Let $A$ be a finite dimensional algebra over an algebraically closed field $K$. The Auslander-Reiten quiver $\Gamma\_A$ of $A$ is a means of presenting the category of finitely generated right $A$-modules. The Auslander-Reiten quiver is a locally finite quiver whose vertices are indecomposable modules (up to isomorphism) and whose arrows are irreducible morphisms between indecomposable modules. It is given the structure of a translation quiver via the Auslander-Reiten translate $\tau$.


Two distinct vertices/modules of $\Gamma\_A$ are said to belong to the same component if there exists a finite path/composition of irreducible morphisms between them. If $A$ is representation-finite (i.e. the category of finitely generated right $A$-modules contains finitely many non-isomorphic indecomposable objects), then $\Gamma\_A$ consists of finitely many components, which each have finitely many vertices. However if $A$ is representation-infinite, then necessarily $\Gamma\_A$ is infinite.


It is easy to construct an example of a representation-infinite algebra $A$ for which there exists a finite component of $\Gamma\_A$. (**Edit:** the example I provide in this question is false, as indicated by the answer given by Jeremy Rickard.) For example, consider the algebra $KQ/I$, where $KQ$ is the path algebra of the quiver
$$ Q\colon \; 1 \begin{smallmatrix} \alpha \\ \rightarrow \\ \rightarrow \\ \beta \end{smallmatrix} 2 \begin{smallmatrix} \gamma \\ \rightarrow \\ \color{white} \gamma \end{smallmatrix} 3 \begin{smallmatrix} \delta \\ \rightarrow \\ \color{white} \delta \end{smallmatrix} 4$$
and $I$ is the ideal generated by the set $\{\beta\gamma, \gamma\delta\}$. This is representation-infinite because it contains the Kronecker quiver as a subquiver. But the relation $\gamma\delta$ 'cuts off' the right-hand side of the quiver, which means we have a finite Auslander-Reiten component
$$\begin{matrix} S(4) & & & & S(3) \\ & \searrow & & \nearrow & \\ & & P(3)=I(4) & & \end{matrix}$$
(which is essentially $\mathrm{mod}\;K\mathbb{A}\_2$) in $\Gamma\_A$, where $S(v)$, $P(v)$ and $I(v)$ are the simple, indecomposable projective, and indecomposable injective modules corresponding to the vertex $v$ in $Q$ respectively. (One can also use the fact that this is a string algebra to easily compute $\Gamma\_A$.)


**My question is this:** When is it the case that for a representation-infinite algebra $A$, every component of $\Gamma\_A$ contains infinitely many vertices? Is this for example true if the algebra is self-injective? (I imagine the proof of the latter question is straightforward if true, but please point me to a reference if one exists.)



Given a representation-infinite algebra, when is every AR component infinite?
If $A$ is connected and has infinite representation type, then every component of its Auslander-Reiten quiver is infinite. See, for example, Theorem 5.4 in Assem, Simson and Skowronski’s *Elements of the Representation Theory of Associative Algebras*, Volume 1.


The example you give is not a complete component. It continues (infinitely) to the right, starting with an irreducible map from $S(3)$ to $P(2)$.


