Let $R$ be an associative ring with unity. I have two questions:
- If $M$ is a left $R$-module such that $\phi:M\mapsto R$ is an $R$-isomorphism, i.e., $M$ and $R$ are isomorphic as modules, does this imply that $M$ can be given a ring structure? I thought about defining multiplication in $M$ simply as $\cdot:M\times M\mapsto M, m_1\cdot m_2 = \phi^{-1}(\phi(m_1)\phi(m_2))$. I did the calculations and this appears to work and guarantee that $M$ with this product is also an associative ring with unity given by $\phi^{-1}(1_R)$. Is this correct? If so, is this a useful construction?
- If $R$ is commutative and $M$ is a simple $R$-module, then $M$ must be $R$-isomorphic to $R/I$, where $I$ is a maximal ideal of $R$. Since $I$ is maximal, this means that $R/I$ is a field. Is there a way of giving $M$ a field structure from this $R$-isomorphism? From the previous questions, I believe that if such isomorphism also happened to be an $R/I$-isomorphism, then yes, but is this true in general? And again, is this a useful construction?
In both cases it is no problem giving the module $M$ a ring (field) structure. As you checked yourself, all ring (field) axioms are true and $\phi$ becomes an isomorphism of rings (fields).
As for usefulness: Usually this is not a very helpful construction. One reason for this is that this ring structure on $M$ is not natural. Unless $1$ is the only unit in $R$ (which is quite a special case) there are multiple isomorphisms $M \to R$ of $R$-modules. Each isomorphism gives a different ring structure on $M$. For example, take the isomorphism $\mathbb{Z} \to \mathbb{Z}, n \mapsto -n$ of $\mathbb{Z}$-modules. The resulting ring structure on $\mathbb{Z}$ is given by the multiplication $(n,m) \mapsto -nm$, which is not the usual multiplication.
The question still left is, if an $R$-module homomorphism of $R/I$-modules is also $R/I$-linear. This is true, as can be easily checked: $$\phi(a x) = \phi((a+I)x) = (a+I)\phi(x) = a\phi(x).$$