Is a non-degenerate finite-dimensional algebra unital?

Let $$A$$ be a finite-dimensional (not necessarily unital) associative algebra over the field of complex numbers $$\mathbb{C}$$ (but I'm also interested in more general fields). Assume the multiplication on $$A$$ is non-degenerate, which means that $$A= AA$$ and if $$a \in A$$ satisfies $$aA = 0$$ or $$Aa = 0$$, then $$a=0$$. Is it true that $$A$$ is unital? If not, what is a counterexample?

Some easy observations:

• If $$A$$ is also simple, then it can be shown that the answer is positive. This follows for instance by the argument here.

• If $$A$$ is a $$C^*$$-algebra (which is automatically non-degenerate), then a finite-dimensional $$C^*$$-algebra is automatically unital.

• There are examples I believe coming from finite semigroups. I'll add an example when I get a chance. I just need to make sure $aA\neq 0$ for the example I have in mind. Oct 3, 2021 at 16:49
• @BenjaminSteinberg I'm looking forward to see your counterexample :)
– user160032
Oct 3, 2021 at 17:19

$$A$$ has basis $$\{e,a,b,c\}$$, with all products of basis elements zero except for $$e^2=e,\quad ab=c,\quad ea=a,\quad ec=c,\quad be=b,\quad ce=c.$$
(This is a codimension one ideal in the path algebra of the quiver with two vertices, two arrows $$a$$ and $$b$$ in opposite directions between the two vertices, modulo the relation $$ba=0$$.)