## July 16, 2010

### Commutative Separable Algebras

#### Posted by John Baez

Now that I’m in Singapore, I’ll probably be thinking more about technology and the environment, and less about pure math. But I still want to keep working with Jim Dolan. It’s just too fun to quit.

For the last few years, we’ve been trying to learn algebraic geometry and number theory and recast certain portions of these subjects in terms more friendly to ($n$-)category theorists. Not by adding fancy new layers on top of the existing work: rather, by making the basics even simpler.

This project is slowly picking up speed. But I don’t really want to talk about it now. It’s not ready for prime time yet. I mention it mainly to explain why I’m asking an elementary — I hope! — question about commutative separable algebras.

A separable algebra $A$ over some field is an (associative unital) algebra that’s not only semisimple, but remains semisimple as we extend the field to any larger field.

More precisely, an algebra $A$ over a field $k$ is separable iff for every extension $K$ of $k$, the algebra $A \otimes_{k} K$ is semisimple. And of course a finite-dimensional algebra — I’m only interested in finite-dimensional things, today — is semisimple iff it’s a product of simple algebras, meaning algebras that have no nontrivial ideals.

Q: What are some easy examples of commutative semisimple algebras that are not separable?

They seem a bit hard to come by, since I hear that if $k$ is a perfect field, all semisimple algebras over $k$ are separable. A field $k$ is perfect if every extension of $k$ is separable, regarded as an algebra over $F$. But more importantly, for me, perfect fields include most of the fields I really care about! For example: fields of characteristic zero, finite fields, algebraically closed fields, and fields algebraic over a perfect field.

Am I making some sort of mistake, or is it really true that over finite fields or algebraic number fields, all semisimple algebras are separable? If so, I can stop worrying about the difference — at least for what we’re doing now. But still, I’d like to know the simplest examples where the difference kicks in… especially in the commutative case.

(Apparently there is a theorem of Eilenberg and Nakayama (maybe in here saying that a separable algebra can be given the structure of a symmetric Frobenius algebra. This is just the beginning of a network of interesting relations between separable algebras and Frobenius algebras, some of which are sketched here. But today I am more interested in how separable algebras show up in algebraic geometry!)

Posted at July 16, 2010 1:02 AM UTC

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### Re: Commutative Separable Algebras

Take an inseparable field extension $K/k$ of degree $d$ with characteristic $p$. I claim that $K$ as a $k$-algebra is not separable, that is that the base extension $K \otimes_k K$ is not semisimple. Since it’s commutative it’s enough to just demonstrate a nilpotent element. So take $x$ an element of $K$ which is not in $k$, and look at $x\otimes 1- 1\otimes x$, a simple calculation shows that raising this element to the $p^d$ gives zero but the element itself is not zero. (This argument is taken from an article of Kuperberg’s but presumably is well-known.)

So your favorite example of an inseparable field extension from Galois theory is also an example here (the usual one is adjoing a $p$th root of $x$ to $F_p(x)$).

Posted by: Noah Snyder on July 16, 2010 2:27 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

As for your main question (do perfect fields have no inseparable algebras?) by Artin-Wedderburn you can immediately reduce to the case of division algebras over k. It isn’t totally obvious to me how to reduce from division algebras to fields…

But anyway wikipedia says it’s true: having no inseparable algebras is equivalent to perfect.

Posted by: Noah Snyder on July 16, 2010 2:31 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

John,
emphasizing/clarifying what Noah wrote before.

structure theorem for semi-simple k-algebras (aka Weddenburn-Artin) : is a direct sum of matrix algebras over division algebras with centers a finite dimensional field extension of k.

structure theorem for separable k-algebras : is a direct sum of matrix algebras over division algebras with centers a finite dimensional SEPARABLE field extension of k.

so indeed, if k is perfect semisimple=separable (even noncommutative).

the best place to learn about this stuff is still the old book by Frank Demeyer and Edward Ingraham ‘Separable algebras over commutative rings’ Springer Lecture Notes in Mathematics 181 (1971). As the title suggests, they are really interested in the theory over rings rather than fields, but that classical case is covered in section II.2.

probably you are interested in the monoidal structure of separable algebras.

in categorical terms, studying the monoidal cat of commutative separable k-algebras is the same as studying the etale site of k.

if you want to extend this to noncommutative separable k-algebras, you’ll have to know something about the Brauer group of fields (they classify the division algebras having center that field). For this also the DeMeyer-Ingraham book may be useful, but there are more classical texts such as Herstein or Pierce.

