This a particular case of § 4, no. 3, Corollary to Proposition 13.

Let $R$ and $Q$ be the kernel and cokernel of $u$; if $g \in A$, the kernel and cokernel of $u_g$ (resp. $u_p$) are $R_g$ and $Q_g$, (resp. $R_p$ and $Q_p$) (§ 2, no. 4, Theorem 1). Then $Q_p = 0$; as $N$ is finitely generated, so is $Q$ and there exists $g' \in A - p$ such that $g'Q = 0$ (§ 2, no. 2, Corollary 2 to Proposition 4), whence $Q_{g'} = 0$. Under the hypotheses of (ii), the sequence $0 \to R_{g'} \to M \to N \to 0$ is exact, hence $R_{g'}$ is finitely generated (Chapter I, § 2, no. 8, Lemma 9). Now,

hence there exists $g_1 \in A, - pA_{g'}$ such that $g_1 R_{g'} = 0$ (§ 2, no. 2, Corollary 2 to Proposition 4). Then $g_1 = g''/{g'}^h$, where $g'' \in A - p$; as $g'/1$ is invertible in $R_{g'}$, $(g''/1)R_{g'} = 0$, whence $R_{g'g''} = (R_{g'})_{g''/1} = 0$. If $f = g'g''$, $f \in A - p$, $Q_f = 0$ and $R_f = 0$, so that $u_f$ is bijective.

There exist by hypothesis $p$ elements $x_i \in \mathbf{N}$ ($1 \leq i \leq p$) such that the $x_i/1$ form a basis of the free $A_,-$-module $N_p$. Consider the homomorphism $u : A^p \to \mathbf{N}$ such that $u(e_i) = x_i$ for $1 \leq i \leq p$, $(e_i)_{1 \leq i \leq p}$ being the canonical basis of $A^p$. As $u_p$ is bijective by hypothesis, there exists $f \in A - p$ such that $u_f$ is bijective, by virtue of Proposition 2.

By § 2, no. 4, Theorem 1, each of the $A_i$ is a flat $A$-module, hence so is $B$ (Chapter I, § 2, no. 3, Proposition 2). On the other hand, if $p$ is a prime ideal of $A$, there exists an index $i$ such that $f_i \notin p$ and $p_{f_i} = pA_{f_i}$ is therefore a prime ideal of $A_{f_i}$. Then $pB \subset pA_{f_i} \times \prod_{j \neq i} A_j \neq B$ since $pA_{f_i} \neq A_{f_i}$; this suffices to imply that $B$ is a faithfully flat $A$-module (Chapter I, § 3, no. 1, Proposition 1).

The condition is obviously necessary (§ 2, no. 4). Conversely, if all the $M_{f_i}$ are finitely generated (resp. finitely presented), $M' = \prod_{i \in I} M_{f_i}$ is a finitely generated (resp. finitely presented) $B$-module, for we can obviously suppose that for each $i$ there is an exact sequence $AT_i \to A^n_{f_i} \to M_{f_i} \to 0$, where $m$ and $n$ are independent of $i$. Now, $M' = M \otimes_A B$. The corollary then follows from Proposition 3 and Chapter I, § 3, no. 6, Proposition 11.

Note that the condition on the $f_i$ means that the open sets $X_{f_i}$ form a covering of $\mathrm{Spec}(A)$ (§ 4, no. 3, Corollary 3 to Proposition 11).

(a) $P$ is a finitely generated projective module. (b) $P$ is a finitely presented module and, for every maximal ideal $m$ of $A$, $P_m$ is a free $A_m$*-module*. (c) $P$ is a finitely generated module, for all $p \in \mathrm{Spec}(A)$, the $A_p$*-module* $P_p$ is free and, if we denote its rank by $r_p$, the function $p \mapsto r_p$ is locally constant in the topological space $\mathrm{Spec}(A)$ (that is, every point of $\mathrm{Spec}(A)$ admits a neighbourhood in which this function is constant).

(d) There exists a finite family $(f_i)_{i \in I}$ of elements of $A$, generating the ideal $A$, such that, for all $i \in I$, the $A,$-module $P_{f_i}$ is free with finite rank.

(e) For every maximal ideal $m$ of $A$, there exists $f \in A - m$ such that $P_f$ is a free $A,$-module of finite rank.

