Concrete example of an endomorphism ring? I've been trying to study endomorphism rings, but I'm having trouble finding any concrete examples. There are plenty examples out there that describe endomorphism rings with certain properties, but nothing quite like what I'm looking for.
From what I think I've learned so far, an endomorphism ring is a ring formed from all the homomorphisms from a group onto itself. The addition operation is "point-wise", which I'm not sure I completely understand. If $f$ and $g$ are endomorphisms, then $(f+g)(x) = f(x)+g(x)$, which I suppose would be another endomorphism, but I'm having trouble seeing that. Also the multiplication operation is the composition of two functions.
What I'm asking for is a very concrete example. Something like $\mathbb{Z}_6$, although I'm not particularly attached to it. If there's a more instructive example, then forget $\mathbb{Z}_6$.
I'm looking for this: a list of the endomorphisms (defined explicitly, either by a general definition, or a list of which elements map to which in each endomorphism), examples of how to add two endomorphisms (a Cayley table might be nice too, but I could probably figure that out with enough examples), and the same for multiplication (which is just composition).
I think seeing a concrete example may take some of the mystery out of it. Right now it baffles me that something so strange could still be a ring.
 A: Here are some more "very concrete" examples of endomorphism rings, namely
$$
{\rm End}(\mathbb{Z})\cong \mathbb{Z},\; {\rm End}(\mathbb{Q})\cong \mathbb{Q},\; {\rm End}(C_n)\cong \mathbb{Z}/n,\; {\rm End}(F_n)\cong M_n(\mathbb{Z}),\;{\rm End}(\mathbb{Q}^+/\mathbb{Z})\cong \widehat{\mathbb{Z}}.
$$
where $C_n\cong \mathbb{Z}/n$ denotes the cyclic group of order $n$, and $F_n$ the free group of rank $n$. The group $\widehat{\mathbb{Z}}$ is explained here. There is a whole book on endomorphism rings of abelian groups, where all examples are explained - see here.
A: Let's start with End$(\mathbb{Z})$:  Suppose $f\colon \mathbb{Z}\to \mathbb{Z}$ is an endomorphism.  Suppose I know the value of $f(1)$.  Then $$f(2)=f(1+1)=f(1)+f(1),$$
so I know the value of $f(2)$.  Similarly $$f(-1)=-f(1),$$ so I know the value of $f(-1)$ too.
More generally, for $n>0$ $$f(n)=f(1+1+1+\cdots+1)=f(1)+f(1)+f(1)+\cdots +f(1)=nf(1),$$
and $$f(-n)=-f(n)=-nf(1).$$
Putting all this together, we see that applying $f$ to some integer $n$ is just multiplying by $f(1)$.  So every element of End$(\mathbb{Z})$ is just multiplication by some fixed integer.  For $a\in \mathbb{Z}$, let us use the notation $[a]$ to denote multiplication by $a$.  Thus  End$(\mathbb{Z})$ is just the set of such $[a]$.
Let us see what arithmetic on End$(\mathbb{Z})$ looks like.  First addition:
$$([a]+[b])n=[a]n+[b]n=an+bn=(a+b)n=[a+b]n,$$
for all integers $n$.
The first equality here is just the definition of "pointwise" addition.  Namely to evaluate the sum of two endomorphisms on some integer $n$, we just evaluate them each on $n$ and then add.
The second equality is unpacking our notation: $[a]$ means multiplication by $a$.  The third equality is just bracketing, and the last one is again the definition of $[a+b]$.
So addition on End$(\mathbb{Z})$ is the same as addition on the integers:$$[a]+[b]=[a+b]$$
Now consider multiplication:$$([a][b])n=[a]([b]n)=[a](bn)=abn=[ab]n.$$
The first equality here says that the composition of the function $[a]$ with the function $[b]$, applied to $n$, is simply what you get if you first apply $[b]$ to $n$, then apply $[a]$ to the result.  In other words: Do $[b]$ then do $[a]$.
(Note: This may seem a bit backwards, but it is just how composition is defined, so that the argument can go on the right.)