Posted by: lievenlb on July 16, 2010 7:41 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Lieven wrote:

structure theorem for semi-simple $k$-algebras (aka Weddenburn-Artin): is a direct sum of matrix algebras over division algebras with centers a finite dimensional field extension of $k$.

structure theorem for separable $k$-algebras: is a direct sum of matrix algebras over division algebras with centers a finite dimensional SEPARABLE field extension of $k$.

Thanks! I guess you’re assuming that all algebras here are finite-dimensional? I’m not usually nitpicky, but I am feeling that way today.

In categorical terms, studying the monoidal cat of commutative separable $k$-algebras is the same as studying the etale site of $k$.

Hey! Stop peeking into my brain!

Where’s a nice reference for this?

Also: I guess you’re again assuming these algebras are finite-dimensional?

if you want to extend this to noncommutative separable k-algebras, you’ll have to know something about the Brauer group of fields (they classify the division algebras having center that field).

I like Brauer groups because they arise as $\pi_1$ of some very nice homotopy 3-types, or weak 3-groupoids if you prefer. However, I am only interested in commutative algebras today, for the reason you guessed.

Posted by: John Baez on July 16, 2010 8:40 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Separable algebras over a field k are automatically finite dimensional. DeMeyer-Ingraham Corollary 2.2 p.48.

Etale site : probably not the most readable of all but I’d refer to Milne’s ‘Etale cohomology’ PUP (1980) sections 1.3 and 2.1 (thm 1.9 etc.)

Posted by: lievenlb on July 16, 2010 8:59 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

“A theorem of Eilenberg and Nakayama says an algebra A is separable iff it can be given the structure of a symmetric Frobenius algebra.”

Something is wrong here. Take $A$ to be the commutative ring $k[t]/t^2$, and $\epsilon(a+b t)=b$. This seems to satisfy all of the axioms of a symmetric Frobenius algebra, according to either $n$-Lab or Wikipedia, and it is very far from semi-simple/separable.

Posted by: David Speyer on July 17, 2010 2:26 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras

I think the right statement should be that a finite dimension $k$-algebra $A$ is separable if and only if its trace map makes it into a Frobenius algebra. By “trace”, I mean to embed $A$ into the algebra of all $k$-linear endomorphisms of $A$, by the left action of $A$ on itself, and take trace in that setting.

Noah and I discussed a similar issue at MO.

Posted by: David Speyer on July 17, 2010 2:49 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Nakayama-Eilenberg only says that separable implies symmetric Frobenius, not the other way around.

I’ve given the reference to DeMeyer-Ingraham II.2 before.

Here’s another : Knus-Ojanguren ‘theorie de la descente et algebres d’Azumaya’ Springer Lecture Notes in mathematics 389 (1974). section III.3 Thm 3.1 gives the structure result for sepable algebras over a field as i stated it above.

Posted by: lievenlb on July 17, 2010 4:43 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras

David wrote:

John wrote:

“A theorem of Eilenberg and Nakayama says an algebra A is separable iff it can be given the structure of a symmetric Frobenius algebra.”

Something is wrong here. Take $A$ to be the commutative ring $k[t]/t^2$, and $\epsilon(a+b t)=b$. This seems to satisfy all of the axioms of a symmetric Frobenius algebra, according to either $n$-Lab or Wikipedia, and it is very far from semi-simple/separable.

Whoops, you’re right. I’ll delete that false claim in my blog entry.

