Exercise 4, § 1

Let $Q = (S,F, o, t)$ be a quiver. We denote by I the sum set of the family $(S,F)$; we identify S and F with subsets of I. Let $k$ be a commutative ring. We call the $k$-algebra of the quiver Q the quotient $A_Q$ of the free associative $k$-algebra on the set I by the two-sided ideal generated by the elements $X^2_s-X_s$ (for $s\in S$), $X_uX_v-X_vX_u$ (for $u, v\in S$), $X_fX_{t(f)}-X_f$ and $X_{o(f)}X_f-X_f$ (for $f\in F$). For $i\in I$, we denote by $x_i$ the image of $X_i$ in the algebra $A_Q$.

(a) Prove that one has $\sum_{s\in S}x_s= 1$.

(b) Prove that one has $x_fx_g= 0$ if $f$ and $g$ are arrows of Q which are not composable.

(c) For every path $c= (a_0, f_1, . . . , f_n, a_n)$ in Q, put $x_c$ = $x_{a_0}x_{f_1}. . . x_{f_n}$. Let $c$ and $c'$ be paths in Q; prove that one has $x_cx_{c'}=x_{cc'}$ if $c$ and $c'$ are composable, and $x_c=x_{c'}$ otherwise.

(d) Let C be the set of paths in the quiver Q; prove that the family $(x_c)_{c\in C}$ is a basis of the $k$-module $A_Q$.

(e) Let $J_Q$ be the two-sided ideal of $A_Q$ generated by the elements of the form $x_f$, for $f\in F$. Prove that the algebra $A_Q/J_Q$ is isomorphic to $k^{(S)}$.

(f) In order that the $k$-algebra $A_Q$ be a finitely generated $k$-module, it is necessary and sufficient that the set of vertices of Q be finite and that every loop in Q be of length zero. The ideal $J_Q$ is then nilpotent.

We retain these notations for the remainder of the exercises of §1. When we speak of an $A_Q$-module (without further specification), it is a right module. If A is a ring, an A-module M (without further specification) will be a right A-module; if $a$ is an element of A, we denote by $a_M$ or $(a)_M$ the homothety $x\mapsto xa$ of M.