Let $A$ be an algebra, in the sense that $A$ is a set with some $n$-ary operations defined on it. Let $B$ be a sub-algebra of $A$, in the sense that $B$ is a subset of the set of elements of $A$, closed w.r.t. the $A$'s operations. My question is: when is the quotient algebra $A/B$ trivial, i.e. reduced to the zero element?
My guess is that this happens in exactly two cases: 1) $A=B$ and 2)$A=0$. Is this correct? Have you some comments or precisations or corrections to tell me?
First of all, the notion of sub-algebra that you are giving is not sufficient to define a quotient algebra. For example, subgroups of groups are closed with respect to the operation, but the quotient may not be a group at all.
But let's back up a step and just suppose that you're choosing $B$'s appropriately so that the quotient is indeed an algebra of the same type.
Secondly, your use of "$0$" seems to presume something about your algebras, like maybe $0$ is the neutral element for one of the operations. If you want to make some assumption about that, you should really put it forward, because it may make the general answer easier.
Thirdly, in your 'two cases', the second one implies the first. So it really only seems sensible to ask if $A=B$ iff $A/B$ is trivial. Actually I don't know if there's some weird edge case of an algebra such that this is not true.
I'm pretty sure you need to be more specific about how the congruence that defines $A/B$. For example, in the case of groups, modules, rings, algebras, etc., the underlying congruence is defined as $xy^{-1}\in B$.
For semigroups, you need to get more detailed and ask for an equivalence relation on $S\times S$ which is stable under left and right applications of the binary operation. This case subsumes the above, and it would say that $A/B$ is trivial iff the associated congruence is $A\times A$, because that indicates every element is related to every other element, and there's only one equivalence class.