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.