We show the theorem by proving the following scheme of implications

(a) ⇔ (d)

(b) ⇌ (e)

(c)

(a) ⇒ (b): We know that a finitely generated projective module is finitely presented (Chapter I, § 2, no. 8, Lemma 8 (iii)); if $P$ is a projective $A$-module, $P_m = P \otimes_A A_m$ is a projective $A_m$-module (\emph{Algebra}, Chapter II, § 5, no. 1, Corollary to Proposition 4); finally, as $A_m$ is a local ring, every finitely presented projective $A_m$-module is free (§ 3, no. 2, Corollary to Proposition 5).

(b) ⇒ (e): This follows from the Corollary to Proposition 2 of no. 1.

(c) ⇒ (e): Let $m$ be a maximal ideal of $A$; write $r_m = n$ and let $(x_i)_{1 \leq i \leq n}$ be a basis of $P$. We can assume that the $x_i$ are canonical images of elements $p_i \in P$ $(1 \leq i \leq n)$ to within multiplication by an invertible element of $A,$. Let $(e_i)_{1 \leq i \leq n}$ be the canonical basis of $A^n$ and $u : A^n \to P$ the homomorphism such that $u(e_i) = p_i$ for $1 \leq i \leq n$. As $P$ is finitely generated, it follows from Proposition 2 of no. 1 that there exists $f \in A - m$ such that $u_f$ is surjective. We conclude that $u_{f,g}$ is also surjective for all $g \in A - m$ and by hypothesis there exists $g \in A - m$ such that $r_p = n$ for $p \in X,$. Then, replacing $f$ by $f g$, we may assume that $r_p = n$ for all $p \in X,$. Then $u_p : A_p^n \to P_p$ is a surjective homomorphism and $P_p$ and $A_p$ are both free $A,$-modules of the same rank; hence (§ 3, no. 2, Corollary to Proposition 6) $u_p$ is bijective for all $p \in X_f$. Let $p'$ be a prime ideal of $A,$ and let $p$ be its inverse image in $A$ under the canonical mapping; if $(A_f^{n})_{p'}$ and $(P_f)_{p'}$ are identified with $A_p^n$ and $P_p$ under the canonical isomorphisms, $(u_f)_{p'}$ is identified with $u_p$ and is consequently bijective. We conclude that $u_f$ is bijective (§ 3, no. 3, Theorem 1), which establishes (e).

(e) ⇒ (d): Let $E$ be the set off $\in A$ such that $P_f$ is a finitely generated free $A_f$-module. The hypothesis implies that $E$ is contained in no maximal ideal of $A,$ hence $E$ generates the ideal $A$ and there therefore exist a finite family $(f_i)_{1 \leq i \leq n}$ of elements of $E$ and $a_i \in A$ $(1 \leq i \leq n)$ such that $1 = \sum_{i=1}^n a_i f_i$; whence (d).

(d) ⇒ (c): It follows from no. 1, Corollary to Proposition 3 that $P$ is finitely generated. On the other hand, for every prime ideal $p$ of $A$, there exists an index $i$ such that $p \in X_{f_i}$; if $p' = p_{f_i}$, then $P_p = (P_{f_i})_{p'}$ (§ 2, no. 5, Proposition

10) and hence by hypothesis $P_p$ is free and of the same rank as $P_{f_i}$, which proves (c).

(d) ⇒ (a): Consider the ring $B = \prod_{i \in I} A_i$ and the B-module

For every index $i$, there exists a free $A_i$-module $L$, such that $P_i$ is a direct factor of $L$, and it may be assumed that all the $L$, have the same rank; then $L = \prod_{i \in I} L_i$ is a free B-module of which M is a direct factor, in other words M is a finitely generated projective B-module. As B is a faithfully flat A-module (no. 1, Proposition 3), we conclude that P is a finitely generated projective A-module (Chapter I, §3, no. 6, Proposition 12).

Let p be a prime ideal of A; set $r = r_p$ and let $(y_j)_{1 \leq j \leq r}$ be a basis of the free $A_p$-module $P_p$. There exist elements $x_j$ ($1 \leq j \leq r$) of P and $s \in A - p$ such that $y_j = x_j / s$ for all $j$. Then for every family $(a_j)_{1 \leq j \leq r}$ of elements of A such that $\sum_{j=1}^r a_j x_j = 0$, $\sum_{j=1}^r (a_j/1) y_j = 0$ in $P_p$, whence $a_j/1 = 0$ for $1 \leq j \leq r$. As $A - p$ does not contain 0, this shows that the $a_j$ are all divisors of zero in A (§ 2, no. 1, Remark 3), hence of necessity $r \leq m$.

