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Let $A$ be a finite dimensional algebra. A field $K$ is a splitting field for an indecomposable $A$-module $M$ in case the local algebra $End_A(M)/(rad(End_A(M))$ is 1-dimensional. $K$ is called a global splitting field for an algebra $A$ in case every indecomposable $A$-module splits.

Question: Is there a concrete example of a representation-infinite algebra $A$ over a finite field that is a global splitting field for $A$?

Answer by Jeremy Rickard: No.

This motivates the follow up question:

Question: Is a field $k$ algebraically closed if and only if it is the global splitting field of a representation-infinite $k$-algebra $A$?

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Let $k$ be a finite field and $A$ a representation-infinite finite dimensional $k$ algebra. By the second Brauer-Thrall conjecture, $\bar{k}\otimes_kA$ has infinitely many nonisomorphic indecomposable modules of some dimension, and so has some that are not defined over $k$ (i.e., not of the form $\bar{k}\otimes_kM$ for any $A$-module $M$). However, such a module is defined over some finite field extension of $k$: namely, the extension generated by its structure constants.

So we have a finite extension $K$ of $k$, and an indecomposable $K\otimes_kA$-module $N$ that is not of the form $K\otimes_kM$ for any $A$-module $M$.

Let $X$ be any indecomposable direct summand of the restriction $\text{Res}^{K\otimes_kA}_A(N)$.

Since $K\otimes_k\text{Res}^{K\otimes_kA}_A(N)$ is the direct sum of $|K:k|$ Galois conjugates of $N$, $K\otimes_kX$ is the direct sum of some of the Galois conjugates of $N$, and more than one, since $N$ is not defined over $k$.

Hence (using the fact that $K$ is a separable extension of $k$, so that extending scalars to $K$ commutes with taking radicals of algebras) $$\dim_k\text{End}_A(X)/\text{rad}(\text{End}_A(X)) =\dim_K\text{End}_{K\otimes_kA}(K\otimes_kX)/\text{rad}(\text{End}_{K\otimes_kA}(K\otimes_kX))$$ is greater than $1$.

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  • $\begingroup$ Thanks, I asked the question for general fields now (that was a question in brackets before). It seems your proof works for any perfect field (equivalently fields where any finite extension is seperable)? $\endgroup$
    – Mare
    Commented May 8, 2020 at 10:59
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    $\begingroup$ @JeremyRickard What does your first sentence mean when A is specialized to A=k? I am not understanding something. $\endgroup$ Commented May 8, 2020 at 17:14
  • $\begingroup$ @NicholasKuhn The question has a hypothesis "representation-infinite". Sorry, I forgot to carry that into the answer: I'll fix that. The second Brauer-Thrall conjecture, now a theorem, states that, for a representation infinite algebra over an infinite field, there is some $d$ for which there are infinitely many nonisomorphic indecomposable modules of dimension $d$. $\endgroup$ Commented May 8, 2020 at 17:23

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