Let $A$ be an arbitrary (not necessarily finite-dimensional) associative algebra over an algebraically closed field $K$, and let $\mathrm{fin\,}A$ denote the category of finite-dimensional $A$-modules. The algebra $A$ is said to be wild if for any finite-dimensional $K$-algebra $B$, there exists an exact $K$-linear functor $F:\mathrm{fin\,} B \rightarrow \mathrm{fin\,} A$ such that $F$ maps indecomposable $B$-modules to indecomposable $A$-modules and $F(X)\cong F(Y) \Rightarrow X \cong Y$ ($F$ respects iso-classes). Such a functor is called a representation embedding.

To prove an algebra $A$ is wild, it is sufficient to construct a representation embedding $F:\mathrm{fin\,} C\rightarrow \mathrm{fin\,} A$ for some other wild algebra $C$. The prototypical example of a wild (actually, strictly wild) algebra is the free algebra on two variables $K\langle x,y\rangle$. The commutative algebra $K[x,y]$ is also wild (but not strictly wild).

My question is what happens if we have an algebra with $n>1$ commuting variables, with some being non-trivial units? For example, is the algebra $K[x,x^{-1},y,y^{-1}]$ wild? (Commutative algebras are never strictly wild, so this is not my concern.) The only difference in terms of the indecomposable modules is that the actions of $x$ and $y$ on $K[x,x^{-1},y,y^{-1}]$-modules correspond to non-singular matrices. To me, this does not seem like a particularly heavy restriction, and so my instinct says that the algebra is still wild. However, I've not found any references on the matter, and I've struggled to come up with a suitable representation embedding as a proof. Any help would be much appreciated.

EDIT: I'm of course excluding algebras such as $K[x,(x-\lambda_1)^{-1},\ldots,(x-\lambda_m)^{-1}]$ ($\lambda_i \in K$ for all $i$), as these are obviously tame.

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.