Remark

(1) Let $K$ be a field and $A$ a subring of $K$ with the following property: for every finite family $(\xi_i)_{1 \leq i \leq n}$ of elements of $K$, there exists a $\gamma \in A$ which is non-zero and such that $\gamma \xi_i \in A$ for $1 \leq i \leq n$ (a hypothesis which is always satisfied when $A$ is commutative and $K$ is the field of fractions of $A$). Let $E$ be a vector space over $K$ and $E_{[A]}$ the $A$-module obtained by restricting the ring of scalars to $A$. Then, if a family $(x_\lambda)_{\lambda \in L}$ is free in $E_{[A]}$ it is also free in $E$. Attention may be confined to the case where $L = \{1, n\}$; if there were a relation $\sum_{i=1}^n \xi_i x_i = 0$ with $\xi_i \in K$, the $\xi_i$ not all zero, it would follow that for all $\beta \in A$, $\sum_{i=1}^n (\beta \xi_i) x_i = 0$. By hypothesis we can suppose $\beta \neq 0$ in $A$ such that $\beta \xi_i = \alpha_i$ belongs to $A$ for all $i$; but the relation $\sum_{i=1}^n \alpha_i x_i = 0$ is contrary to the hypothesis, the $\alpha_i$ being not all zero.

(2) If the ring homomorphism $\rho : B \to A$ is surjective and $b$ is its kernel (so that $A$ is canonically identified with $B/b$), then, for every $A$-module $E$, $b$ is contained in the annihilator of $\rho_*(E)$ and $E$ is the $A$-module derived from $\rho_*(E)$ by the process defined in no. 12.

Let $A, B$ be two rings and $\rho : B \to A$ a homomorphism. Let $E$ be an $A$-module and $F$ a $B$-module; a B-linear mapping $u : F \to \rho_*(E)$ (also called a B-linear mapping of $F$ into $E$ if no confusion arises) is also called a semi-linear mapping (relative to $\rho$) of the $B$-module $F$ into the $A$-module $E$; it is also said that the ordered pair $(u, \rho)$ is a dimorphism of $F$ into $E$; this therefore means that, for $x \in F, y \in F$ and $\beta \in B$,

The set $\mathrm{Hom}_B(F, \rho_*(E))$ of $B$-linear mappings of $F$ into $E$ is also written as $\mathrm{Hom}_B(F, E)$ if no confusion can arise.

When $\rho$ is an isomorphism of $B$ onto $A$, the relation $u(\beta x) = \rho(\beta) u(x)$ for all $\beta \in B$ may also be written as $u(\rho'(\alpha)x) = \alpha x$ for all $\alpha \in A$, where $\rho'$ denotes the inverse isomorphism of $\rho$; to say that $u$ is semi-linear for $\rho$ is then equivalent to saying that $u$ is an A-linear mapping of $\rho'_*(F)$ into $E$.