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Recently I have been studying the representation type of algebras (tame-wild dichotomy). A question that I feel is probably well known but that I cannot answer is the following:

Can I define tame algebras to be those algebras for which any representation may be written (after choosing a basis) in a form "only zeros and ones, and a free parameter lambda"? In other words, no scalars of the base field apart from lambda.

Assume of course that the base field is algebraically closed.

In all the examples I know (tame quivers, all the tame algebras that I have seen for which a classification exists) this is true.

What if I relax the hypotheses to "defined over F(lambda)", where F is the prime field, and lambda the scalar? Maybe it is true if we allow rational functions in lambda?

If this is not true, I would love to see a counterexample, so that I can rest in peace with this question.

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    $\begingroup$ That is certainly the idea, but it is quite difficult to make this precise. One obvious problem is that $k$ is in bijection with $k^n$ for infinite fields, so you have to be more restrictive what you mean by "a free parameter". The only precise formulations I know of are the one with the finitely many $A$-$k[x]$-bimodules $M_i$ for each dimension and the one using $1$-dimensional constructible subsets of the representation variety, see [de la Pena: Functors preserving tameness, Fund. Math., 1991, Theorem 1.3] $\endgroup$
    – Julian Kuelshammer
    Commented Feb 6, 2018 at 9:45
  • $\begingroup$ I would guess one might make some progress on this problem - perhaps in either direction - using Sergeichuk's normal form based on Belitskii's algorithm: Sergeichuk, "Canonical matrices for linear matrix problems", Lin. Alg. Appl. 2000, doi.org/10.1016/S0024-3795(00)00150-6 $\endgroup$
    – Joshua Grochow
    Commented Jun 2, 2023 at 3:55

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