I'm currently learning about modules and am pretty new to the topic. I recently found this question:
Suppose that $R$ is an integral domain and not a field. Give, with proof, an eample of a proper submodule of $R$ that is linearly isomorphic to $R$.
I don't really know how to approach this question. Also I'm struggling with linearly isomorphic. From my notes linearly isomorphic means it only has the scalar multiplication property $(\phi(rm)=r\phi(m))$ but does not necessarily have the addition property $(\phi(m_1+m_2)=\phi(m_1)+\phi(m_2))$. But somehow i can't really find much on that topic either here or in the books that I'm using. Any help would be really appreciated!
I don't think the definition of linearly isomorphic you're writing is standard. Addition is certainly linear! My best guess is that the problem is asking for a proper submodule of $R$ which is isomorphic to $R$ as an $R$-module.
Here's the example I have. As $R$ is not a field, take some nonzero nonunit element $a \in R$. You may make an objection about the zero ring, but this is not considered a domain. Anyways, consider the ideal $(a) \subseteq R$. As $a$ is not a unit this is a proper ideal. Furthermore, it is a submodule of $R$ (in fact, submodules of $R$ are precisely ideals of $R$). Now, consider the map $R \longrightarrow (a)$ via $r \mapsto ra$. This is certainly $R$-linear, and by definition of $(a)$ it is onto. Furthermore, as $a \neq 0$ and as $R$ is a domain, this map is injective. Hence, it is an isomorphism of $R$ with a proper submodule.