TITLE: About the Schur's lemma
QUESTION [6 upvotes]: In modules we have the following result:

If $M$ is a simple $R$-module then $End_R M$ is a division ring. 

This theorem is known as Schur lemma and its proof is not difficult. 
Usually in ungraduate courses the only division ring mentioned is the quaternion ring. The Shur lemma is a "machine" to produce division ring but nothing assure me that the divison ring we get from a simple module is not a field.
I have the following question about this topic: 

How do I calcule the $End_R M$. For example $End_\mathbb{Z} \mathbb{Z}$
Is there a simple $R$-module $M$ such that $End_R M \cong \mathbb{H}$ the quaternion ring?
Where can I find more examples about division ring wich are not fields. There existe any finite division ring?

Thanks a lot!

REPLY [2 votes]: When $\mathbb{R}$ is the field of real numbers, the quaternion group $Q$ of order $8$ has a $4$-dimensional irreducible representation. This yields a simple $4$-dimensional module $M$ for the group ring $R = \mathbb{R}G$, with ${\rm End}_{R}(M)$ being the ring of real quaternions.
In (finite) group representation theory a common way to realise modules whose endomorphism rings are non-commutative division algebras is to find a finite group $G$ and a complex irreducible character $\chi$ of $G$ such that the representation affording $\chi$ can not be realised over the extension of $\mathbb{Q}$ generated by the character values of group elements, but some multiple of it can be. The $4$-dimensional real representation of the 
quaternion group of order $8$ is really (as a complex representation) a direct sum of two copies of a $2$-dimensional irreducible representation of that group. But perhaps you have not yet seen group representation theory.