downarrow & & 1_M \downarrow \\ 0 & \longrightarrow & N & \longrightarrow & R''_n & \longrightarrow & \ldots & \longrightarrow & R''_1 & \longrightarrow & M & \longrightarrow & 0 \\ & & 1_N \uparrow & & \uparrow & & & & \uparrow & & 1_M \uparrow \\ 0 & \longrightarrow & N & \xrightarrow{f'_{n+1}} & R'_n & \longrightarrow & \ldots & \longrightarrow & R'_1 & \xrightarrow{f'_1} & M & \longrightarrow & 0 . \end{array}

Let us prove $a$ . Let $\alpha \in \mathrm{Ext}_A^n(M, N)$ , and let $P$ be a projective resolution of $M$ . Let $a : P(n) \to N$ be a morphism of complexes representing $\alpha$ ; its unique non-zero component is an $A$ -homomorphism $u : P_n \to N$ which satisfies $u \circ d_{n+1} = 0$ , hence factors as $u = \overline{u} \circ \delta_n$ , where $\delta_n : P_n \to Z_{n-1}$ is the mapping induced by $d_n$ (we put $Z_{n-1} = \mathrm{Im}\, d_n$ ) and $\overline{u}$ is an $A$ -homomorphism from $Z_{n-1}$ into $N$ . By Remark 1, p. 117, the class $\theta \in \mathrm{Ext}_A^n(M, Z_{n-1})$ of the exact sequence

0 \to Z_{n-1} \to P_{n-1} \to \ldots \to P_0 \to M \to 0

is equal to the homotopy class of the morphism $(-1)^{n(n+1)/2} \delta_n$ . We therefore have

\alpha = (-1)^{n(n+1)/2} \overline{u} \circ \theta ,

which makes it possible, by Cor. 3, p. 120, to represent $\alpha$ as the class of an exact sequence.

Let us prove $b$ . It follows from Cor. 1 of X, p. 120 that (ii) $\Rightarrow$ (i) and that (iii) $\Rightarrow$ (i). Suppose that (i) is satisfied, and let $P$ be a projective resolution of $M$ . There exists a commutative diagram

\begin{array}{cccccccccccccc} 0 & \rightarrow & N & \xrightarrow{f_{n+1}} & R_n & \rightarrow & R_{n-1} & \rightarrow & \ldots \rightarrow & R_1 & \rightarrow & M & \rightarrow & 0 \\ & & u_n \uparrow & & u_{n-1} \uparrow & & u_{n-2} \uparrow & & & & & & 1_M \uparrow \\ & & P_n & \xrightarrow{d_n} & P_{n-1} & \rightarrow & P_{n-2} & \rightarrow & \ldots \rightarrow & P_0 & \rightarrow & M & \rightarrow & 0 \\ & & u'_n \downarrow & & u'_{n-1} \downarrow & & u'_{n-2} \downarrow & & & & & & 1_M \downarrow \\ 0 & \rightarrow & N & \xrightarrow{f'_{n+1}} & R'_n & \rightarrow & R'_{n-1} & \rightarrow & \ldots \rightarrow & R'_1 & \rightarrow & M & \rightarrow & 0 . \end{array}

The morphisms from $P(n)$ into $N$ defined by $u_n$ and $u'_n$ are homotopic, since they both belong to the class $(-1)^{n(n+1)/2} \theta$ , hence $u'_n - u_n$ is of the form $w \circ d_n$ , where $w : P_{n-1} \to N$ is an $A$ -homomorphism. Replacing $u'_{n-1}$ by $u'_{n-1} - f'_{n+1} \circ w$ and $u'_n$ by $u_n$ , we reduce to the case where $u_n = u'_n$ . This makes it possible to construct a new commutative diagram with exact rows:

\begin{array}{cccccccccccccc} 0 & \rightarrow & N & \rightarrow & R_n & \rightarrow & R_{n-1} & \rightarrow & \ldots \rightarrow & R_1 & \rightarrow & M & \rightarrow & 0 \\ & & 1_N \uparrow & & v \uparrow & & u_{n-2} \uparrow & & & & & & 1_M \uparrow \\ 0 & \rightarrow & N & \rightarrow & N' & \rightarrow & P_{n-2} & \rightarrow & \ldots \rightarrow & P_0 & \rightarrow & M & \rightarrow & 0 \\ & & 1_N \downarrow & & v' \downarrow & & u'_{n-2} \downarrow & & & & & & 1_M \downarrow \\ 0 & \rightarrow & N & \rightarrow & R'_n & \rightarrow & R'_{n-1} & \rightarrow & \ldots \rightarrow & R'_1 & \rightarrow & M & \rightarrow & 0 \end{array}

where $N'$ is the quotient of $P_{n-1} \oplus N$ by the submodule formed by the couples $(d_n(x), -u_n(x))$ for $x \in P_n$ , and where $v$ (resp. $v'$ ) is defined by passing to the quotient from the mapping $u_{n-1} \oplus f_{n+1}$ (resp. $u'_{n-1} \oplus f'_{n+1}$ ). Condition (ii) is therefore satisfied.

