Why Jacobson, but not the left (right) maximals individually?

Question

I firstly asked the following question on MathStackExchange a couple of months ago. I did not receive any answers, but a short comment. So, I decided to post it here, hoping to receive answers from experts. It ended in a nice argument, again about the Jacobson Radical, proposed an proved in the following, but the main question remained untouched.

Working with Path Algebras, it does not need sophisticated tools to prove for a finite, connected, acyclic quiver $Q$, the Jacobson Radical of $KQ$ is the arrow ideal.

But, I have never seen any description of the left (right) maximal ideals of the path algebra for a given quiver, even under the assumptions above (finite, connected and acyclic). Expect for some simple examples, which repeatedly appear in the literature and talks, I am inclined that textbooks and notes intentionally skip this classification, may be due to complexity.

It is puzzling to me why this question is not even addressed! In the aforementioned setting, intersection of a certain class of ideals (left maximals) of $KQ$ is the arrow ideal. What about an explicit description of each element of this class, in the sense of the description we have for simples, indecomposable projectives and injectives? i.e., could one classify all the maximal (right) ideals of such a path algebra, in the above or a bit more general setting? Any reference which might address this question would be highly appreciated.

Answer 1

Dag has already answered the case where the quiver is finite and acyclic, and given a conjecture in the case that cycles are allowed. I will prove his conjecture.

Suppose we have an element $x$ of the Jacobson radical. We want to show that it is generated by arrows not lying on any cycle. We can therefore throw away term in $x$ which includes such an arrow, leaving some $x'$. Suppose that $x'$ is non-zero. We want to show that $x'$ is, in fact, not in the Jacobson radical.

Choose some term in $x'$, which we can identify as a path $p$ in $Q$. Choose it so that $p$ is maximal length among paths appearing in $x'$. Extend $p$ to a cycle $c$. (This is possible because every arrow of $p$ lies on a cycle, so if $p$ is $a_1\dots a_r$, for each $a_i$ there exists a path $f_i$ which completes $a_i$ to a cycle, and then $a_1\dots a_rf_r \dots f_1$ is a cycle.) Note that this cycle may pass more than once through some vertices. Denote this cycle by $b_1\dots b_m$.

Define a representation of dimension $m$, where we define the vector space at vertex $v$ to be the sum of one-dimensional vector space $V_i$ for $i$ such that $b_i$ starts at vertex $v$. Then, we define $b_i$ to be the identity map from $V_i$ to $V_{i+1}$ and the zero map on all other $V_k$. (We let $V_{m+1}$ stand for $V_1$.)

This representation has no subobjects (any non-zero element of the representation generates the whole thing), so it is simple. $p_i$ acts non-trivially on it, so $x'$ acts non-trivially on it, because no shorter path than $p_i$ could act like $p_i$ to cancel it out, and $x'$ contains no longer paths.

Since this representation is simple, it is isomorphic to the algebra modulo a maximal ideal, and since $x'$ acts non-trivially on the representation, $x'$ is not in the maximal ideal. This contradicts our assumption that $x'$ was in the Jacobson radical.