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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.

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    $\begingroup$ If the left module decomposition of the ring is $KQ=\bigoplus_i P_i$, then the maximal left ideals are of the form $(\bigoplus_{i \neq j} P_i) \oplus {\rm{rad}} (P_j)$. $\endgroup$ Mar 22 '15 at 19:29
  • $\begingroup$ @DagOskarMadsen, Thanks for your comment. I assume your comment addresses the case where $Q$ is finite, connected and acyclic. In this case, I consider an algebra of lower triangular matrices, associated to $KQ$ and studying the left maximal ideals of this algebra is easier for me, and coincides with your description. I was wondering if there is any kind of stronger versions of the description you gave, if we relax one of the conditions. $\endgroup$
    – Kaveh
    Mar 22 '15 at 22:22
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    $\begingroup$ If there are oriented cycles in the quiver, then there are many more simple $KQ$-modules than the ones corresponding to vertices and therefore many more maximal ideals. In this case the Jacobson radical is no longer the arrow ideal. $\endgroup$ Mar 22 '15 at 22:33
  • $\begingroup$ @DagOskarMadsen, The important point you mentioned about the existence of oriented cycles and consequently the change in the number of simple modules is clear to me. The easiest example I know is just one vertex and one loop, for which the Jacobson becomes zero. But, what we can conclude out of this argument is the importance of the condition ($Q$ is acyclic). Cannot we hope to say a bit more? $\endgroup$
    – Kaveh
    Mar 22 '15 at 22:50
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    $\begingroup$ The two-sided ideal $I$ generated by all arrows not belonging to a cycle is a nil ideal. We know that all (left, right or two-sided) nil ideals are contained in the Jacobson radical. Perhaps the Jacobson radical is equal to $I$? $\endgroup$ Mar 24 '15 at 17:18
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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.

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    $\begingroup$ This statement is also proved in a slightly greater generality in [Coelho, Liu: Generalized path algebras. Interactions between ring theory and representations of algebras (Murcia), 53–66, Lecture Notes in Pure and Appl. Math., 210, Dekker, New York, 2000. MR1758401]. $\endgroup$ Apr 2 '15 at 8:56
  • $\begingroup$ @Hugh Thomas, Thanks for the clear and neat proof you wrote for Dag's conjecture. $\endgroup$
    – Kaveh
    Apr 2 '15 at 12:41
  • $\begingroup$ @JulianKuelshammer Thanks a lot for the reference you introduced. $\endgroup$
    – Kaveh
    Apr 2 '15 at 13:01
  • $\begingroup$ By the way the book Introduction to the Representation Theory of Algebras by Barot has got this wrong. In section 3.7 it is claimed that the Jacobson radical is equal to the arrow ideal, which is clearly not true in general. $\endgroup$ Nov 7 '18 at 11:31

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