Suppose again that condition (i) is satisfied, and let $E$ be an injective resolution of $N$ . There exists a commutative diagram

\begin{array}{cccccccccccc} 0 & \rightarrow & N & \rightarrow & R_n & \rightarrow & \ldots \rightarrow & R_1 & \rightarrow & M & \rightarrow & 0 \\ & & 1_N \downarrow & & v_0 \downarrow & & & & & & & \\ 0 & \rightarrow & N' & \rightarrow & E^0 & \xrightarrow{\delta^0} & \ldots \rightarrow & E^{n-1} & \xrightarrow{\delta^{n-1}} & E^n \\ & & 1_N \uparrow & & v'_0 \uparrow & & & & & & & \\ 0 & \rightarrow & N & \rightarrow & R'_n & \rightarrow & \ldots \rightarrow & R'_1 & \rightarrow & M & \rightarrow & 0 \end{array}

and we show as above that we can assume $v'_n = v_n$ . We then have a commutative diagram with exact rows

\begin{array}{cccccccccccccc} 0 & \rightarrow & N & \rightarrow & R_n & \rightarrow & \ldots \rightarrow & R_2 & \rightarrow & R_1 & \rightarrow & M & \rightarrow & 0 \\ & & 1_N \downarrow & & \downarrow & & & & & & & & 1_M \downarrow \\ 0 & \rightarrow & N & \rightarrow & E^0 & \rightarrow & \ldots \rightarrow & E^{n-2} & \rightarrow & M' & \rightarrow & M & \rightarrow & 0 \\ & & 1_N \uparrow & & \uparrow & & & & & & & & 1_M \uparrow \\ 0 & \rightarrow & N & \rightarrow & R'_n & \rightarrow & \ldots \rightarrow & R'_2 & \rightarrow & R'_1 & \rightarrow & M & \rightarrow & 0 \end{array}

with $M' = M \times_{\mathbf{R}_1} E_{n-1}$ (cf. X, p. 120, cor. 3, b)). Condition (iii) is therefore satisfied, which completes the proof of the theorem.

Remark 1

If the ring $A$ is Noetherian, and if the $A$ -modules $M$ and $N$ are of finite type, it follows from the proof of a) that every element of $\mathrm{Ext}_A^n(M, N)$ is the class associated with an exact sequence $0 \to N \to R_n \to \ldots \to R_1 \to M \to 0$ where the $R_i$ are of finite type.

Corollary

Let* $0 \to N \xrightarrow{f} R \xrightarrow{g} M \to 0$ and $0 \to N \xrightarrow{f'} R' \xrightarrow{g'} M \to 0$ be two exact sequences, $\theta$ and $\theta'$ their associated classes in $\mathrm{Ext}^1(M, N)$ . For $\theta = \theta'$ to hold, it is necessary and sufficient that there exists an A-homomorphism $h : R \to R'$ *making the diagram

\begin{array}{ccc} & & R \\ N & \xrightarrow{f} & R \\ & \downarrow h & \downarrow g \\ & & M \\ & \xrightarrow{f'} & R' \\ & & \xrightarrow{g'} \end{array}

commutative. Such a homomorphism is necessarily an isomorphism.

The condition is sufficient according to cor. 1 of prop. 4. If $\theta = \theta'$ , there exists a commutative diagram with exact rows:

\begin{array}{ccccccccc} 0 & \longrightarrow & N & \longrightarrow & R & \longrightarrow & M & \longrightarrow & 0 \\ & & ^{1_N} & & ^{h'} & & ^{1_M} & & \\ 0 & \longrightarrow & N & \longrightarrow & R'' & \longrightarrow & M & \longrightarrow & 0 \\ & & ^{1_N} & & ^{h''} & & ^{1_M} & & \\ 0 & \longrightarrow & N & \longrightarrow & R' & \longrightarrow & M & \longrightarrow & 0 . \end{array}

The homomorphisms $h'$ and $h''$ are isomorphisms according to X, p. 7, cor. 3, and $h = h'' \circ {h'}^{-1}$ answers the question. The last assertion follows from loc. cit.