The remaining equalities follow from the definitions of $[b],[a], [ab]$.
Thus we have $$[a][b]=[ab].$$
That is multiplication in End$(\mathbb{Z})$ is just ordinary multiplication of the corresponding integers.
Thus End$(\mathbb{Z})$ can be identified with $\mathbb{Z}$, under the correspondence $$[a]\leftrightarrow a.$$ Further addition and multiplication are the same in End$(\mathbb{Z})$ and $\mathbb{Z}$.  We write End$(\mathbb{Z})\cong\mathbb{Z}$.
The same approach can calculate End$(\mathbb{Z}/6\mathbb{Z})$.  Again the endomorphisms are multiplication by integers, and pointwise addition corresponds to ordinary addition, whilst composition corresponds to ordinary multiplication.  The only difference this time is that multiplication by $a$ and multiplication by $a+6$ are the same: $$[a]=[a+6]$$ Thus we have End$(\mathbb{Z}/6\mathbb{Z})\cong\mathbb{Z}/6\mathbb{Z}$.
For a third example let us calculate End$(\mathbb{Z}^3)$.  We can write elements of $\mathbb{Z}^3$ as column vectors: $$\left(\begin{array}{c}u\\v\\w\end{array}\right)$$
Given an endomorphism $f\colon \mathbb{Z}^3 \to \mathbb{Z}^3$ we have
\begin{eqnarray*}f\left(\begin{array}{c}u\\v\\w\end{array}\right)&=&f\left(u\left(\begin{array}{c}1\\0\\0\end{array}\right)+v\left(\begin{array}{c}0\\1\\0\end{array}\right)+w\left(\begin{array}{c}0\\0\\1\end{array}\right)\right)\\\\&=&
uf\left(\begin{array}{c}1\\0\\0\end{array}\right)+vf\left(\begin{array}{c}0\\1\\0\end{array}\right)+wf\left(\begin{array}{c}0\\0\\1\end{array}\right)\\\\&=&
u\left(\begin{array}{c}a\\b\\c\end{array}\right)+v\left(\begin{array}{c}d\\e\\f\end{array}\right)+w\left(\begin{array}{c}g\\h\\i\end{array}\right)\\\\&=&
\left(\begin{array}{ccc}
a&d&g\\
b&e&h\\
c&f&i
\end{array}\right)
\left(\begin{array}{c}u\\v\\w\end{array}\right)
\end{eqnarray*}
Here $$
\left(\begin{array}{c}a\\b\\c\end{array}\right)=f\left(\begin{array}{c}1\\0\\0\end{array}\right),\qquad 
\left(\begin{array}{c}d\\e\\f\end{array}\right)=f\left(\begin{array}{c}0\\1\\0\end{array}\right),\qquad
\left(\begin{array}{c}g\\h\\i\end{array}\right)=f\left(\begin{array}{c}0\\0\\1\end{array}\right)$$
So we see that the endomorphisms of $\mathbb{Z}^3$ are precisely multiplication by $3\times 3$ matrices with integer entries.  For a $3\times 3$ matrix $M$, let $[M]$ denote the endomorphism of $\mathbb{Z}^3$ given by multiplication by $M$.
What about addition?  We have $$([M]+[N])v=[M]v+[N]v=Mv+Nv=(M+N)v=[M+N]v.$$
That is pointwise addition of endomorphisms corresponds to addtion of matrices.
For multiplication we have $$([M][N])v=[M]([N]v)=[M](Nv)=M(Nv)=(MN)v=[MN]v.$$
That is composition of endomorphisms corresponds to multiplication of matrices.
In summary End$(\mathbb{Z}^3)=M_3(\mathbb{Z})$.
A: If we are speaking about endomorphisms, the endomorphism ring of a finite dimensional, $\mathbb{k}$-vector space is the algebra of all $n\times n$ real matrices with elements from $\mathbb{k}$:
$$
\mathcal{E}nd_{\mathbb{k}}(V)\cong M_{n}(k)
$$
(Here the endomorphisms of the vector space $V$ are the linear maps).
