Algebra projective over subalgebra Let $A$ be an algebra, $S \subset A$ a subalgebra.*
I was wondering:

*

*Is $A$ projective as an $S$-module?

I have a feeling that this cannot be true in this generality (the feeling comes from the statement being false on the level of rings), hence I should actually be asking

*

*When is $A$ projective as an $S$-module?


*When is it not aka what are some counterexamples?
It's true for group algebras because of Lagrange's theorem.
More generally, the Nichols-Zoeller theorem shows that it's true for Hopf algebras and Hopf subalgebras.

*EDIT: I should add "over some ring", and the answers can/should also involve conditions on the ring. Also, in my examples I'm really thinking "finite-dimensional and over a field" just to be safe.
 A: No it is not true. I don't think one can get a satisfactory answer to your question, because it is too general.
For example, if $A$ is a ring with prime characteristic $p$, it contains $\mathbb{F}_p$, and your question  translate as:let $S$ be a subring. Is $A$ a projective $S$-module ?
I would be surprised if there was a reasonable answer to that.
However, one can answer (more or less) if you pick $S$ wisely:
Let $A$ be a finitely generated (commutative associative unital) $k$-algebra, where $k$ is a field.
Noether normalization lemma states that there is a polynomial subalgebra $S$ (that is $S\simeq k[X_1,\ldots,X_n]$) such that $A$ is integral over $S$, hence in particular a finitely generated $S$-module.
By Quillen -Suslin theorem (https://en.wikipedia.org/wiki/Quillen%E2%80%93Suslin_theorem), such a module is $S$-projective if and only if it is $S$-free.
This should give you plenty of examples/counterexamples. For an explicit counter-example, you can take  $S=k[x^4,y^4]$ and $A=k[x^4,x^3y,xy^3,y^4]$ (see When an integral extension of integral domains is flat?)
