Classification of simple modules for the free algebra Let $A=K\langle x,y\rangle$ be the free associative algebra in two generators over a field $K$ (we can assume  that the field is algebraically closed or even $K=\mathbb{C}$ first if that helps)

Question: Is there a classification of finite dimensional simple $A$-modules or can one show that this is "impossible" as in the word problem for groups?

 A: This is an extended version of my comments.   For more details one should look at my very old blogpost at my now discontinued blog.
Very roughly, a $K$-algebra $A$ is of wild representation type if classifying the finite dimensional indecomposable representations of $A$ is as hard as classifying the finite dimensional indecomposable representation of any $K$-algebra.  There is a precise technical definition of what this means.
A commonly given explanation as to why wild classification problems are hard is that one can encode the word problem for groups into the first order theory of the module category of wild algebra.    Prest has proved more precise statements of this sort.  To be honest, this doesn't seem all that related to me to the difficulty of the classification problem and so I'm not sure why people always bring it up in this context.
It's pretty clear what is the first order theory of two automorphisms of a vector space over a field.   You can talk about the two automorphisms, vectors, scalar multiplication and addition.  Now if one wants to work with infinite dimensional vector spaces, then it is easy to encode the word problem for groups into the first order theory.  Just take a finite presentation of a group $G$ with $2$-generators with undecidable word problem.  To test if $w=1$ in this group you write a sentence saying that if the finitely many relations defining $G$ are satisfied by the automorphisms, then subbing the automorphisms into $w$ gives you the identity map.    By considering the regular representation of $G$, we see that this sentence is true iff $w=1$ in the group $G$.
The problem is that this uses an infinite dimensional vector space because $G$ is an infinite group. So this is totally irrelevant to the question of finite dimensional representations.   But there is a beautiful result of Slobodoskoi that says that the uniform word problem for finite groups is undecidable.  What this means is given a finite set of generators $X$ and relations $R$, it is undecidable given a word $w$ if the image of $w$ is $1$ under all homomorphisms from the free group on $X$ to a finite group such that the words in $R$ map to $1$.   A theorem of Mal'cev says that any finitely generated group of matrices over a field is residually finite and hence it is also undecidable whether $w$ maps to $1$ under all finite dimensional representations of the free group on $X$ over the field $K$ for which $R$ maps to $1$.  This question can then be rephrased like above in the first order theory of two automorphisms  (and hence of two endomorphisms) by first embedding the free group on $X$ into the two generated free group.
Now if $K$ is of characteristic $0$, then since finite group has semisimple algebras, it follows that $w=1$ is a consequence of $R$ in all finite groups iff it is consequence of $R$ under all irreducible representations of the free group on $X$.  So at least in characteristic $0$ this problem can be encoded into the first order theory of irreducible representations of the free algebra.
