THE QUIVER OF A FINITE-DIMENSIONAL ALGEBRA
SIMON CRAWFORD
Throughout these notes, we assume that A is a finite-dimensional algebra over an algebraically
closed field k. We first recall some definitions and results of Matt’s talk.
Definition 1. Assume that {e1 , . . . , en } is a complete set of primitive orthogonal idempotents
for A. Then A is said to be basic if ei A ej A for all i 6= j. Equivalently, A is basic if and only
if 1
A/ rad(A) ∼
× ··· × k
=k
| {z }
n copies
Since we are interested in the representation theory of finite-dimensional k-algebras (that is,
their module categories) it suffices to consider k-algebras up to Morita equivalence. It turns out
that every finite-dimensional k-algebra to be Morita equivalent to one which is basic.
Definition 2. Assume that {e1 , . . . , en } is a complete set of primitive orthogonal idempotents
for A. A basic algebra associated to A is an algebra of the form Ab = eA AeA , where eA =
ej1 + · · · + ejr and ej1 , . . . , ejr are chosen so that the ejk A are pairwise non-isomorphic, and each
ei A is isomorphic to one of ej1 A, . . . , ejr A.
We have then have the following theorem:
Theorem 3. Let Ab be a basic algebra associated to A, and use the notation of Definition 2.
(1) The k-algebra Ab does not depend on the choices of a complete set of primitive orthogonal
idempotents {e1 , . . . , en } or the subset {ej1 , . . . , ejr }, up to k-algebra isomorphism.
(2) Ab is a basic k-algebra.
(3) Ab is Morita equivalent to A.
Proof. See [ASS06, I.6.5 and I.6.10]. For (2), one checks that the functor − ⊗Ab eA A : mod-Ab →
mod-A is fully faithful and essentially surjective.
Before defining the quiver associated to a finite-dimensional algebra, we need to introduce the
notion of an admissible ideal of the path algebra of a quiver.
Definition 4. Let Q be a finite quiver and let RQ be the two-sided ideal of kQ generated by
the arrows in Q (RQ is called the arrow ideal of kQ). A two-sided ideal I of Q is said to be
admissible if
m
2
RQ
⊆ I ⊆ RQ
for some m > 2. If I is admissible, then kQ/I is called a bound quiver algebra.
In words, a two-sided ideal of kQ is admissible if it doesn’t contain any arrows of Q, and it
contains all paths of length > m.
Recall that the path algebra of a finite quiver Q is finite-dimensional if and only if Q is acyclic.
Since we are interested in finite-dimensional algebras, the following two results are useful.
1This equivalence crucially depends on k being algebraically closed.
1
THE QUIVER OF A FINITE-DIMENSIONAL ALGEBRA
2
Lemma 5. If Q is a finite quiver and I is an admissible ideal of kQ, then kQ/I is finitedimensional and basic.
Proof. [ASS06, II.2.6 and II.2.10].
Lemma 6. Let Q be a finite quiver and I an admissible ideal of kQ. Then there exists a finite
set of relations {r1 , . . . , rn } such that I = hr1 , . . . , rn i.
Proof. [ASS06, II.2.9].
We are now in a position to define the quiver associated to a finite-dimensional algebra. As
noted earlier, because we care about the representation theory of a finite dimensional k-algebra,
we may assume that A is basic.
Definition 7. Let A be a basic finite-dimensional k-algebra and {e1 , . . . , en } a complete set of
primitive orthogonal idempotents of A. The (ordinary) quiver associated to A, denoted by QA ,
has vertices labelled by the idempotents e1 , . . . , en , and dimk ei (rad A/ rad2 A)ej arrows between
the vertices labelled by ei and ej .
Remarks.
(1) We note that radn A = (rad A)n for all n ∈ N.
(2) Since A is finite dimensional, so are the ei (rad A/ rad2 A)ej , so QA is a finite quiver.
(3) In [Ben98, §4.1], the quiver QA is equivalently defined to have vertices labelled by the simple
A-modules Si , and having dimk Ext1A (Si , Sj ) arrows between the vertices labelled by Si and Sj .
Example 8.
(1) Let A = k[x]/hxm i, where m > 1. Direct calculation shows that the only nonzero idempotent
of A is 1, and this also implies that A is basic. I claim that rad A is the ideal I = hxi. By [ASS06,
I.1.4], it suffices to show that I is nilpotent and that A/I is isomorphic to a product of copies of
k, both of which are clear. Thus rad2 A = hx2 i. It follows that dimk rad A/ rad2 A = 1, so QA is
the quiver
1
α
(
)
k k[x]/hx2 i
1 0
0 0
(2) Let A =
. One checks that e1 =
, e2 =
is a complete
0 k[x]/hx2 i
0 0
0 1
set of primitive orthogonal idempotents. The corresponding indecomposable projective modules correspond to the rows of A, and these modules
have different
dimensions so are not
0 k[x]/hx2 i
isomorphic. Consider the two-sided ideal I =
. Then I is nilpotent and
0 xk[x]/hx2 i
k 0 ∼
0 xk[x]/hx2 i
2
∼
A/ rad A =
,
= k × k, so that rad A = I. Also, we have rad A =
0 k
0
0
so dimk rad A/ rad2 A = 2, and so QA has two arrows. Direct calculation gives
1 if (i, j) = (1, 2) or (i, j) = (2, 2)
dimk ei (rad A/ rad2 A)ej =
.
0 otherwise
Therefore, QA is the quiver
1
α
2
β
At this stage, one should ask whether QA depends on our choice of a complete set of primitive
orthogonal idempotents. Reassuringly, it doesn’t.
THE QUIVER OF A FINITE-DIMENSIONAL ALGEBRA
3
Lemma 9. Let A be a finite-dimensional basic k-algebra.
