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Multiplication is not defined coordinatewise, rather $(n,r)(m,s)=(nm,rs+rm+ns)$. In particular $(1,0)(n,s)=(n,s)$. Writing $(n,r)$ as $n+r$ should help in understanding why we have defined multiplication in such a way.

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I'm not $100\%$ I'm the right person to speak for "modern algebraists," but here's some thoughts. Dietrich mentions semisimple elements in his comment: these generalize diagonalizeable operators to elements of an algebra in the presence of an algebra representation - either an associative algebra or alternatively a Lie algebra. Indeed, given such an element ...

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(Revised) Step 1:It is easy to prove that $E_{ij}\in [A,A]$ for all $i\neq j$. Also $E_{ii}-E_{jj}\in [A,A]$. Step 2: Using step 1, let's prove that $E_{ii}\in ([A,A])$. Step 3: Using step 2, it is easy to prove that $I\in ([A,A])$. Just please use different $E_{ij}$ to obtain all of these..Also we have two sided ideals.:)

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Given any $R$-modules $M$ and $N,$ there is at least one $R$-module homomorphism $M\to N.$ In particular, there is always the trivial homomorphism $m\mapsto 0_N.$

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If $M$ is a nontrivial projective module and $N=R^n$ for $n\geq 1$ and you want a nontrivial homomorphism $M\to N$, you could do the following. Since every projective module is (isomorphic to) a direct summand of a free module $F$, you would have an injective homomorphism $g:M\to F$. If $F$ has lower rank than $M$, then it is easy to come up with another ...

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I'd recommend the book Elements of the Representation Theorey of Associative Algebras by Assem, Simson, & Skowroński. It covers both Auslander-Rieten Theory, Tilting Theory, and almost all of the background required to understand those chapters.

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Well, if we assume that $[A, A] = \{ [x, y] \mid x, y \in A \}, \tag{1}$ $[\lambda, A ] =\{ [\lambda, x ] \mid x \in A \}, \tag{2}$ and in general, for $C, D \subset A$, $CD = \{ cd \mid c \in C, d \in D \}, \tag{3}$ then I think we can proceed as follows: first of all, we describe $(C)$ for $C \subset A$; we have: $(C) =$ \$S = \{ \sum_1^n r_i c_i ...

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