## July 22, 2010

### 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:

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’.

Posted at July 22, 2010 6:07 AM UTC

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### Re: Commutative Separable Algebras II

John wrote:

Question: is a finite separable commutative ring the same as a finite product of finite fields?

Chris Schommer-Pries gave a hint regarding this question, as follows:

In this book [DeMeyer and Ingraham’s Separable Algebras over Commutative Rings] you’ll find a characterization that an $R$-algebra $A$ is separable iff $A/m$ is separable over $R/m$ for all maximal ideals $m$. So being separable can be “checked at points”. So over $\mathbb{Z}$, $A$ is separable if and only if $A/p$ is a finite semi-simple $\mathbb{Z}/p$-algebra for all primes $p$.

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?

Posted by: John Baez on July 22, 2010 1:52 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras II

I assume what you want follows from the fact that a finite Abelian group is the product of its p-parts.

But then, as to separability, the problem resides in the ‘other’ primes, that is those q such that A has no q-torsion. Then A/qA=0 and do you consider this to be a separable extension of Z/qZ?

I guess ‘most’ people would claim that the only separable extensions of the integers Z are of the form ZxZx…xZ.(there’s a small issue of finite generation here).

This follows from the fact that there is always ramification in rings of integers, or equivalently, that the fundamental group of Spec(Z) is trivial (that old idea that if one would view Spec(Z) as a 3-dml manifold it must be the 3-sphere).

as to my previous comment : all i meant was that over a field commutative separable algebras coincide with etale extensions, and separable algebras are closed under tensor-product just as etale covers are closed under fiber-products and clearly they correspond to each other.

Posted by: lievenlb on July 22, 2010 8:24 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras II

“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.”

The answer is no. Consider the ring Z/4. When we reduce mod an odd prime we get zero, which is certainly a finite sum of finite fields. It is a sum of zero many such fields. At the prime 2 we get (Z/4)/2 = Z/2 is also a finite sum of finite fields.

But Z/4 is not a finite sum of fields.

You can also see that Z/4 is separable directly by looking at Z/4 tensored with itself over Z. We get Z/4 again. This is free over Z/4, so in particular it is projective. Hence Z/4 is a separable ring over Z.

Posted by: Chris Schommer-Pries on July 22, 2010 10:54 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras II

I wasn’t looking hard enough for counterexamples.

Thanks!

Posted by: John Baez on July 23, 2010 1:04 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras II

Hi John,

I hope you’re having fun in Singapore! I just happened onto your post while searching for something on separable algebras. I can’t write too long, but I thought I’d point out one fact that you’ll like, if you haven’t seen it before. Possibly with some reasonable assumptions, like $A$ being a finitely-generated $k$-algebra, $A$ is separable if and only if

$\Omega_{A/k}=0.$

In particular, this is true if $A$ is a quotient ring of $k$. You see, the separability corresponds to not having relative tangent vectors. But if $A$ is a quotient, that is like the ring of functions on a subspace of $Spec(k)$. So, of course, there will be no relative tangent vectors.

Etale, by the way, is stronger than separability. It further requires that the algebra be flat. It sounds like you’re interested in a situation where $A$ is finitely generated as a $k$-module. In this case, flat is the same as projective, or locally-free. This is why etale algebras really are like covering spaces: evenly spread out over our space with no funny infinitesimal variation. Separability by itself allows the algebra to just live over a subspace.

In case of $Z$, the only finitely-generated (as a module) etale $Z$-algebras are direct sums of $Z$. This is an elementary, but somewhat hard-core fact from algebraic number theory.

Posted by: Minhyong Kim on July 27, 2010 1:48 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras II

Hi, Minhyong! Singapore is great.

Thanks for the nice story. If you bump into this: I have another question that’s up your alley, here. James Borger gave a nice answer, but I’m still wondering where people are supposed to learn the basics of Hasse–Weil zeta functions…

Posted by: John Baez on July 29, 2010 12:43 PM | Permalink | Reply to this

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