Remark 2

Theorem 1 provides a description of $\mathrm{Ext}_A^n(M, N)$ as a set of equivalence classes of exact sequences; it is easy to describe the group law obtained on this set by transport of structure. Indeed, let $\theta$ (resp. $\theta'$ ) be the class of an exact sequence $0 \to N \xrightarrow{f_{n+1}} R_n \xrightarrow{f_n} \ldots \to R_1 \to M \to 0$ (resp. $0 \to N \xrightarrow{f'_{n+1}} R'_n \xrightarrow{f'_n} \ldots \to R'_1 \to M \to 0$ ). Let $\Delta : M \to M \oplus M$ and $\nabla : N \oplus N \to N$ be the $A$ -linear maps defined by $\Delta(x) = (x, x)$ for $x \in M$ and $\nabla(y, z) = y + z$ for $y, z \in N$ . Consider the mapping

m : \mathrm{Ext}_A(M, N) \oplus \mathrm{Ext}_A(M, N) \to \mathrm{Ext}_A(M \oplus M, N \oplus N)

defined in the remark, p. 121. With the notations of loc. cit., one has $\nabla \circ i_N = 1_N$ and $q_M \circ \Delta = 1_M$ , and thus $\theta + \theta' = \nabla \circ m(\theta, \theta') \circ \Delta$ . Taking into account loc. cit. and cor. 3, p. 120, this provides an exact sequence of class $\theta+\theta'$ : for example, if $n \geq 2$ , one can take the sequence

0\longrightarrow \mathbf{N}\longrightarrow R_n''\longrightarrow R_{n-1}\oplus R'_{n-1}\xrightarrow{\,f_{n-1}\oplus f'_{n-1}\,}\cdots\longrightarrow R_2\oplus R'_2\longrightarrow R'_1\longrightarrow M\longrightarrow0

where $R_n''$ is the quotient of $R_n\oplus R'_n$ by the submodule formed by the couples

\left(f_{n+1}(x),-f'_{n+1}(x)\right)\quad\text{pour }x\in\mathbf{N},

and where $R''_1=R_1\times_M R'_1$ .

6. Composition Product and Connecting Homomorphisms of Extension Modules

Proposition 5

Let

(\mathscr{E})\qquad 0\longrightarrow M'\xrightarrow{\,f\,}M\xrightarrow{\,g\,}M''\longrightarrow0

be an exact sequence of left $A$ -modules, $\theta\in\operatorname{Ext}^1_A(M'',M')$ the associated class, $N$ a left $A$ -module, $n$ an integer.

a) The connecting homomorphism $\delta^n(N,\mathscr{E}):\operatorname{Ext}^n_A(N,M'')\longrightarrow\operatorname{Ext}^{n+1}_A(N,M')$ is the composition product $\alpha\mapsto\theta\circ\alpha$ with $\theta$ .

Corollaire 1

a) The connecting homomorphism $\mathrm{Hom}_A(M'', M'') \to \mathrm{Ext}_A^1(M'', M')$ sends $1_{M''}$ to $\theta$ .

b) The connecting homomorphism $\mathrm{Hom}_A(M', M') \to \mathrm{Ext}_A^1(M'', M')$ sends $1_{M'}$ to $-\theta$ .

Corollaire 2

Consider two short exact sequences of left A-modules

\begin{align*} 0 &\to M' \to M \to M'' \to 0 \\ 0 &\to N' \to N \to N'' \to 0 . \end{align*}

Then the composite connecting homomorphisms

\operatorname{Ext}_A^n(M',N'') \to \operatorname{Ext}_A^{n+1}(M'',N'') \to \operatorname{Ext}_A^{n+2}(M'',N')

and

\operatorname{Ext}_A^n(M',N'') \to \operatorname{Ext}_A^{n+1}(M',N') \to \operatorname{Ext}_A^{n+2}(M'',N')

are opposite.

Indeed, let $\theta_1$ , $\theta_2$ be the classes associated with the given exact sequences, and let $\alpha \in \operatorname{Ext}_A^n(M', M'')$ . The images of $\alpha$ are respectively

\theta_2 \circ \bigl((-1)^{n+1} \alpha \circ \theta_1\bigr) \quad \text{and} \quad (\theta_2 \circ \alpha) \circ \bigl((-1)^{n+2} \theta_1\bigr).

Consider a short exact sequence of left A-modules

(\mathscr{S}) \qquad 0 \to N \to R_n \xrightarrow{f_n} R_{n-1} \xrightarrow{f_{n-1}} \cdots \to R_1 \xrightarrow{f_1} M \to 0

and set $K_0=M$ , $K_i=\operatorname{Ker}f_i$ , $i=1,\ldots,n-1$ , $K_n=N$ . We therefore have exact sequences

(9)\qquad 0\to K_i\to R_i\to K_{i-1}\to0,\qquad 1\leq i\leq n,

to which are associated for every $A$ left module $P$ , connecting homomorphisms

\operatorname{Ext}_A^m(P,K_{i-1})\to\operatorname{Ext}_A^{m+1}(P,K_i),

\operatorname{Ext}_A^m(K_i,P)\to\operatorname{Ext}_A^{m+1}(K_{i-1},P),

whence by composition of the iterated connecting homomorphisms, associated with $(\mathscr{S})$

\delta^m(P,\mathscr{S}):\operatorname{Ext}_A^m(P,M)\to\operatorname{Ext}_A^{m+n}(P,N)

\delta^m(\mathscr{S},P):\operatorname{Ext}_A^m(N,P)\to\operatorname{Ext}_A^{m+n}(M,P).

