It is known that a finite dimensional basic algebra over an algebraically closed field is isomorphic to the path algebra of a finite quiver modulo an admissible ideal.

Question 1: Is the same true for algebras over $\mathbb{Q}$? If not, are there suitable further assumptions that would guarantee this?

Question 2: Are there some general useful tools to compute the dimension of the algebra, given the quiver and the ideal? (In concrete, easy cases it might be doable by hand, but I am looking for methods that are more broadly applicable)


1 Answer 1


Question 1: No, take any finite field extension of $\mathbb{Q}$. It is basic but has a simple module that is not 1-dimensional and thus it is not of the form $KQ/I$ for $I$ admissible (since all simple modules are 1-dimensional for algebras of the form $KQ/I$).

For any field $K$, a basic algebra is isomorphic to a quiver algebra with admissible ideal if and only if the algebra $A$ is split (also called elementary sometimes), meaning that $A/J$ is isomorphic to matrix algebras over $K$. You can find this result in most representation theory books (of finite dimensional algebras) such as in the book of Auslander,Reiten and Smalo or the book by Kirichenko and Drozd.

Question 2: The best (computer) tool is the GAP-package qpa: https://folk.ntnu.no/oyvinso/QPA/ . Given a quiver and admissible ideal, you can calculate many information for the algebra, including the dimension.

  • $\begingroup$ Thanks a lot for your answer. With the sentence "$A/J$ is isomorphic to matrix algebras over $K$" do you mean that the quotient of $A$ by its Jacobson radical is isomorphic to some subalgebra of $M^{n \times n}(K)$ for some $n$? $\endgroup$
    – 57Jimmy
    Mar 3, 2020 at 13:37
  • 3
    $\begingroup$ He meant "a direct product of matrix algebras", i.e. the skew fields $D_i$ in the Wedderburn decomposition $A/J \cong \prod_{i=1}^n D_i^{k_i\times k_i}$ are all equal to $K$ itself. This is always the case for algebraically closed $K$ (because the $D_i$ are of finite dimension over $K$) but not in general. $\endgroup$ Mar 3, 2020 at 14:07

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