is an exact sequence of commutative groups.

This is the Corollary to Proposition 5 of Algebra, Chapter II, § 3, no. 6.

It follows that, for any left A-module homomorphism $u : M \to N$,

is canonically identified with $\operatorname{Coker}(1_E \otimes u)$, as Lemma 1 shows when applied to the exact sequence

In the notation of Lemma 1, we know (loc. cit.) that if v is injective, that is if the sequence $0 \to M' \xrightarrow{v} M \xrightarrow{w} M'' \to 0$ is exact, it does not necessarily follow that $1_E \otimes v$ is injective and so $E \otimes_A M'$ cannot in general be identified with a

(*) We inform readers already familiar with Homological Algebra that they will find other characterizations of flat modules in § 4.

subgroupof $E \otimes_A M$. Recall however (Algebra, Chapter II, § 3, no. 7, Corollary 5 to Proposition 7) the following result:

(*j being the canonical injection $M' \to M$) be injective.

Suppose that this condition holds and let $N$ be any submodule of $M$. Suppose that the canonical image in $E \otimes_A M$ of an element

is zero and let $M'$ be the finitely generated submodule of $N$ generated by they.; as by hypothesis the composite mapping $E \otimes_A M' \to E \otimes_A N \to E \otimes_A M$ is injective, the sum $z' = \sum x_i \otimes y_i$, considered as an element of $E \otimes_A M'$, is zero. As $z$ is the image of $z'$, we have also $z = 0$, whence the lemma.

The case in which $N$ is a submodule is easy, as, if $N'$ is a submodule of $N$, the composite homomorphism

is injective, hence so is $E \otimes_A N' \to E \otimes_A N$. Suppose then that $N$ is a quotient module of $M$, that is there exists an exact sequence $0 \to R \xrightarrow{i} M \xrightarrow{p} N \to 0$.

Let $N'$ be a submodule of $N$ and $M' = \overline{p}(N')$. Let $i'$ denote the mapping of $R$ to $M'$ with the same graph as $i$, $p'$ the surjection $M \to N'$ with the same graph as the restriction of $p$ to $M'$, $r$ the identity mapping of $\mathbf{R}$ to $\mathbf{R}$, $m$ the canonical injection $M' \to M$ and $n$ the canonical injection $N' \to N$. The diagram

is commutative and its rows are exact.

To simplify the writing, we set $T(Q) = E \otimes_A Q$ for every left $A$-module $Q$ and $T(v) = 1_E \otimes v$ for every left $A$-module homomorphism $v$. The diagram

is commutative and its rows are exact by Lemma 1 of no. 1. Moreover, since $E$ is $M$-flat, the homomorphism $T(m)$ is injective. As $T(r)$ and $T(p')$ are surjective, it follows from § 1, no. 4, Corollary 2 to Proposition 2 that $T(n)$ is injective, which proves the lemma.

(a) Suppose first that $I = \{1, 2\}$, and let $M'$ be a submodule of $M = M_1 \oplus M_2$, $M$, and $M$, being canonically identified with submodules of $M$. Denote by $M'_1$ the intersection $M' \cap M_1$, and by $M'_2$ the image of $M'$ in $M_2$ under the canonical projection $p$ of $M$ onto $M_2$. We have a diagram

where $v_1, v, v_2, i, i'$ are the canonical injections and $p'$ is the mapping with the same graph as the restriction of $p$ to $M'$, which is surjective. It is immediately verified that this diagram is commutative and that its rows are exact. With $T(Q)$ and $T(v)$ used in the same sense as in the proof of Lemma 4, we have a commutative diagram

By Lemma 1 of no. 1 the two rows of this diagram are exact; as E is flat for $M_1$ and $M_n$, $T(v_1)$ and $T(v_2)$ are injective; moreover, by Lemma 2 of no. 1, $T(i)$ is injective. Corollary 2 to Proposition 2 of § 1, no. 4 then shows that $T(v)$ is injective and consequently E is M-flat.

(b) If I is a finite set with n elements, the lemma follows by induction on n using (a).

(c) In the general case let $M'$ be a finitely generated submodule of M. Then there exists a finite subset J of the indexing set I such that $M'$ is contained in the direct sum $M_J = \bigoplus_{i \in J} M_i$. By (b) E is flat for $M_J$; the canonical homomorphism $E \otimes_A M' \to E \otimes_A M$, is therefore injective. On the other hand, as $M_J$ is a direct factor of M, the canonical homomorphism $E \otimes_A M_J \to E \otimes_A M$ is injective (no. 1, Lemma 2). Taking the composition, it follows that $E \otimes_A M, \to E \otimes_A M$ is injective and E is flat for M by Lemma 3.

