### Grothendieck–Galois–Brauer Theory (Part 3)

#### Posted by John Baez

Last time we saw that Galois theory is secretly all about covering spaces. Among other things, I told you that a field $k$ gives a funny kind of space called $\mathrm{Spec}(k)$, and a separable commutative algebra over $k$ gives a covering space of $\mathrm{Spec}(k)$. I gave you a lot of definitions and stated a few big theorems. But I didn’t prove anything, and some big issues were left untouched.

One of these is: *why* is a separable commutative algebra over $k$ like a covering space of $\mathrm{Spec}(k)$? Today I’ll talk about that, and actually prove a few things.

I’ll start with some hand-wavy geometry. For any field $k$, the space $\mathrm{Spec}(k)$ is a lot like a point. So, its covering spaces should be zero-dimensional in some sense. That’s what the concept of ‘separable’ commutative algebra tries to formalize. Separability is a kind of zero-dimensionality.

That seems hard to understand at first, because the definition of a separable algebra doesn’t seem to be about geometry! An algebra $A$ over a field $k$ is **separable** if there exists a map $\Delta \colon A \to A \otimes A$ with

$\qquad \qquad m(\Delta(a)) = a \qquad \qquad \qquad \qquad (1)$

and

$\qquad a \Delta(b) = \Delta(a b) = \Delta(a) b \;\;\;\qquad \qquad (2)$

I motivated these laws in Part 1 using the relation between sets and vector spaces. But what do they have to do with zero-dimensionality — apart from the fact that sets are topologically zero-dimensional, which is ultimately the whole point?

Today I want to show that separable commutative algebras can be described in a more geometrical way, using what geometers call 1-forms. On a manifold, a 1-form is something that eats a vector field and spits out a smooth function. If all the 1-forms on a manifold are *zero*, that means all the vector fields are zero. And that happens if and only if the manifold is zero-dimensional!

So, what we want is an algebraic way of talking about 1-forms for any commutative algebra $A$. These are called ‘Kähler differentials’. I’ll explain them, and prove that an algebra is separable iff all its Kähler differentials are zero.

By the way: while I’ll mainly talk about commutative algebras $A$ over a field $k$, the whole subject gets even more interesting when we let $k$ be a more general *commutative ring*. I don’t want to get into that too much yet. But many of the things I’ll be doing work just as easily for commutative rings. So I’ll work at that level of generality when it’s easy, but switch to demanding that $k$ be a field whenever that helps. Please think of $k$ as a field whenever it helps *you*.

### Kähler differentials

For any commutative algebra $A$ over any commutative ring $k$ we define the **Kähler differentials**, $\Omega^1(A)$, to be the module over $A$ freely generated by symbols $d a$, one for each $a \in A$, modulo relations

$d(a + b) = d a + d b , \qquad d(\alpha a) = \alpha d a$ $d(a b) = a \, d b + b \, d a , \qquad d 1 = 0$

where $a,b \in A$ and $\alpha \in k$. These relations are just the laws obeyed by derivatives. If you’ve studied 1-forms, you should have seen these relations when $A$ is the algebra of smooth functions on a manifold.

We can also describe Kähler differentials more abstractly. Say a **derivation on $A$** is any map

$d \colon A \to M$

obeying the four equations above, where now $M$ is any $A$-module. Then $d \colon A \to \Omega^1(A)$ is the **universal** derivation on $A$, meaning any derivation on $A$ factors uniquely through this one!

For example, suppose $k = \mathbb{R}$. If $A$ is the algebra of smooth real-valued functions on a manifold $X$, vector fields on $X$ give derivations $v \colon A \to A$, and the map sending vector fields to such derivations is bijective, so if we like we can say vector fields *are* such derivations. But the Kähler differentials $\Omega^1(A)$ are an algebraic variant of the 1-forms on $X$. By its universal property there’s a map sending Kähler differentials to 1-forms, and this map is surjective — but it’s not injective.

All these facts are more tricky than you’d expect: for example, the only known proof of the last one uses the axiom of choice, but nobody knows if this is truly necessary! But I digress. Today I just want to show this:

**Theorem 14.** Suppose $A$ is a finite-dimensional commutative algebra over a field $k$. Then $A$ is separable iff $\Omega^1(A) \cong 0$.

To prove this, we’ll need to relate the definition of separability to Kähler differentials.

First note that any algebra $A$ over a commutative ring has a multiplication map

$m \colon A \otimes A \to A$

which is a map of $A,A$-bimodules. Call the kernel of this map $I$.