Corollary 3

If $\theta\in\operatorname{Ext}_A^n(M,N)$ is the class of the exact sequence $(\mathscr{S})$ , we have

\delta^m(P,\mathscr{S})(\alpha)=\theta\circ\alpha,\qquad \delta^m(\mathscr{S},P)(\beta)=(-1)^{mn+n(n+1)/2}\beta\circ\theta.

If $\theta_i\in\operatorname{Ext}_A^1(K_{i-1},K_i)$ is the class associated with the exact sequence (9), we have by Proposition 5

\delta^m(P,\mathscr{S})(\alpha)=\theta_n\circ\cdots\circ\theta_2\circ\theta_1\circ\alpha

\delta^m(\mathscr{S},P)(\beta)=(-1)^{(m+1)+\cdots+(m+n)}\beta\circ\theta_n\circ\cdots\circ\theta_1.

Moreover, by Proposition 3 (X, p. 118), we have $\theta=\theta_n\circ\cdots\circ\theta_1$ . The corollary follows immediately from this, and from the relation (E, III, p. 44)

(m+1)+\cdots+(m+n)=mn+n(n+1)/2.

Corollary 4

If each module $R_i,\ i = 1, ..., n,$ is injective (resp. projective), the mapping $\alpha \mapsto \theta \circ \alpha$ (resp. $\alpha \mapsto \alpha \circ \theta$ ) from $\mathrm{Ext}_A^m(P, M)$ to $\mathrm{Ext}_A^{m+n}(P, N)$ (resp. from $\mathrm{Ext}_A^m(N, P)$ to $\mathrm{Ext}_A^{m+n}(M, P)$ ) is bijective for every $A$ -module $P$ and every integer $m > 0$ .

This indeed follows from Corollary 3 and the exact sequences

\mathrm{Ext}_A^{m+i-1}(P, R_i) \to \mathrm{Ext}_A^{m+i-1}(P, K_{i-1}) \to \mathrm{Ext}_A^{m+i}(P, K_i) \to \mathrm{Ext}_A^{m+i}(P, R_i)

(resp.

\mathrm{Ext}_A^{m+i-1}(R_i, P) \to \mathrm{Ext}_A^{m+i-1}(K_i, P) \to \mathrm{Ext}_A^{m+i}(K_{i-1}, P) \to \mathrm{Ext}_A^{m+i}(R_i, P)

), whose extreme terms are zero by assumption.

Remark

The definitions and propositions of Nos. 3 to 6 apply to $A$ right modules, considered as left modules over the ring $A^\circ$ opposite to $A$ .

7. The homomorphism $\mathrm{Ext}_A(P, Q) \otimes \mathrm{Tor}^A(P, M) \to \mathrm{Tor}^A(Q, M)$

Let $M$ be a $A$ left module, $P$ and $Q$ two $A$ right modules. Consider the homomorphism $\mathrm{Homgr}_A(L(P), L(Q)) \otimes_k (L(P) \otimes_A L(M)) \to L(Q) \otimes_A L(M)$ which associates with $f \otimes (x \otimes y)$ the element $f(x) \otimes y$ . By X, p. 99, it is a morphism of complexes. We deduce from this a graded $k$ -linear mapping of degree 0

H(\mathrm{Homgr}_A(L(P), L(Q))) \otimes_k \mathrm{Tor}^A(P, M) \to \mathrm{Tor}^A(Q, M)

then by the isomorphism $\varphi(L(P), L(Q))$ of § 6 (X, p. 100, th. 1), a graded $k$ -linear mapping of degree 0

(10)

\mathrm{Ext}_A(P, Q) \otimes_k \mathrm{Tor}^A(P, M) \to \mathrm{Tor}^A(Q, M),

corresponding to

k

-bilinear mappings

(11)

c_{P, Q; M}: \mathrm{Ext}_A^n(P, Q) \times \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-n}^A(Q, M);

The image of the pair

(\alpha, \gamma)

under

c_{P, Q; M}

is called the composition product of

\alpha

and

\gamma

and is denoted by

\alpha \circ \gamma

By construction, $\alpha \circ \gamma$ is obtained as follows: one represents $\alpha$ by a morphism of complexes $f : L(P) \to L(Q)(-n)$ , $\gamma$ by an element $z \in Z_m(L(P) \otimes_A L(M))$ , and $\alpha \otimes \gamma$ is the class of the element

(f \otimes 1)(z) \in Z_m(L(Q)(-n) \otimes_A L(M)) = Z_{m-n}(L(Q) \otimes_A L(M)).

For example, if $\alpha \in \mathrm{Hom}_A(P, Q)$ , then $\alpha \circ \gamma = \mathrm{Tor}(\alpha, 1)(\gamma)$ .

