The relation $x \in I$ is equivalent to $1 \notin xA$ and hence $xA \neq \mathbf{A}$. If $a$ is an ideal of $\mathbf{A}$ distinct from $\mathbf{A}$ and $x \in a$, then $xA \subset a$, hence $xA \neq \mathbf{A}$ and $x \in I$. Hence every ideal of $\mathbf{A}$ distinct from $\mathbf{A}$ is contained in $I$ and every element $x \in I$ belongs to a principal ideal $xA \neq \mathbf{A}$. This proves the first assertion, which immediately implies the equivalence of (a), (b) and (c).

Remark (1). Note that, if (c) holds, $I$ is the Jacobson radical of the ring $\mathbf{A}$ (Algebra, Chapter VIII, §6, no. 3, Definition 3).

Let $S = A - p$ and $j : A \to A_v$, be the canonical homomorphism; the hypothesis that $p$ is prime implies that $p$ is saturated with respect to $S$, hence $j^{-1}(pA_p) = p$ ($§ 2$, no. 4, Proposition 10) and, as the ideals of $A$ not meeting $S$ are those contained in $p$, the first two assertions are special cases of $§ 2$, no. 5, Proposition 11 (ii). Moreover, iff is the canonical homomorphism $A \to A/p$, $f(S)$ is the set of elements $\neq 0$ of the integral domain $A/p$ and hence the last assertion is a special case of $§ 2$, no. 5, Proposition 11 (i).

(i) Let $f$ be the canonical homomorphism $A \to A/b$. The homomorphism from $A$, to $(A/b)_{p/b}$ canonically associated with $f$ ($§ 2$, no. 1, Proposition 2) is surjective and its kernel is $b'$, which defines, by taking quotients, a canonical isomorphism of $A_p/b'$ onto $(A/b)_{p/b}$.

(ii) The mapping $b' \to b = (i_A^S)^{-1}(b')$, restricted to the set of prime ideals of $A_p$, is an isomorphism (with respect to inclusion) of this set onto the set of prime ideals of $A$ contained in $p$. If $b'$ is prime in $A_p$, there exists an isomorphism of the ring $A_b$ onto the ring $(A_p)_{b'}$, which maps $a/s$ to $(a/1)/(s/1)$ for all $a \in A, s \in A - b$.

This is just a particular case of $§ 2$, no. 5, Proposition 11.

Remarks (3) If $a$ is an ideal of $A$ not contained in $p$, then $aA_p = A$, and $(A/a)_p = 0$ ($§ 2$, no. 5, Remark). (4) Let $A, B$ be two rings, $\rho : A \to B$ a homomorphism, $q$ a prime ideal of $B$ and $p$ the prime ideal $\rho^{-1}(q)$ of $A$. As $\rho(A - p) \subset B - q$, a canonical homomorphism $\rho_q : A \to B_q$ is derived from $\rho$ ($§ 2$, no. 1, Proposition 2) and it is immediate that $\rho_q(pA_p) \subset qB_q$, hence $\rho$ is a local homomorphism.

The assertion with respect to hypothesis (i) is precisely Corollary 3 to Proposition 6 of Algebra, Chapter VIII, $§ 6$, no. 3. On the other hand, the relation $(A_d/m) \otimes_A M = 0$ is equivalent to $M = mM$ and hence implies $M = m^nM$ for every integer $n > 0$; whence the assertion with respect to hypothesis (ii).

$(A_d/m) \otimes_A (N/u(M))$ is canonically isomorphic to

(Algebra, Chapter II, , no. 6, Corollary 1 to Proposition 6); then the hypothesis implies $(A_d/m) \otimes_A (N/u(M)) = 0$, hence $N/u(M) = 0$ by Proposition 4

Let $(e_i)_{i \in I}$ be the canonical basis of the left $A$-module $A_s^{(I)}$: it is sufficient to apply Corollary 1 to the $A$-linear mapping $u : A_s^{(I)} \to M$ such that $u(e_i) = x_i$ for all $i \in I$.

If $a \in A$, $x \in M$ and $\bar{a}$ is the class of $a$ in $A/m$, then $\bar{a} \otimes x = 1 \otimes (ax)$ and hence the hypothesis implies that there exists a family $(x_i)_{i \in I}$ of elements of $M$ such that $(1 \otimes x_i)$ is a basis of the $(A/m)$-module $(A/m) \otimes_A M$. We already know that the $x_i$ generate $M$ (Corollary 2 to Proposition 4); we shall see that they are linearly independent over $A$. To this end, let us consider the free $A$-module $L = A_s^{(I)}$; let $(e_i)$ be its canonical basis and $u : A_s^{(I)} \to M$ the $A$-linear mapping such that $u(e_i) = x_i$ for all $i \in I$; if $R$ is the kernel of $u$, we shall prove that $R = 0$. Under hypothesis (i), $(A/m) \otimes_A M$ is a finitely generated $(A/m)$-module, hence $I$ is necessarily finite and $R$ is a finitely generated $A$-module by Chapter I, § 2, no. 8, Lemma 9. Then by Proposition 4 it will be sufficient to prove (under either hypothesis) that $R = mR$.

