By an algebra $A$ over some field $F$ I mean a finite dimensional vector space over $F$ with an $F$-bilinear multiplication. That $A$ has a unit with respect to its multiplication is not assumed. A subalgebra $B$ is called integral if there exists a complement $C$ such that $A = B + C$ and $BC = CB = 0$. Now suppose $$ A = B + C $$ with $BC = CB = 0$ and $B \cap C = 0$, where $B$ has unit $e_1$ and $C$ has no unit and no integral subalgebra which contains a unit. Then I have a question on an argument that $C$ is uniquely determined.
If $A$ has a unit $e$, then as $e_1 \in B$ we have $(e - e_1)C(e - e_1) = C$ and $(e-e1)A(e-e_1) = (e - e_1)C(e - e_1)$ so that if $A = B + C'$ like above we get $$ C' = (e - e_1)C'(e - e_1) = (e - e_1)A(e - e_1) = (e - e_1)C(e - e_1) = C $$ and $C$ is unique. If $A$ does not contain a unit, we can adjoint one, call it also $e$, built the span $C'$ of $C$ and $e - e_1$ and find like above that $C'$ is unique. Now the argument goes on and claims that this gives uniqueness of $C$. But I do not see that, clearly if we just have a vector space, for two subspaces if we adjoin a single vector and built the span, then equality of those does not yield equality of the original subspaces. So I must have overlooked something?
The argument is taken from J.H. Wedderburn, On hyperomplex numbers, the famous paper in which the Wedderburn structure theorem for finite-dimensional algebras is proven. It is in the proof of Theorem 10 on page 86. What I call a unit he calls a modulus.
The point is that $C \subset A$ and $(e - e_1) \notin A$ (and $A$ is a subspace of $A$-with-a-unit-adjoined):
Let $V$ be a vector space, and suppose $W \subset V$ (a subspace) and $v \in V \setminus W$ are such that $W + v = V$. Suppose $C, C' \subset W$ are such that $C + v = C' + v$. Then $C = (C + v) \cap W = (C' + v) \cap W = C'$.