Lemma 10

Let $(f_\lambda)_{\lambda \in L}$ be a family of generators of $F$ and $(e_\lambda)_{\lambda \in L}$ a family of elements of $E$ of finite support. For $\sum_{\lambda \in L} e_\lambda \otimes f_\lambda = 0$, it is necessary and sufficient that there exist a finite set $J$, a family $(x_j)_{j \in J}$ of elements of $E$ and a family $(a_{j,\lambda})$ ($j \in J, \lambda \in L$) of elements of $A$ with the following properties:

(1) the family $(a_{j,\lambda})$ has finite support;

(2) $\sum_{\lambda \in L} a_{j,\lambda} f_\lambda = 0$ for all $j \in J;$

(3) $e_\lambda = \sum_{j \in J} x_j a_{j,\lambda}$ for all $\lambda \in L.$

Loosely speaking, the system of $e_\lambda$ must be a linear combination with coefficients in $E$ of systems of elements of $A$ which are "relations between the $f_\lambda$".

Consider the free $A$-module $A_s^{(L)}$, its canonical basis $(u_\lambda)$ and the homomorphism $g : A_s^{(L)} \to F$ such that $g(u_\lambda) = f_\lambda$ for all $\lambda \in L$; denoting by $R$ the kernel of $g$, we have (since the $f_\lambda$ generate $F$) an exact sequence

where $i$ denotes the canonical injection. By Lemma 1 of no. 1 we derive the exact sequence

(*) The results of this no. will not be used in the rest of this chapter, except in § 3, no. 7.

Now, $E \otimes_A A_s^{(L)}$ is canonically identified with $E^{(L)}$, a family $e = (e_\lambda) \in E^{(L)}$ being identified with $\sum_{\lambda \in L} e_\lambda \otimes u_\lambda$ (Algebra, Chapter II, , no. 7, Corollary 1 to Proposition 7). For such a family to belong to the kernel of $1_E @ g$, it is necessary and sufficient that $\sum_{\lambda \in L} e_\lambda \otimes f_\lambda = 0$ in $E \otimes_A F$; taking account of the exact sequence (13), this is equivalent to saying that $e$ belongs to the image of $1, \otimes i$, in other words there is a relation of the form

where $x_j \in E, r_j \in R$ and $J$ is finite. Writing $i(r_j) = \sum_{\lambda \in L} a_{j\lambda} u_\lambda$, the hypothesis $r_j \in R$ implies the relation $\sum_{\lambda \in L} a_{j\lambda} f_\lambda = 0$ for all $j \in J$; on the other hand the relation (14) implies $e_\lambda = \sum_{j \in J} x_j a_{j\lambda}$ for all $\lambda \in L$ (Algebra, Chapter II, § 3, no. 7, Corollary 1 to Proposition 7), which completes the proof.