(a) E is flat for A, (in other words, for every left ideal $a$ of A, the canonical homomorphism $E \otimes_A a \to E \otimes_A A, = E$ is injective). (b) E is M-flat for every left A-module M. (c) For every exact sequence of left A-modules and homomorphisms

the sequence

is exact.

It is immediate that (b) implies (a). Conversely, suppose that (a) holds; by Lemma 5 of no. 2, E is flat for every free left A-module; as every left A-module is isomorphic to a quotient of a free module (Algebra, Chapter 11, § 1, no. 11, Proposition 20), it follows from Lemma 4 of no. 2 that E is flat for M.

We show that (c) implies (b). If $v : M' \to M$ is an injective homomorphism, the sequence $0 \to M' \xrightarrow{v} M$ is exact; by (c) the sequence

is exact; this means that $1 \otimes v$ is injective, in other words, E is M-flat.

Finally, the implication (b) $\Rightarrow$ (c) is a consequence of the following more precise lemma:

is exact.

We use the notation $T(Q)$ and $T(v)$ in the same sense as in the proof of Lemma 4 of no. 2. We write $M_1'' = w(M)$ and let $i : M_1'' \to M''$ be the canonical injection and $p$ the mapping of M to $M_1''$ with the same graph as w. The sequence $M' \xrightarrow{v} M \xrightarrow{p} M_1'' \to 0$ is exact and it follows from Lemma 1 of no. 1 that the sequence

is exact. Moreover, as E is $M''$-flat, the mapping $T(i) : T(M_1'') \to T(M'')$ is injective, and as $T(i) \circ T(p) = T(w)$, the sequence

is exact. (§ 1, no. 3.)

(ii) Let $I$ be an ordered set and $(E_\alpha, f_{\beta \alpha})$ a direct system of right $A$-modules (Algebra, Chapter II, § 6, no. 2). If each of the $E_\alpha$ is flat, then $E = \lim_{\longrightarrow} E_\alpha$ is flat.

Let $M' \to M$ be an injective left $A$-module homomorphism.

(i) For the direct sum homomorphism

to be exact, it is necessary and sufficient that each of the homomorphisms $E_t \otimes_A M' \to E_t \otimes_A M$ be so (Algebra, Chapter II, § 1, no. 6, Corollary 1 of Proposition 7), which proves (i) since $\bigoplus_{t \in I} (E_t \otimes_A M)$ is canonically identified with $E \otimes_A M$ (Algebra, Chapter II, § 3, no. 7, Proposition 7).

(ii) By hypothesis each of the sequences

is exact; so then is the sequence since taking the direct limit commutes with tensor products (Algebra, Chapter II, § 6, no. 3, Proposition 7) and preserves exactness (ibid., § 6, no. 2, Proposition 3).

(i) Suppose that $E$ is flat. For every element $a \in A$ which is not a right divisor of $0(*)$, the relations $x \in E,\ xa = 0$ imply $x = 0$.

(*) Recall that a right (resp. left) divisor of 0 in a ring $A$ is an element $b \in A$ such that the mapping $x \mapsto xb$ (resp. $x \mapsto bx$) is not injective.

(ii) Suppose that $A$ is an integral domain in which every finitely generated ideal is principal (for example a principal ideal domain (Algebra, Chapter VII, , no. 1)). Then for $E$ to be flat it is necessary and sufficient that $E$ be torsion-free

We prove (i). Let $v : A, \to A$, be the left $A$-module homomorphism $t \mapsto ta$; the hypothesis implies that $v$ is injective. As $E$ is flat, the homomorphism $1_E \otimes v : E \otimes_A A, \to E \otimes_A A$, is also injective. When $E \otimes_A A$, is canonically identified with $E$, $1, \otimes v$ becomes the endomorphism $x \mapsto xa$ of $E$. Thus the relation $xa = 0$ implies $x = 0$.

We prove (ii). By (i), if $E$ is flat, $E$ is torsion-free. Conversely, let $E$ be a torsion-free $A$-module; we verify that, for every finitely generated ideal $a$ of $A$, the canonical homomorphism $E \otimes_A a \to E$ is injective (no. 3, Remark 1). This assertion is obvious if $a = (0)$; otherwise, by hypothesis $a = Aa$ for some $a \in A, a \neq 0$, and $t \mapsto ta$ is then an isomorphism $v$ of $A$ onto $a$; using $i$ to denote the canonical injection $a \to A$, $i \circ v$ is the homothety with ratio $a$ on $E$ and is injective since $E$ is assumed to be torsion-free. Then $1, \otimes (i \circ v) = (1, \otimes i) \circ (1_E \otimes v)$; as $1_E \otimes v$ is an isomorphism, $1, \otimes i$ is injective, which completes the proof.

