# 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

1. When is $$A$$ projective as an $$S$$-module?

2. 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.

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?)