Let $A$ be an associative $k$-algebra with unit.
Definition. A set $B \subseteq A$ is called a multiplicative basis of $A$ if $B$ is a $k$-basis of $A$ and if $B \cup \{0\}$ is closed under multiplication, meaing $b \cdot b' \in B \cup \{0\}$ for all $b,b' \in B \cup \{0\}$.
Definition. For a multiplicative basis $B$ of $A$, we call a two-sided ideal $I \subseteq A$ compatible if it is generated by elements of the form $b - b'$ for $b,b' \in B \cup \{0\}$.
Let us fix a multiplicative basis $B$ for $A$.
I would like to show the following.
Proposition. Let $I \subseteq A$ be a compatible (two-sided) ideal.
Then $\overline B \setminus \{ 0\} \subseteq A/I$ (the set of residue classes of $B$ without $0$) is a multiplicative basis for $A / I$.
The only thing missing for me to show this is the linear independence. I somehow struggle to prove this, even though it seems like it should be easy to prove (although maybe tedious).
I strongly suspect this is true, because a lot of other things would make sense then. Intuitively, it makes sense because modding out a compatible ideal basically just means that I can swap out some $b$ for a $b'$, so things won't change much, because all we are doing is identifying certain basis elements with each other.