Proposition 25

Let $A, B$ be two rings, $\rho : B \to A$ a ring homomorphism and $E$ an $A$-module. Let $(\alpha_\lambda)_{\lambda \in L}$ be a generating system (resp. free family of elements, basis) of $A$ considered as a left $B$-module. Let $(a_\mu)_{\mu \in M}$ be a generating system (resp. free family of elements, basis) of the $A$-module $E$. Then $(\alpha_\lambda a_\mu)_{(\lambda, \mu) \in L \times M}$ is a generating system (resp. (when $\rho$ is injective) free family of elements, basis) of the $B$-module $\rho_*(E)$.

If $x = \sum_{\mu \in M} \gamma_\mu a_\mu$, where $\gamma_\mu \in A$ and $(\alpha_\lambda)$ is a generating system of $A$, we may write $\gamma_\mu = \sum_{\lambda \in L} \rho(\beta_{\lambda \mu}) \alpha_\lambda$, with $\beta_{\lambda \mu} \in B$, for all $\mu \in M$, whence $x = \sum_{\lambda, \mu} \rho(\beta_{\lambda \mu}) \alpha_\lambda a_\mu$. On the other hand, if $(\alpha_\lambda)$ and $(a_\mu)$ are free families, a relation $\sum_{\lambda, \mu} \rho(\beta_{\lambda \mu}) \alpha_\lambda a_\mu = 0$, with $\beta_{\lambda \mu} \in B$, may be written $\sum_{\mu \in M} \left( \sum_{\lambda \in L} \rho(\beta_{\lambda \mu}) \alpha_\lambda \right) a_\mu = 0$; it thus implies $\sum_{\lambda \in L} \rho(\beta_{\lambda \mu}) \alpha_\lambda = 0$ for all $\mu \in M$ and therefore $\beta_{\lambda \mu} = 0$ for all $\lambda, \mu$ if $\rho$ is injective.