Proposition 27

For every exact sequence

of $A$-linear mappings, the corresponding sequence of $K$-linear mappings

is exact.

Suppose that $g_{(K)}(\lambda^{-1} \otimes x) = 0$, with $\lambda \in A, \lambda \neq 0, x \in E$; this is equivalent to $\lambda^{-1} \otimes g(x) = 0$ in $E''_{(K)}$ and hence also to

by Proposition 26, there exists $\alpha \neq 0$ in $A$ such that $\alpha g(x) = 0$ in $E''$, or also $g(\alpha x) = 0$. By hypothesis, there is therefore an $x' \in E'$ such that $\alpha x = f(x')$ and therefore $\lambda^{-1} \otimes x = f_{(K)}(\alpha^{-1}\lambda^{-1} \otimes x')$, which proves the proposition.