Let $j$ be the canonical injection $R \to L$; then there is a commutative diagram

in Which the two rows are exact, $j$ is injective and $1 \otimes u$ is surjective (Chapter

(Chapter I, § 1, no. 4, Proposition 2); it is sufficient to verify that v is bijective, for then we deduce that Coker(a) = 0, in other words that a is surjective and consequently R = mR. Now, Coker(b) = (A/m) \otimes_A L and

and by definition v(1 \otimes e_i) = 1 \otimes x_i; as (1 \otimes e_i) is a basis of (A/m) \otimes_A L, the definition of the x_i shows that v is bijective.

If this condition holds, it can be assumed that (y_\lambda) is a subfamily of a family (x_i) of elements of M such that (1 \otimes x_i) is a basis of M/mM (Algebra, Chapter II, § 7, no. 1, Theorem 2) and Proposition 5 then proves that (x_i) is a basis of M.

The implications (a) \Rightarrow (b) \Rightarrow (c) \Rightarrow (d) are immediate. As A/m is a field, (A/m) \otimes_A M is a free (A/m)-module and Proposition 5 shows that (d) implies (a).

Finally, we know that Tor_1^A(A, M) = 0 and from the exact sequence 0 \to m \to A \to A/m \to 0 we therefore derive the exact sequence

this proves that Tor$_1^A(A/m, M)$ is isomorphic to the kernel of the canonical homomorphism $m \otimes_A M \to M$; whence the equivalence of (d) and (e). \*

It can be shown that, for every ring $A$ with Jacobson radical $m$ such that $A/m$ is a field, every projective $A$-module is free (Exercise 3).

(a) $u$ is an isomorphism of $M$ onto a direct factor of $N$; (b) $1 \otimes u: (A/m) \otimes_A M \to (A/m) \otimes_A N$ is injective; (c) $u$ is injective and $\operatorname{Coker}(u)$ is a free $A$-module; (d) the transpose homomorphism ${}^t u: N^* \to M^*$ is surjective.

We know (Algebra, Chapter II, § 1, no. 11, Proposition 21) that, if $N/u(M)$ is free, $u(M)$ is a direct factor of $N$, hence (c) implies (a); conversely, (a) implies that $\operatorname{Coker}(u)$, isomorphic to a complement of $u(M)$ in $N$, is a finitely generated projective $A$-module and a fortiori finitely presented (Chapter I, § 2, no. 8, Lemma 8); hence this module is free by Corollary 2 to Proposition 5 and (a) implies (c). On the other hand, (a) obviously implies (b). For simplicity we write $M' = (A/m) \otimes_A M$, $N' = (A/m) \otimes_A N$; as $M$ and $N$ are finitely generated, the duals ${M'}^*$ and ${N'}^*$ of the $(A/m)$-modules $M'$ and $N'$ are canonically identified with $M^* \otimes_A (A/m)$ and $N^* \otimes_A (A/m)$ and ${}^t(1 \otimes u)$ with $({}^t u) \otimes 1$ (Algebra, Chapter II, § 5, no. 4, Proposition 8); as $M'$ and $N'$ are vector spaces over the field $A/m$, the hypothesis that $1 \otimes u$ is injective implies that ${}^t(1 \otimes u)$ is surjective (Algebra, Chapter 11, § 7, no. 5, Proposition 10); Corollary 1 to Proposition 4 then shows that ${}^t u$ is surjective and we have thus proved that (b) implies (d). Finally we show that (d) implies (a). Suppose that ${}^t u$ is surjective; as $M^*$ is free, there exists a homomorphism $f$ from $M^*$ to $N^*$ such that $1_{M^*} = {}^t u \circ f$ (Algebra, Chapter II, § 1, no. 11, Proposition 21); as $M$ and $N$ are free and finitely generated, there exists a homomorphism $g$ from $N$ to $M$ such that $f = {}^t g$; hence ${}^t 1_M = 1_{M^*} = {}^t u \circ {}^t g = {}^t(g \circ u)$, whence $1_M = g \circ u$; this proves that $u$ is an isomorphism of $M$ onto a submodule which is a direct factor of $N$ (Algebra, Chapter 11, § 1, no. 9, Corollary 2 to Proposition 15).

