Let $A$ be a finite dimensional $k$-algebra, $k$ is a field, with a finite global dimension. I wonder if that implies $A$ is tame or finite type? or more generally is there a relation between these two notions?
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3A path algebra is hereditary and it can be of wild representation type: see http://mathoverflow.net/questions/5895/what-are-tame-and-wild-hereditary-algebras – egreg Dec 15 '15 at 14:51
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This is really expanding on egreg's comment: there are wild algebras of any finite global dimension. Indeed, if you take $A$ to be a wild hereditary algebra (these exist: take the path algebra of any connected non-Dynkin or non-extended-Dynkin quiver), and $B$ to be any algebra of global dimension $n$, then the algebra $A\times B$ is wild of global dimension $n$.
Pierre-Guy Plamondon
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Thank you for the answer, does $n$ have to be positive, or this works for zero also? – Math137 Dec 15 '15 at 18:51
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2It has to be positive. A finite-dimensional algebra of global dimension zero is semisimple, and thus of finite representation type. – Pierre-Guy Plamondon Dec 15 '15 at 18:53
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2And how about the converse? Do tame algebras have finite global dimension? – Math137 Dec 15 '15 at 18:54
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3Even representation-finite algebras can be of infinite global dimension. An example would be $k[x]/(x^n)$ for any $n\geq 2$. – Pierre-Guy Plamondon Dec 15 '15 at 18:55