(a) For a sequence $N' \xrightarrow{v} N \xrightarrow{w} N''$ of left $A$-modules to be exact, it is necessary and sufficient that the sequence

be exact.

(b) $E$ is flat and, for every left $A$-module $N$, the relation $E \otimes_A N = 0$ implies $N = 0$.

(c) $E$ is flat and, for every homomorphism $v : N' \to N$ of left $A$-modules, the relation $1_E \otimes v = 0$ implies $v = 0$.

(d) $E$ is flat and, for every maximal left ideal $m$ of $A$, $E \neq Em$.

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 homomorphism $v$ of left $A$-modules.

We prove first the equivalence of (a), (b) and (c).

We prove that (a) implies (b). If (a) holds, clearly $E$ is flat (§ 2, no. 3, Proposition 1). On the other hand, let $N$ be a left $A$-module such that $T(N) = 0$ and consider the sequence $0 \to N \to 0$; the hypothesis $T(N) = 0$ means that the sequence $0 \to T(N) \to 0$ is exact. By (a) the sequence $0 \to N \to 0$ is exact, whence $N = 0$.

We show that (b) implies (c). Suppose that (b) holds and let $u : N' \to N$ be a homomorphism and $I$ its image. As the image of $T(v)$ is identified with $T(I)$ (§ 2, no. 3, Remark 2), the hypothesis $T(v) = 0$ implies $T(I) = 0$, hence $I = 0$ by (b) and consequently $v = 0$.

We show that (c) implies (a). Suppose then that (c) holds and consider a sequence

of homomorphisms of left $A$-modules and the corresponding sequence

If the sequence (1) is exact, so is (2), since $E$ is flat (§ 2, no. 3, Proposition 1). Conversely, if (2) is exact, we have first $T(w \circ v) = T(w) \circ T(v) = 0$, hence $w \circ v = 0$ by hypothesis. Set $I = v(N')$ and $K = \overline{w}(0)$; then $I$ is contained in $K$ by the above. Consider the exact sequence

$i$ and $p$ being the canonical mappings. As $E$ is flat, the sequence is exact, in other words, $T(K/I)$ is isomorphic to $T(K)/T(I)$, which is 0 by hypothesis, since $T(I)$ (resp. $T(K)$) is identified with the image of $T(v)$ (resp. the kernel of $T(w)$) (§ 2, no. 3, Remark 2). But the relation $T(p) = 0$ implies $p = 0$ by hypothesis, hence $K = I$, which proves that the sequence (1) is exact.

Finally we show the equivalence of (b) and (d). If (b) holds, then

since $A_s/m \neq 0$; whence (d). Conversely, suppose that (d) holds; every left ideal $a \neq A$ of $A$ is contained in a maximal left ideal $m$ (Algebra, Chapter I, §8, no. 7, Theorem 2), then the hypothesis $E \neq Em$ implies $E \neq Ea$, in other words, $E \otimes_A (A_s/a) \neq 0$. That is to say, for every monogenous left $A$-module $N \neq 0$, $T(N) \neq 0$. If now $N$ is any left $A$-module \#0, it contains a monogenous submodule $N' \neq 0$; since $E$ is flat, $T(N')$ is identified with a subgroup of $T(N)$; we have just seen that $T(N') \neq 0$, hence $T(N) \neq 0$.

This is an immediate consequence of criterion (a) of Proposition 1.

We know already that $E$ is flat ($§ 2$, no. 5, Proposition 5). We verify that $E$ has property (b) of Proposition 1. Let $N$ be a left $A$-module. As $E''$ is flat, there is an exact sequence

($§ 2$, no. 5, Proposition 4). If $E \otimes_A N = 0$, it follows that $E' \otimes_A N$ and $E'' \otimes_A N$ are zero; as one of the modules $E'$, $E''$ is faithfully flat, this implies that $N = 0$.

(1) If $F$ is flat, $E \otimes_R F$ is flat ($§ 2$, no. 7, Proposition 8).

(2) Suppose that $E \otimes_R F$ is flat and let $\nu : N' \to N$ be an injective left $S$-module homomorphism. The homomorphism

is then injective ($§ 2$, no. 3, Proposition 1). It follows from Proposition 2 of no. 1 that $l_F \otimes \nu : F \otimes_S N' \to F \otimes_S N$ is injective; then $F$ is a flat $S$-module ($§ 2$, no. 3, Proposition 1).