I think the right statement should be that a finite dimension $k$-algebra $A$ is separable if and only if its trace map makes it into a Frobenius algebra. By “trace”, I mean to embed $A$ into the algebra of all $k$-linear endomorphisms of $A$, by the left action of $A$ on itself, and take trace in that setting.

I don’t think that’s right. As I mentioned in my comment to Todd, Aguiar says an algebra is ‘strongly separable’ when its trace map makes it into a Frobenius algebra. Being strongly separable implies being separable, but not conversely. He gives a counterexample: the algebra of $n \times n$ matrices over $F_p$ is separable but not strongly separable when $n$ is divisible by $p$. The trace is zero in this case!

We’ll get this stuff ironed out eventually…

Posted by: John Baez on July 18, 2010 5:09 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

John wrote:
“He gives a counterexample: the algebra of n×n matrices over F p is separable but not strongly separable when n is divisible by p. The trace is zero in this case!”

That’s not really a counter example. The trace of the identity element is zero, but the trace map is not the zero map. In fact a careful check shows that in fact the trace map does give the algebra of p x p matrices over Z/p the structure of a symmetric Frobenius algebra. The bilinear form induced by the trace map is non-degenerate and you can write down a compatible comultiplication.

The case of 2 x 2 matrices over Z/2 is an instructive exercise.

Posted by: Chris Schommer-Pries on July 22, 2010 11:23 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Chris wrote:

John wrote:

He gives a counterexample: the algebra of $n \times n$ matrices over $F_p$ is separable but not strongly separable when $n$ is divisible by $p$. The trace is zero in this case!”

Sorry, that last sentence isn’t right, as Chris noticed. But I stand by the rest.

Chris wrote:

That’s not really a counter example. The trace of the identity element is zero, but the trace map is not the zero map. In fact a careful check shows that in fact the trace map does give the algebra of $p \times p$ matrices over $Z/p$ the structure of a symmetric Frobenius algebra. The bilinear form induced by the trace map is non-degenerate and you can write down a compatible comultiplication.

I agree with everything you say here except the first sentence. I still think $p \times p$ matrices over $F_p$ is separable, but not strongly separable.

It’s separable: one can see this using various equivalent characterizations of separable algebras on the nLab. The theorem of Eilenberg and Nakayama states that a separable algebra can be given the structure of a symmetric Frobenius algebra, and you’ve shown how in this case.

It’s not strongly separable: one can see this using various equivalent characterizations here. Most simply, an algebra is strongly separable if and only if it can be made into a special Frobenius algebra, one where the comultiplication followed by the multiplication is the identity:

This forces the Frobenius algebra’s bilinear form to be

$tr(L_a L_b)$

where $L_a$ means ‘left multiplication by $a$’. So, one can show that an algebra is strongly separable if and only if the above bilinear form is nondegenerate. For an $n \times n$ matrix algebra, we have

$tr(L_a L_b) = n tr(a b)$

where $tr$ at right is the usual matrix trace. So for $p \times p$ matrices over a field of characteristic $p$, the bilinear form $tr(L_a L_b)$ is degenerate — in fact zero! So, this matrix algebra isn’t strongly separable.

There’s also a fun way to think about separable and strongly separable algebras using ‘separability idempotents’, as explained in that nLab article.

Posted by: John Baez on July 23, 2010 1:27 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Ahh, Good. I misunderstood exactly what this was supposed to be a counter example of. You’re right that it is not strongly separable. I’ll check out the n-lab page…

Posted by: Chris Schommer-Pries on July 23, 2010 2:33 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

I’m feeling a little lazy to check this out carefully myself, but I seem to recall that another characterization of $A$ being separable is that $A$, regarded as a $A \otimes_k A^{op}$-module via the algebra map $m: A \otimes_k A^{op} \to A$, is projective.

What is the connection between this and the separability condition of a symmetric Frobenius algebra? Is it something like $m \circ \delta$ is a nonzero scalar multiple of the identity on $A$, where $\delta$ is the comultiplication?