If P is a finitely presented flat A-module and m a maximal ideal of A, the $A_m$-module P, is flat (§ 3, no. 4, Proposition 13) and finitely presented (§ 2, no. 4) and hence free (§ 3, no. 2, Corollary 2 to Proposition 5), Condition (b) of Theorem 1 therefore holds.

Remarks (1) There exist finitely generated flat modules which are not projective (Exercise 7). (2) Corollary 2 to Theorem 1 extends to modules over non-commutative rings (Chapter I, §2, Exercise 15).

(a) $P$ is projective of rank $n$.

(b) $P$ is finitely generated and, for every maximal ideal $m$ of $A$, the $A$-*module* $P$, is free of rank $n$.

(c) $P$ is finitely generated and, for every prime ideal $p$ of $A$, the $A$-*module* $P_p$ is free of rank $n$.

(d) For every maximal ideal $m$ of $A$, there exists $f \in A - m$ such that the $A$-*module* $P_f$ is free of rank $n$.

By Definition 2 and Theorem 1, (a) and (c) are equivalent; (b) implies (c), as, for every prime ideal $p$ of $A$, there exists a maximal ideal $m$ containing $p$ and, writing $p' = p_m$, $P_p$ is isomorphic to $(P_m)_{p'}$ (§ 2, no. 5, Proposition 11); if $P_m$ is free of rank $n$, so then is $P_{p'}$. Property (c) implies (d) by virtue of Theorem 1 and the fact that, iff $\in A - m$ and $m' = m_f$, $P_m$ is isomorphic to $(P_f)_{m'}$ and therefore the ranks of $P_f$ and $P$ are equal. Finally, this last argument and Theorem 1 show that (d) implies (b).

We know that $P_{(B)}$ is projective and finitely generated (Algebra, Chapter II, § 5, no. 1, Corollary to Proposition 4). If q is a prime ideal of B and p its inverse image in A, then

and, as, by hypothesis, $P \otimes_A A_i$ is a free A,-module of rank $n$, $(P_{(B)})_q$ is a free $B_q$-module of rank $n$.

Suppose first that A is isomorphic to a product of fields $K_i$ ($1 \leq i \leq n$). The $K_i$ are then identified with the minimal ideals (Algebra, Chapter VIII, § 3, no. 1) of A and, for all $i$, the sum $p_i$ of the K, of index $j \neq i$ is a maximal ideal of A, the $p_i$ ($1 \leq i \leq n$) being the only prime ideals of A. Every finitely generated A-module P is therefore the direct sum of its isotypical components $P_i$ ($1 \leq i \leq n$), $P_i$ being isomorphic to a direct sum of a finite number $r_i$ of A-modules isomorphic to K, (Algebra, Chapter VIII, § 5, no. 1, Proposition 1 and no. 3, Proposition 11); the ring $A_{r_i}$ is identified with K, and annihilates the $P_j$ of index $j \neq i$, hence $r_i = \mathrm{rg}_{p_i}(P)$; if all the $r_i$ are equal to the same number $r$, P is isomorphic to $A^r$, whence the proposition in this case. In the general case, let $\mathfrak{R}$ be the Jacobson radical of A and $B = A / \mathfrak{R}$; as B is a product of fields, the projective B-module $P_{(B)}$ is free by the remarks preceding Proposition 4. Also P is a flat A-module and the proposition then follows from § 3, no. 2, Proposition 5.

(i) If there exists an $A$-module $N$ such that $M \otimes_A N$ is isomorphic to $A$, the module $M$ is projective of rank 1.

(ii) Conversely, if $M$ is projective of rank 1 and $M^*$ is the dual of $M$, the canonical homomorphism $u : M \otimes_A M^* \to A$ corresponding to the canonical bilinear form $(x, x^*) \to \langle x, x^* \rangle$ on $M \times M^*$ (Algebra, Chapter II, § 2, no. 3) is bijective.

(i) It is required to prove that, for every maximal ideal $m$ of $A$, the $A_m$-module $M$, is free of rank 1 (Theorem 2 (b)); we are free to replace $A$ by $A_m$ and hence may assume that $A$ is a local ring ($§ 2$, no. 7, Proposition 18). Let $k = A/m$. The isomorphism $v : M \otimes_A N \to A$ defines an isomorphism

as the rank over $k$ of $(M/mM) \otimes_k (N/mN)$ is the product of the ranks of $M/mM$ and $N/mN$, these latter are necessarily equal to 1, in other words $M/mM$ is monogenous. It follows that $M$ is monogenous ($§ 3$, no. 2, Corollary 2 to Proposition 4); on the other hand, the annihilator of $M$ also annihilates $M \otimes_A N$ and hence is zero, which proves that $M$ is isomorphic to $A$.