(a) $E$ is flat. (b) For every exact sequence of right $A$-modules of the form

(1)

and every left $A$-module $F$, the sequence

(2)

is exact.

(c) There exists an exact sequence (1), where $H$ is flat, such that the sequence (2) is exact for every left $A$-module $F$ of the form $A_s/a$, where $a$ is a finitely generated left ideal of $A$.

We show first that (a) implies (b). The left $A$-module $F$ is isomorphic to a quotient of a free module (Algebra, Chapter II, § 1, no. 11, Proposition 20); in other words, we have an exact sequence

where L is free. Consider the diagram

(3)

It follows immediately that this diagram is commutative, and its rows and columns are exact by Lemma 1 of no. 1; moreover, as $l_G \otimes p$ and $l_H \otimes p$ are surjective (no. 1, Lemma 1), we have $G \otimes F = \operatorname{Coker}(l_G \otimes i)$,

$w \otimes 1_1$ is surjective (no. 1, Lemma 1); finally, as L is free and hence flat, $v \otimes 1_L$ is injective. Thus the snake diagram (§ 1, no. 4, Proposition 2, (iii)) can be applied to prove the existence of an exact sequence

(4)

This being so, if E is flat, $l_E \otimes i$ is injective, in other words $\operatorname{Ker}(l_E \otimes i) = 0$, and the exact sequence (4) shows that $v \otimes 1_F$ is injective, hence the sequence (2) is exact (taking account of Lemma 1 of no. 1).

As (b) obviously implies (c), it remains to prove that (c) implies (a). Consider the diagram (3) in the case $\mathbf{R} = \mathbf{a}, \mathbf{L} = \mathbf{A}_s, \mathbf{F} = \mathbf{A}_s/\mathbf{a}$ and apply the exact sequence (4). By hypothesis $v \otimes 1_F$ is injective, hence $\operatorname{Im}(d) = 0$; moreover, as H is flat, we have $\operatorname{Ker}(1, @i) = 0$; the exactness of the sequence (4) then implies $\operatorname{Ker}(l_E \otimes i) = 0$, in other words, $1, \otimes i$ is injective and this proves that E is flat (no. 3, Remark 1).

Let $u : F' \to F$ be an injective homomorphism of left $\mathbf{A}$-modules. Consider the diagram

It is commutative and its rows are exact (no. 1, Lemma 1). Since $E''$ is flat, l_{E''} \otimes u is injective; moreover, Proposition 4 proves that v \otimes 1_{F'}, and v \otimes 1, are injective. This being so, if E is flat, l_E \otimes u is injective, hence also

it follows that l_{E'} \otimes u is injective and consequently E' is flat. Conversely, if E' is flat, l_{E'} \otimes u is injective; then it follows from § 1, no. 4, Corollary 1 to Proposition 2 that l_E \otimes u is injective and so E is flat.

Remarks (1) It is possible for E and E' to be flat without E'' being so, as the example of the \mathbf{Z}-modules E = \mathbf{Z}, E' = n\mathbf{Z}, E'' = \mathbf{Z}/n\mathbf{Z} (n \geq 2), shows. (2) A submodule of a flat module is not necessarily a flat module (Exercise 3).

Consider the diagram

where the unspecified arrows are the canonical injections and surjections and j the canonical isomorphism defined in Algebra, Chapter I, § 6, no. 13, Theorem 6. This diagram is commutative and its rows are exact. We derive (since F = F' + F'') a commutative diagram

The rows of this diagram are exact (no. 1, Lemma 1) and 1 \otimes j is an isomorphism. Our assertion is then a special case of § 1, no. 4, Proposition 1, (i) (cf. Exercise 5).

As E is F-flat, $\phi(F' + F'')$ is identified with $E \otimes (F' + F'')$, and the submodules $\phi(F')$, $\phi(F'')$ and $\phi(F' \cap F'')$ are identified with the canonical images of $E \otimes F'$, $E \otimes F''$ and $E \otimes (F' \cap F'')$ in $E \otimes (F' + F'')$ respectively. Proposition 6 then follows from Lemma 7.

Suppose for example that $E/E'$ is flat and consider the diagram

where the arrows are the canonical homomorphisms. This diagram is commutative and its rows are exact (no. 1, Lemma 1). As $E/E'$ is flat, $u$ is injective. Then our assertion is a special case of § 1, no. 4, Proposition 1, (i).

(i) *Suppose that E/E' is flat. Then, for every left ideal a of A,

(ii) Conversely, suppose that E is flat and, for every finitely generated left ideal a of A, relation (5) holds. Then E/E' is flat.