Remark 1

If one uses the isomorphisms $\psi$ of X, p. 69, one can also define the composition product by the commutative diagram

\begin{array}{ccc} \mathrm{Ext}_A^n(P, Q) \times \mathrm{Tor}_m^A(P, M) & \xrightarrow{c_{P,Q;M}} & \mathrm{Tor}_{m-n}^A(Q, M) \\ \bar{a}_{P,Q} \times \psi_{P(M)} \downarrow & & \downarrow \psi_{Q(M)} \\ H^n(\mathrm{Homgr}_A(L(P), L(Q))) \times H_m(L(P) \otimes_A M) & \longrightarrow & H_{m-n}(L(Q) \otimes_A M); \end{array}

in other words, one represents $\alpha$ by a morphism $f$ from $L(P)$ to $L(Q) (-n)$ , $\gamma$ by a cycle $x \in L_m(P) \otimes_A M$ , and $\alpha \circ \gamma$ is the class of the cycle

(f_m \otimes 1_M)(x) \in L_{m-n}(Q) \otimes_A M.

Remark 2

One can also use the resolutions $l(P)$ and $l(Q)$ .

Analogously, if $N$ is a second left $A$ -module, one defines a composition product $(\mu, \gamma) \mapsto \mu \circ \gamma$ denoted

c_{P:M,N}: \mathrm{Ext}_A^r(M, N) \times \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-r}^A(P, N)

by the commutative diagram

\begin{array}{ccc} \mathrm{Ext}_A^r(M, N) \times \mathrm{Tor}_m^A(P, M) & \xrightarrow{c_{P:M,N}} & \mathrm{Tor}_{m-r}^A(P, N) \\ 1 \times \sigma_{P,M,r} \downarrow & & \downarrow \sigma_{P,N,m-r} \\ \mathrm{Ext}_A^r(M^\circ, N^\circ) \times \mathrm{Tor}_m^{A^\circ}(M^\circ, P^\circ) & \xrightarrow{c_{M^\circ,N^\circ,P^\circ}} & \mathrm{Tor}_{m-r}^{A^\circ}(N^\circ, P^\circ) \end{array}

where $\sigma$ denotes the commutation isomorphisms (X, p. 71).

If $\mu \in \mathrm{Ext}_A^r(M, N)$ is the class of the morphism $g : L(M) \to L(N) (-r)$ , and if $\gamma \in \mathrm{Tor}_m^A(P, M)$ is the class of the cycle $z = \sum z_{ij}$ , where $z_{ij} \in L_i(P) \otimes_A L_j(M)$ , then $\mu \circ \gamma$ is therefore the class of the cycle $\sum (-1)^{ir} (1 \otimes g)(z_{ij})$ .

One can also represent $\gamma$ by a cycle $y \in P \otimes L_m(M)$ , and $\mu \circ \gamma$ is the class of the cycle $(1 \otimes g)(y) \in P \otimes L_{m-r}(M)$ .

Proposition 6

Let $K, M, N$ be left $A$ -modules, $P, Q, R$ right $A$ -modules, $\alpha \in \mathrm{Ext}_A^n(P, Q)$ , $\beta \in \mathrm{Ext}_A^p(Q, R)$ , $\lambda \in \mathrm{Ext}_A^r(K, M)$ , $\mu \in \mathrm{Ext}_A^s(M, N)$ , $\gamma \in \mathrm{Tor}_m^A(P, K)$ . Then (13)

(\beta \circ \alpha) \circ \gamma = \beta \circ (\alpha \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{m-p-n}^A(R, K),

(14)

(\mu \circ \lambda) \circ \gamma = \mu \circ (\lambda \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{n-r-s}^A(P, N),

(15)

\alpha \circ (\lambda \circ \gamma) = (-1)^{nr} \lambda \circ (\alpha \circ \gamma) \quad \text{in} \quad \mathrm{Tor}_{m-p-r}^A(Q, M).

Formulas (13) and (14) follow immediately from the definitions. Let us prove (15). Let $z = \sum z_{ij}$ , $z_{ij} \in L_i(P) \otimes L_j(K)$ be a cycle representing $\gamma$ , $f : L(P) \to L(Q) (-n)$ and $g : L(K) \to L(M) (-r)$ morphisms representing $\alpha$ and $\lambda$ . Then $\lambda \circ (\alpha \circ \gamma)$ is the class of $\sum (-1)^{(i-n)r} (f \otimes g)(z_{ij})$ and $\alpha \circ (\lambda \circ \gamma)$ is the class of

\sum (-1)^{ir} (f \otimes g)(z_{ij}), \quad \text{whence (15)}.

