A simple module is also called an irreducible module.
Remember that modules are assumed unitary unless otherwise stated, so 1*M = M.
If g is a nonzero element of a simple module M, g generates M, else it would span a proper submodule. Thus a simple module is cyclic, and any nonzero element acts as generator.
We are using the fact that M is unitary here. Since g is the same as 1*g, we are guaranteed to have g+g, via 2*g. The left multiples of a generator are closed under addition because of the action of R, and the fact that g = 1*g. But I digress.
One way to create a simple left module is to start with a maximal left ideal H in R. The cosets of H define a left R module M that is not trivial. For instance, 1 is not in H, so 1+H and 0+H are distinct cosets. Now a proper submodule of M defines a left ideal between R and H, which is impossible. Thus M is simple.
As we shall see, the converse is also true. If M is simple, let g be any nonzero element of M, which acts as a generator for M. Let H be the left ideal of R that maps g to 0. We will abuse the notation somewhat, and let R/H denote the cosets of H in R. (If H were an ideal, R/H would be the quotient ring.)
Build a map from R/H onto M by selecting a coset representative and applying it to g. Since H*g = 0, this map is well defined. Since 1*g = g, the coset 1+H corresponds to g. Since g spans all of M, the map is onto.
Let x and y represent two cosets in R/H, with images xg and yg. Addition in R/H produces x+y, and addition downstairs gives (x+y)g, M being an R module. Our map respects addition. And if s is in R, s acts on the coset of x to give the coset of sx, while s acts on xg to give sxg. The map respects left multiplication by R, and is a module homomorphism. Finally, if x and y lead to the same element of M, then x-y maps to 0, whence x-y kills g, and x-y lies in H. The map is 1-1, a module isomorphism, and R/H is isomorphic to M. Since R/H produces the simple module M, H must be a maximal ideal.
If R is commutative, every simple R module is R mod a maximal ideal, which is a field.
If we start with R/H = M, then reverse the process to find H, is it the same H? Maybe not. The next page presents a noncommutative example.
Remember that M is isomorphic to R/H, where H is a maximal left ideal. Once this isomorphism is established, something in M is associated with 1 in R. An endomorphism on M defines, and is defined by, the image of 1 in M. If 1 moves to y, then f(x) = f(x*1) = x*f(1) = xy. Yet the same holds for x+w, for every w in H. To avoid a contradiction, wy must lie in H. Assuming Hy lies in H, f is well defined, and is equivalent to right multiplication by y across the cosets of H in R.
If R is commutative then H is an ideal, and Hy automatically lies in H. The endomorphisms are isomorphic to the field M.
If R is not commutative, we can still map 1 to any integer, or anything else in the center of R. Use this to show the characteristic of E equals the characteristic of R/H, which is either prime or 0.
To see where endomorphisms fail, let R be the 2×2 matrices over a division ring D, and let M = R. Let H be the submodule that is 0 down the left column. We'll see later on that this is a maximal left ideal. The quotient M, which is our simple module, can be represented by matrices that are 0 on the right. The element 1 in M looks like [1,0|0,0], with a 1 in the upper left, and this can be mapped to [y,0|0,0] without trouble, because Hy lies in H. In fact Hy = 0. But try mapping 1 to [y,0|z,0]. Premultiply by [0,1|0,0] and find something that is not in H. The only R endomorphisms of M scale M by elements of D, even though M looks like D2. Of course the endomorphisms form a division ring, as they should.