(i) It is sufficient to apply Proposition to the case $F = A$, $F' = a$. (ii) To prove that $E/E'$ is flat, apply criterion (c) of Proposition 4 of no. 5; it is then necessary to establish that the sequence

is exact at $E'/E'a$ for every finitely generated left ideal a of A. Now this is precisely what relation (5) expresses.

Let $G$ be a left $B$-module and $G'$ a submodule of $G$. Since $F$ is flat as a right $B$-module, the canonical homomorphism $F \otimes_B G' \to F \otimes_B G$ is injective. Since $E$ is flat, the canonical homomorphism

is injective. As $E \otimes_A (F \otimes_B G')$ and $E \otimes_A (F \otimes_B G)$ are canonically identified with $(E \otimes_A F) \otimes_B G'$ and $(E \otimes_A F) \otimes_B G$ respectively (Algebra, Chapter II, § 3, no. 8, Proposition 8), the canonical homomorphism

is injective, which proves that $E \otimes_A F$ is a flat $B$-module.

$F$ is a $(C, C)$-bimodule and it is sufficient to apply Proposition 8 with $B = A = C$.

By definition $E_{(B)} = E \otimes_A B$, where $B$ is considered as an $(A,B)$-bimodule by means of $\rho$. As the right $B$-module $B_d$ is flat, it is sufficient to apply Proposition 8.

Recall that $\phi_*(M)$ is the right $R$-module defined by $x.r = x.\phi(r)$ for all $x \in M$ and all $r \in R$ (Algebra, Chapter II, § 1, no. 13). Then apply Proposition 8 with $A = S, B = R, E = M$ and $F = S, S$ having the structure of a $(S, R)$-bimodule defined by $\phi$; the right $R$-module $M \otimes_S S$ is then precisely $\phi_*(M)$.

Let $E'_\alpha = E_\alpha \otimes_{A_\alpha} A$, where $A$ is considered as a left $A_\alpha$-module by means of the canonical homomorphism $A_\alpha \to A$; we know that the right $A$-module $E$ is canonically isomorphic to $\lim E'_\alpha$ (loc. cit., Corollary 2 to Proposition 7). It follows from Corollary 2 to Proposition 8 that $E'_\alpha$ is a flat right $A$-module for all $\alpha$, hence $E$ is a flat $A$-module by Proposition 2 of no. 3.

Assertion (i) follows trivially from the definitions. If $A$ is left Noetherian and there exists a surjective homomorphism $u : L_0 \to E$, where $L_0$ is a free left $A$-module with a finite basis, the kernel $R$ of $u$ is finitely generated (Algebra, Chapter VIII, , no. 1, Proposition 1 and no. 3, Proposition 7), hence there is a surjective homomorphism $u : L_1 \to R$, where $L_1$ is free and has a finite basis, and the exact sequence $L_1 \xrightarrow{v} L_0 \xrightarrow{u} E \to 0$ is a finite presentation of $E$; whence (ii)

Finally, suppose that $E$ is a finitely generated projective module; then it is a direct factor of a finitely generated free module $L_0$ (Algebra, Chapter II, §2, no. 2, Corollary to Proposition 4); the kernel $R$ of the surjective homomorphism $L_0 \to E$ is then isomorphic to a quotient of $L_0$ and hence finitely generated and the proof is completed as for (ii).

where $G$ is finitely generated, the module $F$ is finitely generated.

Let $L, \xrightarrow{r} L_0 \xrightarrow{s} E \to 0$ be a finite presentation; if $(e_i)$ is a basis of $L_0$, there exists for each $i$ an element $g_i \in G$ such that $p(g_i) = s(e_i)$; the homomorphism $u : L_0 \to G$ such that $u(e_i) = g_i$ for all $i$ then satisfies $s = p \circ u$. As $s \circ r = 0$, we have $u(r(L_1)) \subset \mathrm{Ker}\, p$, and as $\mathrm{Ker}\, p$ is isomorphic to $F$, it can be seen that there is a homomorphism $v : L, \to F$ such that the diagram

is commutative. As $j$ is injective and $s$ surjective, we can apply the snake diagram ($\S$ 1, no. 4, Proposition 4), in other words there is an exact sequence

This shows that $\mathrm{Coker}\, v$ is isomorphic to $G / u(L_0)$, which is finitely generated by hypothesis. Moreover we have the exact sequence

and as $v(L_1)$ and $\mathrm{Coker}\, v$ are finitely generated, so is $F$ (Algebra, Chapter II, § 1, no. 7, Corollary 5 to Proposition 9).