8. Composition products and connecting homomorphisms of torsion products

Proposition 7

a) Let

(ε)

0 \to P' \xrightarrow{f} P \xrightarrow{g} P'' \to 0

be an exact sequence of right

A

-modules,

\theta \in \mathrm{Ext}_A^1(P'', P')

the associated class,

M

a left

A

-module. The connecting homomorphism

\delta_n(\mathscr{E}, M) : \mathrm{Tor}_n^A(P'', M) \to \mathrm{Tor}_{n-1}^A(P', M) \text{ is the mapping } \gamma \mapsto \theta \circ \gamma .

b) Let

(ε₁)

0 \to M' \to M \to M'' \to 0

be an exact sequence of left

A

-modules,

\theta_1 \in \mathrm{Ext}_A^1(M'', M')

the associated class,

P

a right

A

-module. The connecting homomorphism

\delta_n(P, \mathscr{E}_1) : \mathrm{Tor}_n^A(P, M'') \to \mathrm{Tor}_{n-1}^A(P, M') \text{ is the mapping } \gamma \mapsto \theta_1 \circ \gamma .

Let

\gamma \in \mathrm{Tor}_n^A(P'', M)

be the class of a cycle

z'' \in \dot{Z}_n(L(P'') \otimes_A L(M))

, and let

\begin{array}{ccccccccc} 0 & \longrightarrow & P' & \xrightarrow{f} & P & \xrightarrow{g} & P'' & \longrightarrow & 0 \\ & & \uparrow u_1 & & \uparrow u_0 & & \uparrow 1 & & \\ L_1(P'') & \xrightarrow{d_1} & L_0(P'') & \xrightarrow{p_0''} & P'' & \longrightarrow & 0 \end{array}

be a commutative diagram. We shall denote by

p' : L(P') \to P'

and

p'' : L(P'') \to P''

the canonical morphisms of complexes. By definition,

\delta(\gamma) \in \mathrm{Tor}_{n-1}^A(P', M)

is obtained as follows: one chooses

x \in P \otimes L_n(M)

such that

(g \otimes 1)(x) = (p'' \otimes 1)(z'')

and

\delta(\gamma)

is the class of cycles

z' \in Z_{n-1}(L(P') \otimes L(M))

such that

(f \otimes 1)(p' \otimes 1)(z') = (1 \otimes d_n)(x) .

For

0 \leq i \leq n

, let

z''_i

be the component of

z''

L_i(P'') \otimes L_{n-i}(M)

; we have

0 = Dz'' = \sum_i (d_i \otimes 1 + (-1)^i \otimes d_{n-i})(z''_i) ,

hence

(d_i \otimes 1)(z''_i) = (-1)^i \otimes d_{n-i+1}(z''_{i-1})

and in particular

(d_1 \otimes 1)(z''_1) = -1 \otimes d_n(z''_0) .

Let us then choose

x = (u_0 \otimes 1)(z''_0)

: we indeed have

(g \otimes 1)(x) = (p_0'' \otimes 1)(z''_0) = (p'' \otimes 1)(z'') .

Since

(1 \otimes d_n)(x) = (u_0 \otimes 1)(1 \otimes d_n)(z''_0) = - (u_0 \otimes 1)(d_1 \otimes 1)(z''_1) = - (f \otimes 1)(u_1 \otimes 1)(z''_1) ,

it follows that $\delta(\gamma)$ is the class of cycles $z' \in Z_{n-1}(L(P') \otimes_A L(M))$ such that $(p' \otimes 1)(z') = - (u_1 \otimes 1)(z''_1)$ . But, by definition, the class $\theta$ corresponds via the isomorphism $\mathrm{Ext}_A^1(P'', P') \to H^1(\mathrm{Homgr}_A(L(P''), P'))$ to the class of the morphism $f : L(P'')(1) \to P'$ defined by $- u_1$ , and the product $\theta \circ \gamma$ is the class of cycles

\overline{z}' \in Z_{n-1}(L(P') \otimes_A L(M)) \quad \text{such that} \quad (p \otimes 1)(\overline{z}') = f(z'') = - (u_1 \otimes 1)(z''_1),

which completes the proof of a). Assertion b) follows from a) by the commutation isomorphisms.

Corollary 1

Let* $0 \to P' \to P \to P'' \to 0$ be an exact sequence of right $A$ -*modules*, $0 \to M' \to M \to M'' \to 0$ be an exact sequence of left $A$ -*modules*. Then the composed homomorphisms of connecting homomorphisms

\mathrm{Tor}_n^A(P'', M'') \to \mathrm{Tor}_{n-1}^A(P'', M') \to \mathrm{Tor}_{n-2}^A(P', M')

and

\mathrm{Tor}_n^A(P'', M'') \to \mathrm{Tor}_{n-1}^A(P', M'') \to \mathrm{Tor}_{n-2}^A(P', M')

are opposite.

Indeed, if $\theta$ and $\theta_1$ are the classes associated with the given exact sequences, and if $\gamma \in \mathrm{Tor}_n^A(P'', M'')$ , the images of $\gamma$ are respectively $\theta \circ (\theta_1 \circ \gamma)$ and $\theta_1 \circ (\theta \circ \gamma)$ , hence are opposite by prop. 6.

