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.

  • $\begingroup$ 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. $\endgroup$ Oct 3, 2021 at 16:49
  • $\begingroup$ @BenjaminSteinberg I'm looking forward to see your counterexample :) $\endgroup$
    – user160032
    Oct 3, 2021 at 17:19

1 Answer 1


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 $ba=0$.)

  • $\begingroup$ Thanks for your answer. Everything checks out nicely. $\endgroup$
    – user160032
    Oct 3, 2021 at 20:49

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