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?




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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](https://mathoverflow.net/q/181839).
* If $A$ is a $C^\*$-algebra (which is automatically non-degenerate), then a finite-dimensional $C^\*$-algebra is automatically unital.



Is a non-degenerate finite-dimensional algebra unital?
There's a four-dimensional counterexample over any field.


$A$ 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 $ba=0$.)


