3
$\begingroup$

The Auslander correspondence said there exists a bijection between the set of Morita-equivalence classes of representation-finite finite-dimensional algebras $\Lambda$ and that of finite-dimensional algebras $\Gamma$ with gl.dim $\Gamma \leq 2 \leq$ dom.dim $\Gamma$. It is given by $\Lambda \mapsto \Gamma:=\operatorname{End}_{\Lambda}(M)$ for an additive generator $M$ of $\bmod \Lambda$.

I don't understand why $\Lambda \mapsto \Gamma:=\operatorname{End}_{\Lambda}(M)$ is a bijection between the set of Morita-equivalence classes. And what is the importance of Morita-equivalence classes in representation theory of Artin algebra?

Thank you.

$\endgroup$

1 Answer 1

3
$\begingroup$

Let me answer your second question about the importance of Morita-equivalence classes first. Two rings $R$ and $S$ are called Morita equivalent if $\operatorname{Mod} R$ and $\operatorname{Mod} S$ are equivalent as categories. If one considers representation theory as the study modules over a ring, then in some sense $R$ and $S$ are indistinguishable, so it makes sense to study rings up to Morita equivalence. (Of course, there are many other aspects of representation theory that people are insterested in, which differ a lot if one changes the Morita representative: monoidal structure, dimension of simple modules, existence of subrings, just to name a few).

Now with your particular question, you said that the map is defined by sending $\Lambda$ to $\operatorname{End}_\Lambda(M)$ where $M$ is a additive generator of $\operatorname{mod} \Lambda$. If you would just define that on isomorphism classes, instead of Morita equivalence classes, then this would not be well-defined, as there are many different such additive generators, yielding non-isomorphic algebras. Just to give a trivial example: If $R=\Bbbk$ is the ground field, then $\Bbbk^2$ is an additive generator of $\operatorname{mod}\Lambda$ and $\operatorname{End}_\Bbbk (\Bbbk^2)\cong \operatorname{Mat}_{2\times 2}(\Bbbk)$ is Morita equivalent, but not isomorphic to $\Bbbk\cong \operatorname{End}_\Bbbk(\Bbbk)$. A different way to fix this uses the basic representative on each side. Then you would send (a basic representation-finite) $\Lambda$ to $\operatorname{End}_\Lambda(M)$ where $M$ has precisely one summand from each isomorphism class of indecomposable modules.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .