Commutative Separable Algebras II
Posted by John Baez
Thanks to everyone who helped me with my questions about commutative separable algebras! I have taken my new-found wisdom, meager as it may be, and placed it here:
- nLab, Separable algebra.
But now I have two more questions — one precise, one more open-ended.
First: I’ve learned a lot about separable algebras over fields, but I need to understand them over commutative rings — especially the integers.
A separable algebra $A$ over the commutative ring $k$ is one such that $A$ is projective as an $A \otimes_k A^{op}$ module. There are also some equivalent characterizations in the nLab article. But I want to know what finite separable $k$-algebras are actually like when $k$ is the integers. Of course an algebra over the integers is usually called a ring, so I might as well talk about “separable rings”.
I know I should read this:
- F. DeMeyer and E. Ingraham, Separable Algebras over Commutative Rings, Lecture Notes in Mathematics 181, Springer, Berlin, 1971.
… but I’m having trouble getting ahold of it right now. And I feel like chatting with you. So, I’ll just throw out this question:
Question: is a finite separable commutative ring the same as a finite product of finite fields?
Second, Lieven le Bruyn said that “in categorical terms, studying the monoidal category of commutative separable $k$-algebras is the same as studying the étale site of $k$”.
I would like to understand this comment better. I know I should read this:
- Milne, Étale Cohomology, Sections 1.3 and 2.1 (esp. Thm. 1.9), Princeton U. Press, Princeton, 1980.
… but again I’m having trouble getting ahold of it, and I think it might be more to just chat a bit about these ideas.
Here’s my vague guess. When we’re working over a commutative ring $k$, a commutative separable $k$-algebra acts like ‘the algebra of functions on a finite 0-dimensional space’. For example, when $k$ is the complex numbers, any commutative separable $k$-algebra is just the algebra of complex functions on a finite set.
In topology, when we have a finite covering space of a space, the fibers are finite sets. So, when we go over to algebraic geometry, commutative separable $k$-algebras play the role of ‘fibers of covering spaces’. But instead of saying ‘covering space’, we say ‘étale map of schemes’.
So, there’s a relation between the étale topology and commutative separable $k$-algebras. And I’m hoping that in the case where $k$ is the integers, these algebras are just finite products of finite fields. After all, a finite field acts like a ‘point’, so a finite product of finite fields should act like a ‘finite 0-dimensional space’.
Re: Commutative Separable Algebras II
John wrote:
Chris Schommer-Pries gave a hint regarding this question, as follows:
So, every finite product of finite fields is indeed a commutative separable algebra over $\mathbb{Z}$, i.e. a finite commutative separable ring.
How about the converse? The answer to my question about will be yes if we can show this:
Suppose $A$ is a finite commutative ring such that for every prime $p$, $A/p$ is a finite product of finite fields of characteristic $p$. Then $A$ is a finite product of finite fields.
(Why? Well, combine what Chris said with this: a finite commutative semi-simple $\mathbb{Z}/p$-algebra is the same as a finite product of finite fields of characteristic $p$.)
Can anyone show this?