Let $R$ be a ring, and Define $S:= \operatorname{End}(M_R)$ and $U:= \operatorname{End}(M_{\mathrm{End}(_SM)}).$
I need to show $S=U$
So I am having trouble even decoding this, so elements of $S$ are $M$-endomorphisms over some ring $R$
But elements of $U$ are also endomorphisms of $M$ but the scalar field is an endomorphism field of Where the field is also endomorphisms? Really can't wrap my head around this, I need a small hint.
Let $T={\rm End}({}_SM)$. Then $S,T,U$ are all subsets of the set of abelian group homomorphisms $M\to M$.
By convention for $s_1,s_2 \in S$ we denote "do $s_1$ then $s_2$" by $s_2s_1$. Similarly for $u_1,u_2 \in U$ we denote "do $u_1$ then $u_2$" by $u_2u_1$. However it is the other way round for $T$: for $t_1,t_2 \in T$ we denote "do $t_1$ then $t_2$" by $t_1t_2$. These conventions allow us to apply elements of $S$ and $U$ from the left to elements of $M$, whilst elements of $T$ (like elements of $R$ may be applied from the right). This means that the equations below all come down to rearranging brackets.
There is also a natural map from $R$ to abelian group homomorphisms $M\to M$, given by right multiplication. Let $R^*$ denote the image of this map. Thus $R^*, S,T,U$ are all subsets of a common set, and to show that $S=U$ it suffices to show that $S\subseteq U$ and $U \subseteq S$.
You know (from definition of endomorphism):
$(fx)\lambda=f(x\lambda) \qquad \forall f\in S, \,\,\lambda \in R$
$(fx)\alpha=f(x\alpha) \qquad \forall f\in S, \,\,\alpha \in T$
$(\Gamma x)\alpha=\Gamma(x\alpha) \qquad \forall \Gamma\in U,\,\, \alpha \in T$
From these equations you must deduce:
1) $S \subseteq U$,
2) $R^* \subseteq T$,
3) (2) implies that $U \subseteq S$.
First write down what it means to be in each of $S,T,U$ and then you should see that each of these 3 statements follows from one of the 3 equations.