# Finite dimensional algebras over $\mathbb{Q}$

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)

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.
• 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 $n$? Mar 3, 2020 at 13:37
• 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. Mar 3, 2020 at 14:07