(ii) It is sufficient to prove that, for every maximal ideal $m$ of $A$, $u_m$ is an isomorphism ($§ 3$, no. 3, Theorem 1). As $M$ is finitely presented (Chapter I, $§ 2$, no. 8, Lemma 8), $(M^*)_m$ is canonically identified with the dual $(M_m)^*$ ($§ 2$, no. 7, Proposition 19) and, as $M_m$ is free of rank 1 like its dual $(M_m)^*$, clearly the canonical homomorphism $u_m : (M_m) \otimes_{A_m} (M_m)^* \to A_m$ is bijective, which completes the proof.

Remark (1). If $M$ is projective of rank 1 and $N$ is such that $M \otimes_A N$ is isomorphic to $A$, then $N$ is isomorphic to $M^*$: there are isomorphisms

This follows immediately from formulae (2), (3) and (4).

Let us now note that every finitely generated $A$-module is isomorphic to a quotient module of $L = \mathbf{A}^{(\mathbf{N})}$; we may therefore speak of the set $F(A)$ of classes of finitely generated $A$-modules with respect to isomorphism (Set Theory, Chapter I, §6, no. 9); we denote by $P(A)$ the subset of $F(A)$ consisting of the classes of projective $A$-modules of rank 1 and by $\operatorname{cl}(M)$ the image in $P(A)$ of a projective $A$-module $M$ of rank 1. It is immediate that, for two projective $A$-modules $M$,

N of rank 1, cl(M \otimes_A N) depends only on cl(M) and cl(N); as definition we set

(6)

and an internal law of composition is thus defined on $P(A)$.

(7)

The associativity and commutativity of the tensor product show that the law of composition (6) is associative and commutative; the isomorphism between $A \otimes_A M$ and M prove that $\mathrm{cl}(A)$ is the identity element under this law and, by virtue of Theorem 3, $\mathrm{cl}(M) + \mathrm{cl}(M^*) = \mathrm{cl}(A)$, whence the proposition.

Let B be a commutative A-algebra and M a projective A-module of rank 1; then $M_{(B)} = B \otimes_A M$ is a projective B-module of rank 1 (no. 3, Proposition 4). Then there exists a mapping called canonical $\phi : P(A) \to P(B)$ such that

(8)

The equation $M_{(B)} \otimes_B N_{(B)} = (M \otimes_A N)_{(B)}$ for two A-modules M, N proves that the mapping $\phi$ is a commutative group homomorphism.

Remark (2). Condition (e) of Theorem 1 (equivalent to the fact that P is projective and finitely generated) may also be expressed by saying that the sheaf of modules $\tilde{P}$ over $X = \mathrm{Spec}(A)$ associated() with P is locally free and of finite type* and may therefore be interpreted as the sheaf of sections of a vector bundle over X. Conversely, every vector bundle over X arises from a finitely generated projective module, which is determined to within a unique isomorphism; the projective modules of rank $n$ thus correspond to the vector bundles all of whose fibres have dimension $n$. In particular, the vector bundles of rank 1 correspond to the projective modules of rank 1. If we denote by $\mathscr{O}_X$ the structure sheaf $\tilde{A}$ and by $\mathscr{O}_X^*$ the sheaf of units of $\mathscr{O}_X$ (whose sections over an open set U of X are the invertible elements of the ring of sections of $\mathscr{O}_X$ over U), it follows that the group $P(A)$ is isomorphic to the first cohomology group $H^1(X, \mathscr{O}_X^*)$.*

(* See A. Grothendieck, Éléments de géométrie algébrique, I (§ 1) (Publ. Math. I.H.E.S., no. 4, 1960).

(a) implies (b), for if $B.M = B$, there exists $a \in A, s \in S$ and $x \in M$ such that $(a/s)x = 1$, hence $ax = s$ belongs to $S \cap M$. To see that (b) implies (c), note that $u$ is already injective ($§ 2$, no. 4, Theorem 1); moreover, if $x \in M \cap S$, the image under $u$ of $x/x \in S^{-1}M$ in $B$ is equal to 1 and $u$ is therefore surjective. Finally, (c) clearly implies (a).

The assertion is trivial for $M + N$; on the other hand if $s \in S \cap M$ and $t \in S \cap N$, then $st \in S \cap (M.N)$ and $st \in S \cap (M \cap N)$.