(1) The quiver QA does not depend on the choice of a complete set of primitive orthogonal
idempotents in A.
(2) If A is also connected (meaning that its only central idempotents are 0 and 1), then QA is
connected.
Proof. [ASS06, II.3.2 and II.3.4].
The next result provides some evidence that this definition is a sensible one:
Lemma 10. Let Q be a finite connected quiver, I an admissible ideal of kQ, and A = kQ/I.
Then QA = Q.
Proof. [ASS06, II.3.6].
Before we are able to prove the main result of this section, we must state two more lemmas.
In what follows, we write εi for the lazy path at vertex i.
Lemma 11 (Universal property). Let Q be a finite connected quiver and A a k-algebra. Then,
if φ0 : Q0 → A and φ1 : Q1 → A are maps satisfying
P
2
(i)
i∈Q0 φ0 (i) = 1, φ0 (i) = φ0 (i), and φ0 (i)φ0 (j) = 0 whenever i 6= j; and
(ii) if α : i → j then φ1 (α) = φ0 (i)φ1 (α)φ0 (j),
then there exists a unique k-algebra homomorphism φ : kQ → A with φ(εi ) = φ0 (i) for all i ∈ Q0
and φ(α) = φ1 (α) for all α ∈ Q1 .
Proof. [ASS06, II.1.8].
Lemma 12. For each arrow α : i → j in (QA )0 , let xα ∈ rad A be such that {xα + rad2 A | α :
i → j} is a basis of ei (rad A/ rad2 A)ej . Then for any two vertices a, b ∈ (QA )0 , any element
x ∈ ea (rad A)eb can be written as
X
x=
λα1 ...α` xα1 xα2 . . . xα`
where λα1 ...α` ∈ k and the sum runs over all paths xα1 xα2 . . . xα` in QA from a to b.
Proof. [ASS06, II.3.3].
We are finally able to prove that any basic and connected finite-dimensional k-algebra over
an algebraically closed field is isomorphic to a bound quiver algebra.
Theorem 13. Let A be a basic and connected finite-dimensional k-algebra. Then there exists
an admissible ideal I of kQA such that A ∼
= kQA /I.
Proof. For each arrow α : i → j in (QA )0 , let xα ∈ rad A be such that {xα + rad2 A | α : i → j}
is a basis of ei (rad A/ rad2 A)ej . Define maps
φ0 : (QA )0 → A, φ0 (i) = ei , i = 1, . . . , n
φ1 : (QA )1 → A, φ1 (α) = xα .
By the universal property of path algebras (Lemma 11) there exists a (unique) k-algebra homomorphism φ : kQA → A extending φ0 and φ1 .
I claim that φ is surjective. First note that we have a k-vector space isomorphism A ∼
=
A/ rad A ⊕ rad A. Now, since k is algebraically closed, the first factor is isomorphic to n copies
of k, and φ maps the each of the n vertices of QA to each of these factors, so it remains to show
that φ surjects onto rad A. But this follows from Lemma 12.
It remains to show that I := ker φ is admissible. Let R be the arrow ideal of kQA . Since the
xα lie in rad A, φ(R) ⊆ rad(A). But rad A is nilpotent, so radm A = 0 for some m, and then
THE QUIVER OF A FINITE-DIMENSIONAL ALGEBRA
4
φ(Rm ) ⊆ radm A = 0; that is, Rm ⊆ ker φ = I. On the other hand, we also need to show that
I ⊆ R2 . Given any x ∈ kQ, we can always write
X
X
x=
λi εi +
µα xα + y,
i∈(QA )0
α∈(QA )1
2
where λi , µα ∈ k and y ∈ R . If we also assume that x ∈ ker φ, then applying φ and rearranging
yields
X
X
λi ei = −
µα xα − φ(y) ∈ rad A.
i∈(QA )0
α∈(QA )1
But rad A is nilpotent and the ei are orthogonal idempotents, and so λi = 0 for all i ∈ (QA )0 .
We then have
X
µα xα = −φ(y) ∈ rad2 A.
α∈(QA )1
Hence we have
X
µα (xα + rad2 A) = 0
α∈(QA )1
2
2
in rad A/ rad A. But {xα + rad A | α ∈ (QA )1 } is a basis of rad A/ rad2 A. This forces µα = 0
for all α ∈ (QA )1 , and so x = y ∈ R2 .
Example 14. We return to the examples of Example 8.
(1) We have A = k[x]/hxm i and QA is
1
α
.
Noting that x = x + rad2 A is a basis of rad A/ rad2 A, the map φ of Theorem 13 sends ε1 7→ 1
m
and α 7→ x. Clearly,
φ surjects2 (as
the theorem promises) and the ker φ = hα i.
k k[x]/hx i
so that QA is the quiver
(2) Now let A =
0 k[x]/hx2 i
1
α
2
0 1
0 0
+ rad2 A and
+ rad2 A
0 0
0 x
e1 (rad A/ rad2 A)e2 , it follows that φ sends
0
ε1 7→ e1 , ε2 7→ e2 , α 7→
0
2
∼
and I = ker φ = hβ i, so that A = kQ/I.
Noting that
β
.
span, respectively, e1 (rad A/ rad2 A)e2 and
1
0
, β 7→
0
0
0
,
x
References
[ASS06] Ibrahim Assem, Andrzej Skowronski, and Daniel Simson, Elements of the representation theory of associative algebras: Volume 1: Techniques of representation theory, vol. 65, Cambridge University Press,
2006.
[Ben98] David J Benson, Representations and cohomology: Volume 2, cohomology of groups and modules, vol. 2,
Cambridge university press, 1998.