Let us resume the notations of X, p. 127 and consider the sequence $(\mathscr{S})$ of left $A$ -modules and the connecting homomorphisms associated with the exact sequences (9)

\mathrm{Tor}_m^A(P, K_{i-1}) \to \mathrm{Tor}_{m-1}^A(P, K_i);

we deduce from this by composition of the iterated connecting homomorphisms

\hat{\partial}_m(P, \mathscr{S}) : \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-n}^A(P, N).

Then by prop. 7 and prop. 3 of X, p. 118 :

Corollary 2

If $\theta \in \mathrm{Ext}_A^n(M, N)$ is the class associated with the exact sequence $(\mathscr{S})$ , we have $\hat{\partial}_m(P, \mathscr{S})(\alpha) = \theta \circ \alpha$ for all $\alpha \in \mathrm{Tor}_m^A(P, M)$ .

Corollary 3

If all the modules $R_i$ , $i = 1, ..., n$ , are flat, the mapping $\alpha \mapsto \theta \circ \alpha$ from $\mathrm{Tor}_{m+n}^A(P, M)$ to $\mathrm{Tor}_m^A(P, N)$ is bijective for every right $A$ -module $P$ and every integer $m > 0$ .

This follows from Cor. 2 and the exact sequences

\mathrm{Tor}_{m+n-i+1}^A(P, R_i) \to \mathrm{Tor}_{m+n-i+1}^A(P, K_{i-1}) \xrightarrow{\hat{\partial}} \mathrm{Tor}_{m+n-i}^A(P, K_i) \to \mathrm{Tor}_{m+n-i}^A(P, R_i)

where the extreme terms are zero by hypothesis.

Analogously, if

(\mathscr{S}_1)

0 \to Q \to S_n \to S_{n-1} \to \ldots \to S_1 \to P \to 0

is an exact sequence of right $A$ -modules, and $M$ a left $A$ -module, we define iterated connecting homomorphisms

\partial^m(\mathscr{S}_1, M) : \mathrm{Tor}_m^A(P, M) \to \mathrm{Tor}_{m-n}^A(Q, M)

and we have:

Corollary 4

If $\theta_1 \in \mathrm{Ext}_A^n(P, Q)$ is the class associated with the exact sequence $(\mathscr{S}_1)$ , we have $\partial^m(\mathscr{S}_1, M)(\alpha) = \theta_1 \circ \alpha$ for all $\alpha \in \mathrm{Tor}_m^A(P, M)$ .

9. Computation of Composition Products by Resolution Shifts

Let

(16)

0 \to M \xrightarrow{\iota} K_n \to K_{n-1} \to \ldots \to K_1 \xrightarrow{\rho} M' \to 0

be an exact sequence of left $A$ -modules and $\theta \in \mathrm{Ext}_A^n(M', M)$ the associated class.

Let $a : (R, d) \to M$ be a left resolution of $M$ ; hence there is an exact sequence

\to R_k \xrightarrow{d_k} R_{k-1} \to \ldots \xrightarrow{d_1} R_0 \xrightarrow{a_0} M \to 0.

and by shifting $n$ (X, p. 26) an exact sequence

(17)

\to R_k \xrightarrow{(-1)^n d_k} R_{k-1} \to \ldots \xrightarrow{(-1)^n d_1} R_0 \xrightarrow{(-1)^n a_0} M \to 0.

We deduce from (16) and (17) an exact sequence

\to R_k \xrightarrow{(-1)^n d_k} R_{k-1} \to \ldots \xrightarrow{(-1)^n d_1} R_0 \xrightarrow{(-1)^n f \circ a_0} K_n \to K_{n-1} \to \ldots \to K_1 \to M' \to 0

thereby obtaining a resolution $R'$ of $M'$ ; let $\varphi : R' \to R(-n)$ be the morphism such that $\varphi_k = 1_{R_{k-n}}$ for $k \geq n$ .

If $N$ is a left $A$ -module and $P$ a right $A$ -module, we therefore have mappings

\mathrm{H}(1_P \otimes \varphi) : \mathrm{H}(P \otimes_A R') \to \mathrm{H}(P \otimes_A R)(-n)

\mathrm{H}(\mathrm{Homgr}_A(\varphi, 1_N)) : \mathrm{H}(\mathrm{Homgr}_A(R, N))(n) \to \mathrm{H}(\mathrm{Homgr}_A(R', N)).

Let $k$ be an integer.

Proposition 8

a)* The following diagram, where $h_\theta(\alpha) = \theta \circ \alpha$ , is commutative

\begin{array}{ccc} \mathrm{Tor}_{k+n}^A(P, M') & \xrightarrow{h_\theta} & \mathrm{Tor}_k^A(P, M) \\ \psi_{k+n}(P, R') \downarrow & & \downarrow \psi_k(P, R) \\ \mathrm{H}_{k+n}(P \otimes_A R') & \xrightarrow{H_{k+n}(1 \otimes \varphi)} & \mathrm{H}_k(P \otimes_A R) \end{array}

b) The following diagram, where $\delta_\theta(\beta) = \beta \circ \theta$ , is commutative

\begin{array}{ccc} H^k(\mathrm{Homgr}_A(R, N)) & \xrightarrow{H^{k+n}(\mathrm{Homgr}_A(\varphi, 1))} & H^{k+n}(\mathrm{Homgr}_A(R', N)) \\ \varphi^k(R, N) \downarrow & & \varphi^{k+n}(R', N) \downarrow \\ \mathrm{Ext}_A^k(M, N) & \xrightarrow{\delta_\theta} & \mathrm{Ext}_A^{k+n}(M', N). \end{array}