Given two sub-A-modules $M$ and $N$ of $B$, let us denote by $N : M$ the sub-A-module of $B$ consisting of those $b \in B$ such that $bM \subset N$ (Chapter I, $§ 2$, no. 10, Remark). If every $b \in N : M$ is mapped to the homomorphism $h_b : x \mapsto bx$ of $M$ to $N$, a canonical homomorphism $b \mapsto h_b$ is obtained from $N : M$ to $\operatorname{Hom}_A(M, N)$.

Let $s \in S \cap M$. If $b \in N : M$ is such that $bx = 0$ for all $x \in M$, then $bs = 0$, whence $b = 0$ since $s$ is not a divisor of 0 in $B$. On the other hand, let $f \in \mathrm{Hom}_A(M, N)$ and set $b = f(s)/s$; for all $x \in M$, there exists $t \in S$ such that $tx \in A$. Then

whence $b \in N : M$ and $f = h_b$, which proves the proposition.

If $M.N = A$, then $B.M = B.(B.M) \supset B.(M.N) = B.A = B$, hence $M$ is non-degenerate. Similarly $N$ is non-degenerate. If $t \in S \cap M$ and $u \in S \cap N$ (no. 5, Proposition 8 ), the element $s = tu$ belongs to $S \cap M \cap N$, whence $Ms \subset M.N = A$ and therefore $As \subset M \subset As^{-1}$. On the other hand, obviously $N \subset A : M$, whence

and $M.(A : M) = A$; multiplying the two sides by $N$, we deduce $A : M = N$, which completes the proof.

Let us show first the equivalence of properties (a), (b) and (c). If (a) holds and N is sub-A-module of B such that M . N = A, then there is a relation

For all $x \in M$, set $v_i(x) = n_i x$; the $v_i$ are linear forms on M and by (9) $x = \sum_{i=1}^{n} m_i v_i(x)$ for all $x \in M$; this proves (Algebra, Chapter 11, §2, no. 6, Proposition 12) that M is projective and generated by the $m_i$; hence M is a finitely generated projective module.

Let m be a maximal ideal of A; we show that the integer $r = \mathrm{rg}_m(M)$ is equal to 1. Let S' be the image of S in $\mathbf{A}_{m'}$; as the elements of S are not divisors of 0 in A, those of S' are not divisors of 0 in $\mathbf{A}_{m'}$ since $\mathbf{A}_{m'}$ is a flat A-module (§ 2, no. 4, Theorem 1 and Chapter I, §2, no. 4, Proposition 3); then ${S'}^{-1}A_m \neq 0$ and, as $M_m$ is a free $A_m$-module of rank $r$, ${S'}^{-1}M_m$ is a free ${S'}^{-1}A_m$-module of rank $r$. But if T' is the image of A $-m$ in $S^{-1}A$, ${S'}^{-1}A$, (resp. ${S'}^{-1}M_m$) is canonically identified with ${T'}^{-1}(S^{-1}A)$ (resp. ${T'}^{-1}(S^{-1}M)$) (§ 2, no. 3, Proposition 7). Now $S^{-1}M = B$ (Proposition 8 (c)) and hence ${T'}^{-1}(S^{-1}M)$ is a free A-module of rank 1 over ${T'}^{-1}(S^{-1}A)$, which proves that $r = 1$ and shows the implication (a) $\Rightarrow$ (c).

The implication (c) $\Rightarrow$ (b) is trivial. Let us show that (b) $\Rightarrow$ (a). There exists by hypothesis a family (not necessarily finite) $(f_\lambda)_{\lambda \in L}$ of linear forms on M and a family $(m_\lambda)_{\lambda \in L}$ of elements of M such that, for all $x \in M$, the family $(f_\lambda(x))$ has finite support and $x = \sum_{\lambda \in L} m_\lambda f_\lambda(x)$ (Algebra, Chapter II, §2, no. 6, Proposition 12). Since M is non-degenerate, $f_\lambda(x) = n_\lambda x$ for some $n_\lambda \in A$: M by virtue of Proposition 9 of no. 5. Taking x as an element of M $\cap S$ (no. 5, Proposition 8), it is seen that of necessity $n_\lambda = 0$ except for a finite number of indices and $\sum_{\lambda \in L} m_\lambda n_\lambda = 1$. This obviously implies M . (A: M) = A, whence (a).

By virtue of Definition 2 of no. 3, (c) implies (d). Let us show the converse. As M is non-degenerate, its annihilator is zero (Proposition 8 (b)), then so is the annihilator of $M_m$ (§ 2, no. 4, formula (9)). As M, is assumed to be a monogenous &-module, it is therefore free of rank 1 and it then follows from no. 3, Theorem 2 that M is projective of rank 1.