An $R$-algebra $A$ is simple if and only if $A$ is a simple right $A^{\text{opp}}\otimes A$-module A simple $R$-algebra $A$ is a unital right $R$-module on which is defined a bilinear mapping that is associative and where $A$ has no non-zero proper submodules ($A$ is simple). We define an opposite algebra $A^{\text{opp}}$ to be just the same as $A$ but with reversed multiplication $\square$; i.e. $a\square b=ba,\;a,b\in A^{\text{opp}}$. And $A^e$ is defined as $A^{\text{opp}}\otimes A$. My question is about the following proof from the book Associative Algebras from R.S. Pierce:

I do not understand how they got to the conclusion of (i) so fast and I would like to know whether someone could help me out in understanding this very small proof or providing proper reading material that clarifies what is related to this. Thanks in advance!
 A: The two-sided ideals of $A$ are precisely the right $A^e$-submodules of $A$.
A simple algebra is one without non-trivial two-sided ideals.
A simple module is one without non-trivial submodules.
A: An $A^e$ submodule is the same thing as a submodule of the regular bimodule ${}_{A}A_A$, which in turn is the same thing as a two-sided ideal of the algebra $A$ . 
So, part (i) of the lemma could be stated as: 

$A$ is a simple algebra $\Leftrightarrow$ $A$ is a simple right $A^e$-module $\Leftrightarrow$ the ${}_{A}A_A$-bimodule is simple. 

Since you are asking for some further reading, maybe it would be helpful to take a look at the book of Drozd-Kirichenko: See par. 4.1, p. 69-70, up to  proposition 4.1.1 and the comments of p. 73 (right before theorem 4.2.4). 
(In general an $A$-$B$-bimodule ${}_A M_B$ is the same thing with a right $B\otimes A^{op}$ module:  Given an $A$-$B$-bimodule $M$ we can always view this as a right $B\otimes A^{op}$-module setting $m(b\otimes a^{op})=amb$. Conversely, if $M$ is a right  $B\otimes A^{op}$-module then we get an $A$-$B$-bimodule by setting $am=m(1\otimes a^{op})$ and $mb=m(b\otimes 1)$.
 In the same manner ${}_A M_B$ may be viewed as a right $A^{op}\otimes B$ module via $m(a^{op}\otimes b)=amb$ (see the comment in user's @Lord Shark the Unknown answer). 