Let $\alpha : L(M) \to R$ be a morphism of complexes such that $a \circ \alpha = p_M$ and let

\begin{array}{ccccccccc} L_n(M') & \longrightarrow & L_{n-1}(M') & \longrightarrow & \ldots & \longrightarrow & L_0(M') & \longrightarrow & M' \longrightarrow 0 \\ u_n \downarrow & & u_{n-1} \downarrow & & & & u_0 \downarrow & & 1 \downarrow \\ 0 & \longrightarrow & M & \xrightarrow{f} & K_n & \longrightarrow & \ldots & \longrightarrow & K_1 \longrightarrow M' \longrightarrow 0 \end{array}

a commutative diagram, let us choose a homomorphism $v_n : L_n(M') \to L_0(M)$ such that $p_M \circ v_n = (-1)^n u_n$ ; by X, p. 47, Proposition 1, a), $v_n$ extends to a morphism of complexes $v : L(M') \to L(M)$ $(-n)$ , and $\theta$ is the image under the canonical isomorphism $H^n(\mathrm{Homgr}_A(L(M'), L(M))) \to \mathrm{Ext}_A^n(M', M)$ of the class of $v$ (X, p. 117, Remark 1). We define a morphism of complexes $\beta : L(M') \to R'$ by $\beta_p = u_p$ for $p \leq n - 1$ , $\beta_p = \alpha_{p-n} \circ v_p$ for $p \geq n$ , and we have

\varphi \circ \beta = \alpha(-n) \circ v.

On the other hand, by definition of the mappings $\varphi$ and $\psi$ , we have

\begin{aligned} & \psi_k(P, R) = H_k(p_P \otimes \alpha), \quad \varphi^k(R, N) = H^k(\mathrm{Homgr}_A(\alpha, e_N)), \\ & \psi_{k+n}(P, R') = H_{k+n}(p_P \otimes \beta), \quad \varphi^{k+n}(R', \tilde{N}) = H^{k+n}(\mathrm{Homgr}_A(\beta, e_N)). \end{aligned}

Finally, by definition of the composition product, we have

h_\theta = H(1_{L(P)} \otimes v), \quad \delta_\theta = H(\mathrm{Homgr}_A(v, 1_{L(N)})) .

Consequently, we have the equalities

\begin{aligned} \psi_k(P, R) \circ h_\theta &= H_k(p_P \otimes \alpha) \circ H_k(1_{L(P)} \otimes v) = H_k(p_P \otimes (\alpha \circ v)) = H_{k+n}(p_P \otimes (\varphi \circ \beta)) \\ &= H_{k+n}(1 \otimes \varphi) \circ H_{k+n}(p_P \circ \beta) = H_{k+n}(1 \otimes \varphi) \circ \psi_{k+n}(P, R'), \end{aligned}

from which a) follows; the proof of b) is analogous.

Remark

Via the commutation isomorphisms, we deduce from a) an analogous statement in the case of an exact sequence (16) of right A-modules.

Let $b : M' \to E'$ be a right resolution of $M'$ ; we thus have an exact sequence

0 \to M' \xrightarrow{b^0} {E'}^0 \xrightarrow{\delta^0} {E'}^1 \to \ldots \to {E'}^k \xrightarrow{\delta^k} {E'}^{k+1}

corresponding to a right resolution $E$ of $M$ ; let $\sigma : E'(n) \to E$ be the morphism such that $\sigma^k = 1_{{E'}^{k-n}}$ for $k \geq n$ . We thus have homomorphisms

H(\mathrm{Homgr}_A(1_N, \sigma)) : H(\mathrm{Homgr}_A(N, E'))(n) \to H(\mathrm{Homgr}_A(N, E)) .

Proposition 9

The following diagram, where $\gamma_\theta(\alpha) = \theta \circ \alpha$ , is commutative*:

\begin{array}{ccc} H^k(\mathrm{Homgr}_A(N, E')) & \xrightarrow{H^{k+n}(\mathrm{Homgr}_A(1_N, \sigma))} & H^{k+n}(\mathrm{Homgr}_A(N, E)) \\ \downarrow \varphi^k(N, E') & & \downarrow \varphi^{k+n}(N, E) \\ \mathrm{Ext}_A^k(N, M') & \xrightarrow{\gamma_\theta} & \mathrm{Ext}_A^{k+n}(N, M) \end{array}

This is proved in a way analogous to Proposition 8.

Exercises

See the exercises for § 7.