(3) Suppose that $F$ is faithfully flat and let $N$ be a left $S$-module such that $E \otimes_R F \otimes_S N = 0$. Since $E$ is faithfully flat, this implies that $F \otimes_S N = 0$, whence $N = 0$ since $F$ is faithfully flat; this proves that $E \otimes_R F$ is faithfully flat.

(4) Suppose that $E \otimes_R F$ is faithfully flat and let $N$ be a left $S$-module such that $F \otimes_S N = 0$. Then $E \otimes_R F \otimes_S N = 0$, whence $N = 0$, which shows that $F$ is faithfully flat.

Apply Proposition 4 with $R = S = C$.

Apply Proposition 4 of no. 2 with $R = A, S = F = B$, noting that the $B$-module $B_d$ is faithfully flat.

Apply Proposition 5 with $B = A/a, \rho$ being the canonical homomorphism.

(1) If $E$ is flat (resp. faithfully flat), $E_{(B)}$ is flat (resp. faithfully flat) by § 2, no. 7, Corollary 2 to Proposition 8 (resp. by Proposition 5).

(2) Suppose that $E_{(B)}$ is flat and let $\nu : N' \to N$ be an injective $A$-module homomorphism. By § 2, no. 7, Corollary 3, the $A$-module $E \otimes_A B$ is flat, hence the homomorphism $1, \otimes 1_B \otimes \nu : E \otimes_A B \otimes_A N' \to E \otimes_A B \otimes_A N$ is injective. As the right and left $A$-module structures on $B$ coincide, this homomorphism is identified with

As $B$ is a faithfully flat $A$-module, it follows that $1_E \otimes \nu : E \otimes_A N' \to E \otimes_A N$ is injective (no. 1, Proposition 2), which shows that $E$ is flat.

(3) Suppose finally that $E_{(B)}$ is faithfully flat. First of all $E$ is flat by (2). Also let $N$ be an $A$-module such that $E \otimes_A N = 0$. Then $E \otimes_A N \otimes_A B = 0$, whence, since the right and left $A$-module structures on $B$ coincide, $E \otimes_A B \otimes_A N = 0$, which may also be written $(E \otimes_A B) \otimes_B (B \otimes_A N) = 0$. As $E_{(B)}$ is a faithfully flat $B$-module, this implies that $B \otimes_A N = 0$ (no. 1, Proposition 1), whence $N = 0$ since $B$ is a faithfully flat $A$-module (no. 1, Proposition 1).

Applying Proposition 4 of no. 2 with $B, A, E, B$ in place of $R, S, E, F$ respectively, and the right $A$-module structure on $B$ being defined by $\rho$, it is seen that

B is a flat (resp. faithfully flat) A-module if and only if $E \otimes_B B = \rho^*(E)$ is a flat (resp. faithfully flat) A-module.

Remarks (1) Proposition 7 shows that, for B to be a faithfully flat A-module, it is sufficient that there exist one faithfully flat B-module which is also a faithfully flat A-module. (2) Let A, B, C be three rings and $\rho : A \to B, \sigma : B \to C$ two ring homomorphisms. Proposition 7 shows that if C is a faithfully flat B-module and B a faithfully flat A-module, then C is a faithfully flat A-module. If C is a faithfully flat B-module and a faithfully flat A-module, then B is a faithfully flat A-module (taking the modules as right modules, to fix the ideas). On the other hand B and C may be faithfully flat A-modules without C being a faithfully flat B-module (Exercise 7).

We prove (i). We know (Algebra, Chapter 11, § 5, no. 2, Corollary to Proposition 5) that for every right B-module M, the canonical A-homomorphism $i : M \to \rho_*(M) \otimes_A B = \rho^*(\rho_*(M))$ defined by $i(y) = y \otimes 1$ is injective and that the A-module $i(M)$ is a direct factor of $\rho_*(M) \otimes_A B$. Hence, for every left A-module F,

is injective (§ 2, no. 1, Lemma 2). Taking $M = E$, it follows (since $i \otimes 1_F = 1_M \otimes j$) that $j$ is injective (no. 1, Proposition 2).

