Are polynomial algebras over fields (that are not algebraically closed) tame?

Question

Let $A$ be an algebra over a field $K$. Loosely speaking, an algebra is said to be tame if for each $d \in \mathbb{Z}_{>0}$ all but finitely-many of the indecomposable $A$-modules of $K$-dimension $d$ occur within a finite number of one-parameter families (indexed over $K$). (At least, this is often considered the definition when $K$ is algebraically closed... if there is a different definition for more general fields, please let me know.)

A standard (and extremely important) example of a tame algebra is $K[x]$, where $K$ is algebraically closed. As a principal ideal domain, its finite-dimensional indecomposable modules are isomorphic to quotients $K[x] / (p)$, where each $(p)$ is a primary ideal. In particular, the primary ideals of $K[x]$ are powers of irreducible elements of $K[x]$. Since $K$ is algebraically closed, these are just linear polynomials $x - \lambda$ with $\lambda \in K$. Hence, the indecomposable modules of dimension $d$ are given by the quotients $K[x] / (x - \lambda)^d$. So $K[x]$ is tame whenever $K$ is algebraically closed.

What happens if we remove the assumption that $K$ is algebraically closed? The situation seems to deteriorate somewhat. For example consider $\mathbb{Q}[x]$. This is still a principal ideal domain, so most of the above reasoning applies. However since $\mathbb{Q}$ is not algebraically closed, we have (for example) infinitely many irreducible polynomials $x^2 + p x +q$, where $p,q \in \mathbb{Q}$. Unless I am mistaken, this appears to produce a two-parameter family of indecomposable $\mathbb{Q}[x]$-modules of dimension 2: one can think of this from the perspective of representations of $\mathbb{Q}[x]$ by considering $2 \times 2$ matrices (up to commuting $\mathbb{Q}$-isomorphism/similarity over $\mathbb{Q}$) whose eigenvalues are roots of $x^2 + p x +q$ for each pair of rationals $(p,q)$ that yields an irreducible polynomial.

This appears to indicate that $\mathbb{Q}[x]$ is not tame, and thus (by a theorem of Drozd), is semi-wild (i.e. badly behaved from a representation-theoretic perspective). I find this extremely surprising to the point where I feel like there must be something wrong with my reasoning. This is especially considering that we have a 'classification' of indecomposables in terms of powers of irreducible polynomials of $\mathbb{Q}[x]$... though perhaps the problem here is a lack of classification of the irreducible polynomials themselves? The problem seems to disappear if we consider finite fields, because then there are only finitely many choices of coefficients for each degree of polynomial.

Can anyone with more experience in this area confirm my reasoning here and/or offer further illumination on this matter?

Answer 1

The definition of tame representation type usually assumes that you're dealing with a finite dimensional algebra $A$ over a field $k$. In this case, the definition of a one parameter family of modules is that there is an $A$-$k[T]$-bimodule $X$ which is finitely generated and free over $k[T]$, and then the modules in the family are all required to be of the form $X \otimes_{k[T]} M$ for some $k[T]$-module $M$. The requirement is then that in any particular (finite) dimension over $k$, all but a finite number of indecomposable $A$-modules belong to a finite set of one parameter families. In other words, $k[T]$ is the model for tame representation type.

For an infinite dimensional algebra, I guess one can use the same definition, and then $k[x]$ would be tame. But you should beware that the finite/tame/wild trichotomy has only been proved for finite dimensional algebras, and it's not clear that one should try to apply the same definitions to infinite dimensional algebras. For example, there are algebras with no finite dimensional representations at all, and then do you want to say that those have finite representation type? It's all a bit unsatisfactory.