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This is a phrase Ringel uses a few times in his writing, and I'm not sure exactly what he means by it. The context is that we have a quiver $Q$ with path algebra $\mathbf{k}Q$. If $Q$ is not a finite quiver, then $\mathbf{k}Q$ will not have a unit, but it does have sufficiently many primitive idempotents. I'm not sure what constitutes "sufficiently many". In other writings he refers to a general $\mathbf{k}$-algebra having sufficiently many primitive idempotents, so maybe this just means the algebra is sufficiently like the path algebra of a quiver? I'm only familiar with this sort of a phrasing of a category having sufficiently many projectives/injectives.

The exact paper I'm looking at is Butler and Ringel's Auslander-Reiten Sequences with Few Middle Terms and Applications to String Algebras on page 157.

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    $\begingroup$ I don't know about the "primitive" part, but sometimes (in representation theory of p-adic groups, for example), sufficiently many idempotents means that for every finite collection of idempotents $e_1 , \ldots , e_n$ there exists an idempotent $e$ such that $e e_i = e_i e = e_i$ for all $1 \leq i \leq n$ (a typical thing to imagine is functions with compact support on a space w.r.t. pointwise multiplication - then characteristic functions of compacts are idempotents, and you can take unions of finite collections of compacts...) $\endgroup$
    – Sasha
    Feb 27, 2020 at 15:16
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    $\begingroup$ To fill out @Sasha's remark, probably it should also be required that, for every element $r$ of the ring, there is an idempotent $e$ such that $er=re=r$. (I do not offhand know what "primitive" should mean here.) $\endgroup$
    – paul garrett
    Feb 27, 2020 at 17:28
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    $\begingroup$ For non unital rings this means $R\cong \bigoplus_{i\in I}Re_i$ with the $e_i$ orthogonal primitive idempotents. The index set can be infinite. This is clearly the case for the path algebra of a quiver. $\endgroup$
    – Benjamin Steinberg
    Feb 27, 2020 at 18:12
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    $\begingroup$ Let me add that having the direct sum decomposition as a left module is equivalent to having it as a right module. And it implies Sasha and Paul's conditions which are much weaker. $\endgroup$
    – Benjamin Steinberg
    Feb 28, 2020 at 2:56
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    $\begingroup$ I looked back at the references on rings with enough idempotents and I had misremembered things. One should assume that the ring decomposes as a direct sum of the e_iR as a right module too. You don't get it for free. I got confused because a line in a paper about Leavitt path algebras seemed to imply that. $\endgroup$
    – Benjamin Steinberg
    Mar 22, 2020 at 13:55

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