Assertion (ii) follows from (i) by taking $F = A_s/a$ and (iii) from (ii) by taking $a = \{0\}$.

Finally, if m is a maximal left ideal of A, then $\overline{\rho}^{-1}(Bm) = m$ by (ii), and consequently $Bm \# B$. Then there exists a maximal left ideal n of B containing Bm (Algebra, Chapter I, § 8, no. 7, Theorem 2); then $m \subset \overline{\rho}^{-1}(n)$ and as $\rho(1) \notin n$, $1 \notin \overline{\rho}^{-1}(n)$. Consequently $\overline{\rho}^{-1}(n) = m$.

If $A$ and $B$ satisfy the conditions of Proposition 8, $A$ is usually identified with a subring of $B$ by means of $\rho$.

If (a) is a non-stationary increasing (resp. decreasing) sequence of left ideals of $A$, the sequence $(\mathrm{Ba}_n)$ of ideals of $B$ is non-stationary and increasing (resp. decreasing) since $\rho^{-1}(\mathrm{Ba}_n) = a_n$ contrary to the hypothesis.

Remark (1). If $A$ and $B$ are commutative, we shall see in Chapter II, § 2, no. 5, Corollary 4 to Proposition 11, that the hypothesis of Proposition 8 implies that for every prime ideal $p$ of $A$, there exists a prime ideal $q$ of $B$ such that $\rho^{-1}(q) = p$ (where $p = A \cap q$ when $A$ is identified with a subring of $B$).

An important application of Proposition 8 is when $B$ is itself a faithfully flat $A$-module. But in this case we have the following more precise proposition:

By Proposition 8, (a) implies each of the properties (c), (d), (e). On the other hand, if (e) holds, then $Bm \# B$ for every maximal left ideal $m$ of $A$ (since there exists a maximal left ideal $n$ of $B$ such that $Bm \subset n$), and $B$ is a faithfully flat $A$-module by criterion (d) of Proposition 1 of no. 1; hence (e) implies (a).

We shall now prove that (c) $\Rightarrow$ (d) $\Rightarrow$ (b) $\Rightarrow$ (a), which will complete the proof. In the first place, (c) implies (d) by taking $F = A_s/a$ in (c). If (d) holds, by taking $a = \{0\}$ it follows that $\rho$ is injective; (d) and § 2, no. 6, Corollary to Proposition 7 imply that $B/\rho(A)$ is a flat right $A$-module, that is (d) implies (b). Finally, if (b) holds, Proposition 3 of no. 1 applied to the exact sequence

shows that $B$ is a faithfully flat right $A$-module, since $A_a$ is faithfully flat.

*Remark (2). If $A$ and $B$ are commutative, we shall see in Chapter II, § 2 no. 5, Corollary 4 to Proposition 11 that the conditions of Proposition 9 are equivalent to the following:

(f) B is a flat A-module and, for every prime ideal $\mathfrak{p}$ of $A$, there exists an ideal $q$ of $B$ such that $\bar{\rho}^{-1}(q) = \mathfrak{p}$.

Under the conditions of Proposition 9, let us identify $A$ with a subring of $B$ by means of $\mathfrak{p}$. The relation $\bar{\rho}^{-1}(Ba) = a$ then reads $A \cap Ba = a$. On the other hand if $F$ is a left $A$-module, $F$ is identified with its image in $B \otimes_A F$ under the canonical mapping $x \mapsto 1 @ x$; if $X$ is an additive subgroup of $F$, then we denote by $BX$ the left sub-B-module of $B \otimes_A F$ generated by $X$. With this notation, we have:

As $B$ is a flat right $A$-module, the canonical mapping

is injective; taking account of the identifications made, its image is $BF'$, which proves (i). Assertion (ii) follows from § 2, no. 6, Proposition 7, applied with $E = B$, $E' = A$, and using the formulae $A \otimes_A F = F$ and $A \otimes_A F' = F'$. Assertion (iii) is trivial and (iv) follows from § 2, no. 6, Proposition 6.