Posted by: Todd Trimble on July 17, 2010 5:16 PM | Permalink | Reply to this

### Re: Commutative Separable Algebras

Todd wrote:

I’m feeling a little lazy to check this out carefully myself, but I seem to recall that another characterization of $A$ being separable is that $A$, regarded as a $A \otimes_k A^{op}$-module via the algebra map $m: A \otimes_k A^{op} \to A$, is projective.

I think that’s right, though I haven’t gone through the proof. If your computer is smart enough to read gzipped postscript files, this paper is a pleasant way to learn a bit more these issues:

• Marcelo Aguiar, A note on strongly separable algebras, Boletín de la Academia Nacional de Ciencias (Córdoba, Argentina), special issue in honor of Orlando Villamayor, 65 (2000) 51-60.

Otherwise you can get it here.

For starters, apparently $A$ is projective as a module of

$A^e = A \otimes_k A^{op}$

iff you can split the $A^e$-module map

$m : A^e \to A$

given by

$m( a \otimes b) = a b$

If you can split this map, $A$ is a direct summand of a free module, so it’s projective. Why is the converse true? Aguiar points us to chapter 10 of Pierce’s book Associative Algebras.

What is the connection between this and the separability condition of a symmetric Frobenius algebra? Is it something like $m \circ \delta$ is a nonzero scalar multiple of the identity on $A$, where $\delta$ is the comultiplication?

Rosebrugh, Sabadini and Walters use ‘separable’ for a Frobenius algebra with $m \circ \delta = 1$. But the word ‘separable’ means something quite different for associative algebras, and the Oxford group — for example, Jamie Vicary — has persuaded me that we should instead use the term ‘special’ for a Frobenius algebra with $m \circ \delta = 1$. See the followups to week268, where we had quite a long chat about the terminological morass!

Anyway, here’s what I quickly glean from Aguiar. Suppose we have a separable algebra, and suppose we pick a splitting of the $A^e$-module morphism

$m : A^e \to A$

Then the image of $1 \in A$ under this splitting is an idempotent $p \in A^e$. People call it a separability idempotent.

If we can choose the splitting so that $p$ is symmetric:

$p = \sum_i a_i \otimes b_i = \sum_i b_i \otimes a_i$

then people say $A$ is strongly separable.

You might prefer to think of the splitting as an extra structure, but most algebraists use ‘strong separability’ to mean a property: the existence of a splitting for which $p$ is symmetric. And there’s a reason why…

Every finite-dimensional algebra $A$ comes with a bilinear form

$(a, b) \mapsto tr(L_a L_b)$

where $L_a : A \to A$ means ‘left multiplication by $a$. And, it turns out that an algebra $A$ is strongly separable iff this bilinear form is nondegenerate, in the sense of giving an isomorphism between $A$ and $A^*$ — see Theorem 3.1 in Aguiar’s note.

Furthermore, the above bilinear form is nondegenerate iff $A$ can be made into a special Frobenius algebra. That’s because a Frobenius algebra is special ($m \circ \delta = 1$) iff its counit is given by

$\epsilon(a) = tr(L_a)$

and this counit only works if the above bilinear form is nondegenerate!

In short, an algebra is ‘strongly separable’ if there exists a symmetric separability idempotent, but we can also describe it as the property that

$\epsilon(a) = tr(L_a)$

serves as a counit making $A$ into a special Frobenius algebra.

Posted by: John Baez on July 18, 2010 4:57 AM | Permalink | Reply to this

### Re: Commutative Separable Algebras

John wrote:

For starters, apparently $A$ is projective as a module of

$A^e = A \otimes_k A^{op}$

iff you can split the $A^e$-module map

$m : A^e \to A$

given by

$m( a \otimes b) = a b$

If you can split this map, $A$ is a direct summand of a free module, so it’s projective. Why is the converse true?

Duh, it’s obvious: if $f: Q \to P$ is an epimorphism of modules and $P$ is projective, you can split this map!

Posted by: John Baez on July 19, 2010 4:38 AM | Permalink | Reply to this
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Excerpt: What are some commutative semisimple algebras that aren't separable?
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