This question was asked in my assignment of Commutative Algebra and I am not able to solve it. I am mainly following my class notes.
Question: Let $R$ be an integral domain and $I ⊂ R$ a non-zero principal ideal. Show that $I$, as an $R-$module, is isomorphic to the $R-$module $R$.
The problem I am facing is that I am not able to write ideal $I$ as an $R-$ module. Can you please help me with this and the rest of question if possible?
Suppose $I$ is generated by $a$. Then every element of $I$ is of the form $ra$ where $r \in R$. Since $a \not = 0$ and $R$ is integral domain, $r_1a=r_2a$ implies $r_1=r_2$. So we have a bijection $f:R \to I$ given by $f(r) =ra$.
Its easy to interpet $I$ as an $R$-module. This is because $I$ is closed under addition. Moreover, $rb \in I$ for any $r \in R$ and $b \in I$. (So, its easy to see what is the scalar multiplication)
Just note $f(r_1 + r_2) = (r_1 + r_2)a = r_1a + r_2a = f(r_1) + f(r_2)$ and $f(r'r) = r'ra =r'f(r)$.