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Rings are not necessarily commutative, but associate and unital here. Recall that representation-infinite means that there are infinite non-isomorphic indecomposable modules. For a natural number $m$ define

$r_m:= \inf \{ n \geq 1 | $ there exists a representation-infinite finite connected ring with $n$ elements and $m$ simple modules $\}$ .

Question: What is the sequence $r_m$?

I think we have $r_1=8$, which means that the smallest representation-infinite finite connected ring has 8 elements. It should be $K[x,y]/(x^2,y^2,xy)$, when $K$ is the field with 2 elements.

Question: Is there an elementary argument for $r_1=8$?

Mare
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1 Answers1

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Let's first address your second question: how can we prove that $r_1=8$? I suspect that the classification of small finite rings is the easiest way to do so. According to OEIS sequence A037291, the number of rings witn $n$ elements (for $n=1,2,\ldots$) is $$ 1, 1, 1, 4, 1, 1, 1, 11, \ldots $$ (Comparing with A127707, we see that almost all of these rings are commutative.) This means that, for $n=1,2,3,5,6,7$, the only ring with $n$ elements is $\mathbb{Z}/n\mathbb{Z}$, and these all have finite represenation type. Moreover, the four rings with $4$ elements are listed in this answer: $$ \mathbb{Z}/4\mathbb{Z}, ~\mathbb{F}_2 \times \mathbb{F}_2, ~\mathbb{F}_4, ~\mathbb{F}_2[x]/(x^2). $$ None of these is of infinite representation type. Thus the smallest ring of infinite representation type has $8$ elements. Your example shows that $r_1=8$.

Your first question looks much more difficult, especially because of the "connected" requirement. Here are some upper bounds for $r_m$, with an example to support it:

$r_2 \leq 16$: path algebra of the Kronecker quiver $1\Rightarrow 2$ over $\mathbb{F}_2$.

$r_3 \leq 2^6$: path algebra of a quiver of affine type $A_2$ over $\mathbb{F}_2$.

$r_4 \leq 2^8$: path algebra of a quiver of affine type $A_3$ over $\mathbb{F}_2$.

$r_m \leq 2^{2m-1}$ for $m\geq 5$: path algebra of a star quiver with $m$ vertices over $\mathbb{F}_2$.

Juliana
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