Proposition 9

Let A be a ring, M a finitely generated A-module, N a projective A-module, and $u : M \to N$ an A-linear mapping.

a) Let $\mathfrak{p}$ be a prime ideal of A. The following conditions are equivalent :

(i) there exist $f \in A - \mathfrak{p}$ and $v \in \mathrm{Hom}_{A_f}(N_f, M_f)$ with $v \circ u_f = \mathrm{Id}_{M_f}$; (ii) there exists $v \in \mathrm{Hom}_{A_\mathfrak{p}}(N_\mathfrak{p}, M_\mathfrak{p})$ with $v \circ u_\mathfrak{p} = \mathrm{Id}_{M_\mathfrak{p}}$; (iii) the $\kappa(\mathfrak{p})\text{-linear } 1 \otimes u : \kappa(\mathfrak{p}) \otimes_A M \to \kappa(\mathfrak{p}) \otimes_A N$ mapping is injective; (iv) there exist an integer $m \geq 0$, elements $x_1, \ldots, x_m$ of M and linear forms $y_1, \ldots, y_m$ on N such that the images of the $x_i$ in $M_\mathfrak{p}$ generate the $A_\mathfrak{p}$-module $M_\mathfrak{p}$ and that one has $\det(<y_j, u(x_i)>)\notin \mathfrak{p}$;

If condition (iv) is satisfied, one has $m = [\kappa(\mathfrak{p}) \otimes_A M : \kappa(\mathfrak{p})]$ and the elements $1 \otimes x_i$ form a basis of the $\kappa(\mathfrak{p})$-vector space $\kappa(\mathfrak{p}) \otimes_A M$.

b) The set U of prime ideals $\mathfrak{p}$ of A satisfying the conditions of a) is an open set of Spec(A), and the following conditions are equivalent:

(i) one has $U = \mathrm{Spec}(A)$; (ii) U contains all the maximal ideals of A; (iii) there exists $v \in \mathrm{Hom}_A(N, M)$ with $v \circ u = \mathrm{Id}_M$; (iv) u is injective and Coker(u) is a projective A-module.

Let us prove a).

(i) ⇒ (ii) ⇒ (iii): these implications are clear.

(iii) ⇒ (iv): put $m = [\kappa(\mathfrak{p}) \otimes_A M : \kappa(\mathfrak{p})]$ and let $(x_1, \ldots, x_m)$ be a sequence of elements of M such that the elements $1 \otimes x_i$ form a basis of the $\kappa(\mathfrak{p})$-vector space $\kappa(\mathfrak{p}) \otimes_A M$. The images of the $x_i$ in $M_\mathfrak{p}$ generate the $A_\mathfrak{p}$-module $M_\mathfrak{p}$ (Nakayama lemma). If condition (iii) is satisfied, the elements $1 \otimes u(x_i)$ of the $\kappa(\mathfrak{p})$-vector space $\kappa(\mathfrak{p}) \otimes_A N$ are linearly independent.

Moreover, there exist an A-module N', a set I, and an isomorphism of A-modules $\theta : N \oplus N' \to A^{(I)}$, from which one deduces an isomorphism of $\kappa(p)$-vector spaces

The elements $t_i = \overline{\theta}(1 \otimes u(x_i), 0)$ of $\kappa(p)^{(I)}$ form a finite free family. There therefore exist elements $\alpha_1, \ldots, \alpha_m$ of I such that one has $\det(\mathrm{pr}_{\alpha_j}(t_i)) \neq 0$; the linear forms $y_j : z \mapsto \mathrm{pr}_{\alpha_j}(\theta(z, 0))$ on N are suitable.

Suppose condition (iv) fulfilled. Let $(a_{ij}) \in M_m(A)$ denote the matrix of coefficients $a_{ij} = <y_j, u(x_i)>$. Let g be an element of $A - p$ such that the images of the $x_i$ generate the $A_g$-module $M_g$ (II, § 5, No. 1, Prop. 2), and let $f = g \det(a_{ij})$. Since $\det(a_{ij})$ is invertible in $A_f$, the images of the elements $u(x_i)$ in $N_f$ are linearly independent; consequently the images of the $x_i$ in $M_f$ form a basis of this $A_f$-module. This proves the last assertion of a). Let us now prove (i). Let $w \in \mathrm{Hom}_A(N, M)$ denote the mapping $z \mapsto \sum_j <y_j, z> x_j$. We have

since the images of the $x_i$ form a basis of $M_f$ and the matrix $(a_{ij})$ is invertible in $M_m(A_f)$, the endomorphism $(w \circ u)_f$ of $M_f$ is bijective, and the mapping $v = (w \circ u)_f^{-1} \circ w_f \in \mathrm{Hom}_{A_f}(N_f, M_f)$ satisfies condition (i).

Let us prove b). The fact that U is open follows from condition (i) of a).

(iii) $\Rightarrow$ (i) $\Rightarrow$ (ii): this is clear.

(iv) $\Rightarrow$ (iii): under the hypotheses of (iv), the sequence $0 \to M \xrightarrow{u} N \longrightarrow \mathrm{Coker}(u) \to 0$ is exact and split, whence (iii).

(ii) $\Rightarrow$ (iv): let us introduce, as above, an isomorphism of A-modules $\theta : N \oplus N' \to A^{(I)}$. Let $u'$ denote the mapping of M into $A^{(I)}$ defined by $u'(x) = \theta(u(x), 0)$. There exists a finite subset J of I such that the image of $u'$ is contained in the submodule $A^J$ of $A^{(I)}$. Let $u'' : M \to A^J$ denote the mapping deduced from $u'$. Under hypothesis (ii), for every maximal ideal m of A, the $A_m$-linear mapping $u'_m$ of $M_m$ into $A_m^{(I)}$ admits a retraction, and therefore the same is true of $u''_m$; thus $u''_m$ is injective and its image is a direct factor in $A_m^J$, so that its cokernel is a projective $A_m$-module. The A-module $\mathrm{Coker}(u'')$ is finitely presented by construction; it is therefore projective (II, § 5, No. 2, Th. 1). The homomorphism $u''$ is injective (II, § 3, No. 3, Th. 1); consequently, $u$ is injective. The A-module $\mathrm{Coker}(u')$ is isomorphic, on the one hand to $\mathrm{Coker}(u) \oplus N'$, on the other hand to $\mathrm{Coker}(u'') \oplus A^{(I-J)}$. Since the A-modules $A^{(I-J)}$, $\mathrm{Coker}(u'')$ and $N'$ are projective, the same is true of $\mathrm{Coker}(u)$, which completes the proving of (iv).