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?)


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