Meaning of "$R$ is a free module over $R$ generated by $1$"

Question

I'm reading the proposition below

But I'm stuck at the part that says "Since $R$ is a free module over $R$ generated by $1$, ...". I'd like to understand why and why the part $h(\phi) = \phi(1) = x$ follows.

Answer 1

The module structure on $R$ is where addition is the addition on $R$ and scalar multiplication is multiplication. A free module is a module with a basis. That is, a set $\{e_\alpha : \alpha \in A\}$ such that every element of the module can be written as a finite sum $r_1 e_{\alpha_1} + \dots + r_n e_{\alpha_n}$ in a unique way. For the module $R$, every element can be written as $r = r \cdot 1$ uniquely. Thus $\{1\}$ is a basis for the $R$ as an $R$-module.

The point? Like vector spaces, defining a map from a free module to another module can be uniquely specified by where the elements of the basis are mapped to. In this case, where $1$ gets mapped to. If you know what $\phi(1)$ is then you know what $\phi(r)$ is for any $r \in R$ because $\phi(r) = \phi(r \cdot 1) = r \phi(1)$ by $R$-linearity.