= Solution
Write a finite <quiver> as $Q=(Q_0,Q_1,s,t)$, with vertex and arrow sets and source/target maps. A <representation of a quiver> assigns a <vector space> $X_i$ to each vertex and a <linear map> $f_\rho:X_{s(\rho)}\to X_{t(\rho)}$ to each arrow. A <quiver representation morphism> $h:X\to Y$ is a family satisfying $h_{t(\rho)}f_\rho=g_\rho h_{s(\rho)}$.
The <path algebra> $A=kQ$ has every directed path, including each length-zero path $e_i$, as a <basis>. Multiplication is composition when endpoints match, and zero otherwise; in $pq$, the path $q$ is traversed first. The <orthogonal idempotents> satisfy $1=\sum_ie_i$.
The <path-algebra module equivalence> is explicit. From a representation, form $M=\bigoplus_iX_i$, let $e_i$ project onto $X_i$, and let each path act by the composite of its arrow maps. Conversely, an $A$-<module> gives $X_i=e_iM$ and $f_\rho(x)=\rho x$. An $A$-<module homomorphism> restricts to the required vertex maps, and a compatible family extends by direct sum. These constructions are mutually inverse up to their evident natural identifications.
\b[$kQ$ is finite-dimensional exactly when $Q$ is finite and has no oriented cycle.] For a finite <acyclic quiver>, paths have length at most $|Q_0|-1$. An oriented cycle has arbitrarily many distinct powers, giving infinitely many basis paths. If arbitrary infinite quivers are allowed, finiteness of both vertices and arrows is also necessary; the unital module correspondence above uses finite $Q_0$.
Choose only the orientation $1\to2\to3$. The <interval representations of an equioriented three-vertex quiver> $I[a,b]$ have $k$ at vertices $a,\ldots,b$, zero elsewhere, and identity arrows within that interval. The complete list is
$$
\boxed{I[1,1],\ I[2,2],\ I[3,3],\ I[1,2],\ I[2,3],\ I[1,3]}.
$$
Here is an elementary proof, without the <Gabriel theorem>. For $X_1\xrightarrow fX_2\xrightarrow gX_3$, set $K=\operatorname{im}f\cap\ker g$. Choose $F$ complementing $K$ in $\operatorname{im}f$, $G$ complementing $K$ in $\ker g$, and $H$ complementing $\operatorname{im}f+\ker g$ in $X_2$. Then $X_2=K\oplus F\oplus G\oplus H$, and $g$ is injective on $F\oplus H$. Lift bases of $K,F$ to a complement of $\ker f$ in $X_1$, and extend the bases of $gF,gH$ to $X_3$. These bases split $X$ into precisely the six kinds of interval block. Every block has <endomorphism ring> $k$, hence is indecomposable, and their different supports make them pairwise nonisomorphic. The same basis argument handles arbitrary vertex dimensions; each indecomposable block itself is finite-dimensional.
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