So I'm a post-graduate maths person and self-studying quiver representations. I came across a problem, and before I even lay out the entire problem (which aims to construct a functor $T$ such that $L_0(T)$ is not isomorphic to $T$), I'd like to clarify the certain pieces of it I'm confused by. Part One:
Consider A the quotient of the algebra upper-triangular matrices by the ideal generated by the matrix having only the entry $a_{3,3}$ (bottom right-hand corner) and zeros elsewhere.
(editorial: already a bit confused here. What will elements of this look like? Sure, let's denote by $I$ the ideal by which we are quotient-ting our algebra, and say elements are of the form $M + I$ where $M$ is an upper triangular matrix... but this is hardly enlightening. By considering the definition of the equivalence relation, as usual, two elements (of the original algebra) represent the same coset iff their difference produces a matrix with a sole entry in the lower right hand corner, thus in $I$. Again, I'm having trouble picturing this construction.)
Part Two: Denote by $e_1$, $e_2$, and $e_3$ the cosets of the matrices which are 1 at position $a_{1,1}$, $a_{2,2}$, and $a_{3,3}$ and zero elsewhere. Then $A = e_1A \oplus e_2A \oplus e_3A$.
editorial: again, I'm stuck in this mess. $e_1$, $e_2$, and $e_3$ look like
1 0 0 0 0 0 0 0 0
0 0 0 + I, 0 1 0 + I, 0 0 0 + I
0 0 0 0 0 0 0 0 1
but then in this case, isn't $e_3A$ equal to $I$? Why should this not be trivial? Please forgive me if these are stupid questions, I don't have much "real" experience with matrix algebras.
I suppose I'll just ask about these two things before laying out the rest of the problem, on the off chance I can figure things out from here. Thanks!!
You haven't said what field you're working over. Let's call it $F$. Then, for the first part, for every $a,b,c,d,e\in F$ you have a coset $$ \begin{pmatrix}a&b&c\\0&d&e\\0&0&0\end{pmatrix}+I $$ These are the elements of $A$. I'm not sure how you want to "picture this construction".
For the second part, you are correct that $e_3A=I$ is the zero element in $A$. I'm not sure why the decomposition is written this way, but if you are willing to write $V=V\oplus(0)$ then I guess it's fine. Note that the other two summands are $$ e_1A=\left\{\begin{pmatrix}a&b&c\\0&0&0\\0&0&0\end{pmatrix}+I\mid a,b,c\in F\right\}$$ and $$e_2A=\left\{\begin{pmatrix}0&0&0\\0&d&e\\0&0&0\end{pmatrix}+I\mid d,e\in F\right\}$$ so outside of regarding $I$ as a direct summand, there is nothing strange here.