An $k$-algebra $A$ is tame (or, equivalently, it has tame representation type) ifthere are are, for every dimension $d\geq0$, you can parametrize all isoclasses of indecomposable $A$-modules of dimension $d$, apart from a finite number of them, by a finite number of $1$-parameter families. On the other hand, a finite dimension dimensional $k$-algebra $A$ is wild (or, equivalently, it has wild representation type) if in the category $\mathrm{mod}_A$ of finite dimensional modules contains a copy of the category $\mathrm{mod}_{k\langle X,y\rangle}$x,y\rangle}$ of modules over the free$k$-algebra on two generators. It is an amazing theorem of Drozd that a finite dimensional algebra is either tame or wild; this is the so called dichotomy theory. One of the reasons that make this theorem so amazing is that one can show that if$A$is wild then $\mathrm{mod}_A$ contains copies of the module categories of all finite dimensional algebras; in other words, wild algebras are really wild... In particular, this concepts of tame and wild apply to hereditary algebras, which are those of global dimension$1$. Now, a finite dimensional hereditary algebra is Morita equivalent to the path algebra$kQ$on a quiver without oriented cycles. A well-known theorem of Gabriel and others tells us that such a path algebra$kQ$is tame iff the quiver$Q$is, when you forget the orientation of the arrows, an Dynkin or an extended-Dynkin diagram. In all other cases the parth algebra is wild. Two great references on all of this are the (first volume of the) book by Assem, Skowroński and Simson, or the book by Auslander, Reiten and Smalø. 3 Fix a typo An$k$-algebra$A$is tame (or, equivalently, it has tame representation type) if there are are, for every dimension$d$, d\geq0$, you can parametrize all isoclasses of indecomposable $A$-modules of dimension $k$, d$, apart from a finite number of them, by a finite number of$1$-parameter families. On the other hand, a finite dimension$k$-algebra$A$is wild (or, equivalently, it has wild representation type) if in the category $\mathrm{mod}_A$ of finite dimensional modules contains a copy of the category $\mathrm{mod}_{k\langle X,y\rangle}$ of modules over the free$k$-algebra on two generators. It is an amazing theorem of Drozd that a finite dimensional algebra is either tame or wild; this is the so called dichotomy theory. In particular, this concepts of tame and wild apply to hereditary algebras, which are those of global dimension$1$. Now, a finite dimensional hereditary algebra is Morita equivalent to the path algebra$kQ$on a quiver without oriented cycles. A well-known theorem of Gabriel and others tells us that such a path algebra$kQ$is tame iff the quiver$Q$is, when you forget the orientation of the arrows, an Dynkin or an extended-Dynkin diagram. In all other cases the parth algebra is wild. Two great references on all of this are the (first volume of the) book by Assem, Skowroński and Simson, or the book by Auslander, Reiten and Smalø. 2 Fixed underscore problem An$k$-algebra$A$is tame (or, equivalently, it has tame representation type) if there are are, for every dimension$d$, you can parametrize all isoclasses of indecomposable$A$-modules of dimension$k$, apart from a finite number of them, by a finite number of$1$-parameter families. On the other hand, a finite dimension$k$-algebra$A$is wild (or, equivalently, it has wild representation type) if in the category $\mathrm{mod}A$\mathrm{mod}_A$ of finite dimensional modules contains a copy of the category $\mathrm{mod}{k\langle \mathrm{mod}_{k\langle X,y\rangle}$ of modules over the free $k$-algebra on two generators. It is an amazing theorem of Drozd that a finite dimensional algebra is either tame or wild; this is the so called dichotomy theory.
In particular, this concepts of tame and wild apply to hereditary algebras, which are those of global dimension $1$.
Now, a finite dimensional hereditary algebra is Morita equivalent to the path algebra $kQ$ on a quiver without oriented cycles. A well-known theorem of Gabriel and others tells us that such a path algebra $kQ$ is tame iff the quiver $Q$ is, when you forget the orientation of the arrows, an Dynkin or an extended-Dynkin diagram. In all other cases the parth algebra is wild.