(a) $u$ is an isomorphism of $M$ onto $N$; (b) $M$ and $N$ have the same rank (Algebra, Chapter II, § 7, no. 2) and $u$ is surjective; (c) $1 \otimes u: M/mM \to N/mN$ is bijective.

Clearly (a) implies (b); (b) implies that $1 \otimes u$ is surjective; moreover the hypothesis that M and N have the same rank implies that so do the vector spaces $(A/m) \otimes_A M$ and $(A/m) \otimes_A N$ over $A/m$, hence $1 \otimes u$ is bijective (Algebra, Chapter II, , no. 4, Corollary to Proposition 9) and (b) implies (c). Finally, condition (c) implies, by Proposition 6, that N is the direct sum of $u(M)$ and a free submodule P and $u$ is an isomorphism of M onto $u(M)$; if $P \neq 0$, then $(A/m) \otimes_A P \# 0$ and $1 \otimes u$ would not be surjective; hence (c) implies (a)

The propositions proved above in this no. will usually be applied when $A$ is a local ring and $m$ its maximal ideal. Corollary 2 to Proposition 5 is then completed by

If M is free, clearly the two sides of (1) are equal to the rank of M for all $i \in I$. Suppose now that the condition is satisfied and denote by $n$ the common value of the two sides of (1); by Corollary 2 to Proposition 4 M has a system of $n$ generators $x_j$ ($1 \leq j \leq n$). Suppose first that $A$ is an integral domain, in which case $p_i = 0$ for all $i \in I$. The elements $1 \otimes x_j$ ($1 \leq j \leq n$) generate the vector space $K \otimes M$ over the field of fractions K of $A$; but as by hypothesis this space is of dimension $n$ over K, the elements $1 \otimes x_j$ are linearly independent over K. It follows (Algebra, Chapter II, § 1, no. 13, Remark 1) that the $x_j$ are linearly independent over $A$ and hence form a basis of M.

Passing to the general case, there exists a surjective homomorphism $v$ from $L = A''$ onto M. Consider the commutative diagram

where $u$ (resp. $u'$) is the mapping $x \mapsto (\phi_i(x))$ (resp. $y \mapsto (\psi_i(y))$),

(resp. $\psi_i : M \to (A/p_i) \otimes M$) being the canonical mapping, and $v'$ is the product of the $1_{A/p_i} @ v_e$. Then $(A/p)/(m/p_i) \otimes_{A/p_i} ((A/p_i) \otimes_A M) = (A/m) \otimes_A M$ and, as $A/p_i$ is a local integral domain, it follows from the first part of the argument that each of the $1_{A/p_i} \otimes v$ is an isomorphism; then so is $v'$. On the other hand, as $A$ is reduced, $\bigcap_i p_i = (0)$ ($§ 2$, no. 6, Proposition 13), whence $\bigcap_i (p_i L) = 0$ since $L$ is free (Algebra, Chapter II, $§ 3$, no. 7, Remark); as $p_i L$ is the kernel of $\phi_i$, this shows that $u$ is injective. It follows that $v' \circ u = u' \circ v$ is injective, hence $v$ is injective and, as $v$ is surjective by definition, this shows that $M$ is free.

$A/a$ is then a local ring with maximal ideal $m/a$; $M$ can be considered as an $(A/a)$-module; for all $s \in A - m$ the canonical image of $s$ in $A/a$ is invertible, hence the homothety $x \mapsto sx$ of $M$ is bijective from the definition of $M$, as the solution of a universal problem ($§ 2$, no. 2); whence the proposition.

In particular, if there exists $k \geqslant 0$ such that $m^k M = 0$, the homomorphism $M \to M_m$ is bijective ($§ 1$, no. 1, Corollary to Proposition 1).

Since the case $k = 0$ is trivial, suppose that $k \geqslant 1$. It follows from Proposition 8 that the canonical homomorphism $M/m^k M \to (M/m^k M)_m$ is bijective. On the other hand $(M/m^k M)_m$ is canonically identified with $M_m/(m^k M)_m$ (§ 2, no. 4, Theorem 1) and hence $(m^k M)_m = m^k M_m$ (§ 2, no. 7, Corollary to Proposition 18), whence there is an isomorphism of $M/m^k M$ onto $M_m/m^k M_m$ which maps the class of an element $x \in M$ to the class of $x/1$.

