Let $Q = (Q_0,Q_1)$ be a finite quiver and let $kQ$ be its path algebra, that is the algebra of formal linear combinations of paths on $Q$ with multiplication given by concatenation of paths. I think the definition is standard, but for the details I refer to Chapter 2 in Assem-Simson-Skowronski and to wikipedia for a quick reference.
Suppose $Q$ is acyclic, so that $kQ$ is finite dimensional.
I know that $kQ_+$ (that is the two sided ideal generated by all paths of length at least $1$) is the radical of $kQ$.
I wonder if all two-sided maximal ideals of $kQ$ are as follows. For every vertex $v$, define $M_v = kQ_+ + (1-e_v)$, where $e_v$ is the stationary path at vertex $v$. One can show that $M_v = kQ_+ + (e_w : w \neq v)$: indeed, by multiplying $(1-e_v)$ by $e_w$ one gets $e_w$ (or $0$ if $v = w$).
These ideals $M_v$ are maximal (and two-sided), right?
I claim that all maximal ideals are of this form.
Indeed, suppose $M$ is a maximal ideal which is not of the form $M_v$. Then $M$ must contain a linear combination involving all stationary paths, say $\eta = \sum \lambda_v e_v$ with all $\lambda_v \neq 0$. But then $\eta e_w = \lambda_w e_w$. This shows $e_w \in M$ for every $w$. But then $1 = \sum e_w \in M$, which is a contradiction.
Does this all make sense?
Since I could not find a statement like this, and it seems very simple, I am afraid I am doing something wrong.
One more delicate question: the $M_v$'s are two-sided ideals. But it seems to me that they are maximal also among all left ideals and among all right ideals. Is this right?
And it should also be true that they are all the maximal left ideals (and all the maximal right ideals) as well.
Thank you
Yeah, everything you've said makes sense. Each $M_v$ is maximal both as a left ideal and as a right ideal; you're argument didn't depend of left-ness or right-ness.