(1) Without any hypothesis on $B$, clearly, if $F$ is a finitely generated left $A$-module, $B \otimes_A F$ is a finitely generated left $B$-module. Conversely, if $B \otimes_A F$ is a finitely generated $B$-module, it is generated by a finite number of elements of the form $1 \otimes x_i$ with $x_i \in F$; if $M$ is a sub-A-module of $F$ generated by the $x_i$ and $j$ the canonical injection $M \to F$, $1, \otimes j : B \otimes_A M \to B \otimes_A F$ is a surjective homomorphism, hence $j$ is surjective (no. 1, Proposition 2), which proves that $F$ is finitely generated.

(2) If $F$ admits a finite presentation, so does $B \otimes_A F$ without any hypothesis on B (§ 2, no. 8). It remains to prove that, if $B \otimes_A F$ admits a finite presentation, so does F. We already know from (1) that F is finitely generated, hence there exists a surjective homomorphism $u : L \to F$, where L is a finitely generated free A-module. Let R be the kernel of u, so that $B \otimes_A R$ is identified with the kernel of the surjective homomorphism $l_B \otimes u : B \otimes_A L \to B \otimes_A F$ (§ 2, no. 3, Remark 2). As $B \otimes_A F$ admits a finite presentation by hypothesis, we conclude (§ 2, no. 8, Lemma 9) that $B \otimes_A R$ is finitely generated; then it follows from (1) that R is a finitely generated A-module and consequently F admits a finite presentation.

The condition is obviously necessary without any hypothesis on A or B (Algebra, Chapter II, § 5, no. 1, Corollary to Proposition 4); we prove that it is sufficient. If a finitely generated projective module admits a finite presentation (§ 2, no. 8, Lemma 8), the hypothesis implies that F admits a finite presentation by virtue of Proposition 11, hence, for every A-module M, there is a canonical isomorphism

(§ 2, no. 10, Proposition 11). Then let $v : M \to M''$ be a surjective A-module homomorphism and consider the commutative diagram

As $1_B \otimes v$ is surjective and $B \otimes_A F$ is assumed projective, $\operatorname{Hom}(1_{B \otimes F}, 1_B \otimes v)$ is surjective (Algebra, Chapter II, § 2, no. 2, Proposition 4) and so then is $1_B \otimes \operatorname{Hom}(1_F, v)$. But as B is a faithfully flat A-module, $\operatorname{Hom}(1_F, v)$ is itself surjective (no. 1, Proposition 2), hence F is a projective A-module (Algebra, Chapter II, § 2, no. 2, Proposition 4).

The condition is sufficient. For, as $B$ is a flat $A$-module, every solution with elements in $B$ of the homogeneous linear system associated with (3) is a linear combination with coefficients in $B$ of solutions consisting of elements of $A$ (§ 2, no. 11, Corollary 2, to Proposition 13). The problem then reduces to proving that the existence of a solution of (3) with elements in $B$ implies the existence of one solution with elements in $A$. Now if we set

system (3) is equivalent to the equation $\sum_{k=1}^n y_k \otimes c_k = 1 \otimes \mathrm{din}\, B \otimes_A A_s^m = B_s^m$.

In other words, if $M$ is the sub-$A$-module of $A_s^m$ generated by the $c_k$ $(1 \leq k \leq n)$, the existence of the solution $(y_k)$ of (3) is equivalent (with the identifications made in no. 5) to the relation $d \in BM \cap A_s^m$; but as $BM \cap A_s^m = M$ (no. 5, Proposition 10, (ii)), it implies $d \in M$, that is, the existence of a solution $(x_k)$ of system (3) with elements in $A$.

The condition is necessary. For suppose that $(A, B)$ satisfies the linear extension property; we know already that $B$ is a flat right $A$-module (§ 2, no. 11, Corollary 2 to Proposition 13); we prove that, for every left ideal $a$ of $A$, $Ba \cap A = a$, which shows that $B$ is a faithfully flat right $A$-module (no. 5, Proposition 9, (d)). Now, let $x \in Ba \cap A$; there exists by hypothesis $y_i \in B$ and $a_i \in a$ such that $\sum_i y_i a_i = x$; property (E) applied to this linear equation with coefficients and right hand side in $A$ shows that there exist $x_i \in A$ such that $x = \sum_i x_i a_i$, hence $x \in a$.