This follows easily from Proposition 9 and $§ 1$, no. 2, Proposition 6, the $m_{i_1}^{k_1}$ being relatively prime in pairs ($§ 1$, no. 2, Proposition 3).

In the rest of this no. $A$ will denote a ring and $\Omega(A)$ (or $\Omega$) the set of maximal ideals of $A$.

Each of the $\mathbf{A}_m$ is a flat A-module (§ 2, no. 4, Theorem 1), hence $E = \bigoplus_{m \in \Omega} \mathbf{A}_m$ is flat (Chapter I, § 2, no. 3, Proposition 2). Moreover, for every maximal ideal $m$ of $A$, $mA_m$ is the unique maximal ideal of $\mathbf{A}_m$, hence $mA_m \neq A_m$, whence it follows that $mE \neq E$ and consequently $E$ is faithfully flat (Chapter I, § 3, no. 1, Proposition 1 (d)).

To say that, for all $m \in \Omega$, $u_m$ is injective (resp. surjective, bijective, zero) is equivalent to saying that the homomorphism $\bigoplus_m u_m : \bigoplus M, \to \bigoplus N$, has the same property. But $\bigoplus_m M_m = M \otimes_A E, \bigoplus_m N_m = N \otimes_A E$ and $\bigoplus_m u_m = u \otimes 1$, where $E = \bigoplus_m \mathbf{A}_m$; as $E$ is faithfully flat (Proposition 10), the theorem follows from Chapter I, § 3, no. 1, Proposition 1 (c) and Proposition 2.

Let $\bar{x}$ be the class of $x$ in $M/N$; to say that $x \in N$ means that the A-linear mapping $u : a \mapsto a\bar{x}$ from $A$ to $M/N$ is zero. Now, $(M/N)_m$ is identified with $M_m/N_m$ (§ 2, no. 4, Theorem 1) and $u_m : A_m \to M_m/N_m$ with the mapping $A \mapsto \lambda \bar{x}_m$, where $\bar{x}_m$ is the class mod. $N$, of the canonical image of $x$ in $M$. As the relation $u = 0$ is equivalent to $u_m = 0$ for all $m$ by Theorem 1, this proves the corollary.

Applying Corollary 1 in the case $N = 0$, it is seen that the relation $x = 0$ is equivalent to $f_m(x) = 0$ for all $m \in \Omega$.

(ii) In particular, let b and c be two elements of A. For c to be a multiple of b, it is necessary and sufficient that, for all $m \in \Omega$, the canonical image of c in A, be a multiple of that of b.

As $bA_m = b_m$ (§ 2, no. 7, Corollary to Proposition 18), (i) is a special case of Corollary 1; (ii) follows from (i) applied to the ideal $Ab$.

As M is identified with a submodule of $K \otimes_A M$, M, is identified with a sub-\&-module of $(K \otimes_A M)_m = K_m \otimes_A M$ (§ 2, no. 4, Theorem 1); as $K_m = K$, we see straightway that $M_m$ is torsion-free; moreover, the commutativity of the diagram

proves that the canonical mapping $M \to M_m$ is injective. The corollary then follows from Corollary 1 applied to the A-module $K \otimes_A M$ and its submodule M.

In particular, for every integral domain A,

(2)

Let $(e_i)_{1 \leq i \leq n}$ be the canonical basis of $A''$, $(x_i)_{1 \leq i \leq n}$ a system of generators of $A^n$ with n elements and $u : A^n \to A''$ the A-linear mapping such that $u(e_i) = x_i$ for $1 \leq i \leq n$. By hypothesis $u$ is surjective and it is necessary to show that $u$ is injective. By virtue of Theorem 1 this can immediately be reduced to the case where A is a local ring; if $m$ is the maximal ideal of A, the elements $1 \otimes x_i$ ($1 \leq i \leq n$) in $(A/m)^n$ then form a system of generators of the free $(A/m)$-module $(A/m)^n$; as $A/m$ is a field, this system is a basis of $(A/m)^n$; as $A^n$ is a free A-module, we deduce from Proposition 5 that $(x_i)$ is a basis of $A^n$.

It follows from Theorem 1 that, for $u$ to be surjective, it is necessary and sufficient that $u_m : M_m \to N_m$ be surjective for all $m \in \Omega$. As $A$ is a local ring and $N_m$ is a finitely generated $A_m$-module, this amounts to saying that the homomorphism $u'_m : M_m/mM_m \to N_m/mN_m$, obtained by taking quotients, is surjective (no. 2, Corollary 1 to Proposition 4); but $M_m/mM_m$ (resp. $N_m/mN_m$) is identified with $M/mM$ (resp. $N/mN$) (Proposition9), whence the proposition.