Note that $A \otimes A$ is a commutative algebra in its own right, and $I$ is an ideal in this algebra because $I$ is an $A,A$-bimodule:

$x \in I \implies (a \otimes b)x = a x b \in I$

So, we can square the ideal $I$ and get another ideal $I^2$ in $A \otimes A$. And then we have:

**Lemma 15.** For any commutative algebra $A$ over a commutative ring $k$,

$\Omega^1(A) \cong I / I^2$

as $A$-modules.

I’ll show this fact later. But first let’s use it to prove our theorem! We’ll need an easy lemma:

**Lemma 16.** An algebra $A$ over a commutative ring is separable iff this short exact sequence of $A,A$-bimodules splits:

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

**Proof.** This is immediate from the definition. A splitting is a map $\Delta \colon A \to A \otimes A$ with

$\qquad \qquad m(\Delta(a)) = a \qquad \qquad \qquad \qquad (1)$

and $\Delta$ is an $A,A$-bimodule map iff

$\qquad a \Delta(b) = \Delta(a b) = \Delta(a) b \;\;\; \qquad \qquad (2)$

These are precisely the equations in the definition of a separable algebra. █

Now let’s relate the separability of $A$ to the condition $\Omega^1(A) \cong 0$.

First, let’s assume $\Omega^1(A) \cong 0$. Then Lemma 15 implies $I^2 = I$. To go further we need $I$ to be finitely generated:

**Lemma 17.** If a finitely generated ideal $I$ in a commutative ring $R$ has $I^2 = I$ then $I = p R$ for some idempotent $p \in R$.

I’ll prove this later too. So, let’s assume $I$ is a finitely generated ideal in $A \otimes A$. Then multiplication by $p$ gives an $A,A$-bimodule map

$A \otimes A \stackrel{p}{\longrightarrow} I$

which splits our exact sequence of bimodules

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

So, by Lemma 16, $A$ is separable! Here’s what we’ve shown so far:

**Lemma 18.** Suppose $A$ is a commutative algebra over a commutative ring $k$ and
the kernel of multiplication $m \colon A \otimes A \to A$ is a finitely generated as an ideal in $A \otimes A$. If $\Omega^1(A) \cong 0$, then $A$ is separable.

This is a bit clunky sounding, but *every* ideal of $A \otimes A$ is finitely generated when $k$ is a field and $A$ is finite-dimensional. So in that case, $\Omega^1(A) \cong 0$ implies $A$ is separable.

Now let’s try showing the converse. Suppose $A$ is a separable algebra over a commutative ring $k$. Lemma 15 says our short exact sequence of $A,A$-bimodules

$0 \longrightarrow I \stackrel{i}{\longrightarrow} A \otimes A \stackrel{m}{\longrightarrow} A \longrightarrow 0$

splits. So, there’s an $A,A$-bimodule map

$\pi \colon A \otimes A \to A \otimes A$

projecting $A \otimes A$ onto $I$, with $\pi^2 = \pi$. Let

$p = \pi(1 \otimes 1) \in A \otimes A$

Then a little calculation shows that $\pi$ is multiplication by $p$ in the algebra $A \otimes A$, and $p^2 = p$. This is easy, but best done in the privacy of your own home.

So, our ideal $I$ is of the form $p(A \otimes A)$ for some idempotent $p$. This implies $I^2 = I$. Since $\Omega^1(A) \cong I/I^2$, we get:

**Lemma 19.** Suppose $A$ is a commutative algebra over a commutative ring $k$. If
$A$ is separable then $\Omega^1(A) \cong 0$.

Putting Lemmas 18 and 19 together we get our main result, Theorem 14.

### Conclusions

I’ve shown you something about the geometrical meaning of separability for commutative algebras. But there’s a lot left undone.

For example, while I said $\Omega^1(A) \cong I / I^2$, and I’ll prove it below, I haven’t shown you a *picture* illustrating why 1-forms are like elements of $I/I^2$. I also haven’t shown you a picture illustrating why $I/I^2 = 0$ means $A$ is like the algebra of functions on a zero-dimensional space! Geometry isn’t really geometry without pictures. But I’ve been too busy writing to draw these pictures.

Also, I left off last time with *two* questions:

Why are separable commutative algebras over a field $k$ precisely the ones that give covering spaces of $Spec(k)$?

Why are separable commutative algebras over $k$ the same as finite products of finite separable extensions of $k$?

Using Kähler differentials I’ve made a stab at answering the first question — a tentative stab, which would only be completed if I explained Grothendieck’s whole theory of etale maps. But I haven’t touched the second question at all! I want to do that next time.

I’ve been talking to Tom Leinster about this stuff, and he showed me a nice proof that a separable field extension $K$ of $k$ has $\Omega^1(K) = 0$. Thanks to our work today, that shows finite separable field extensions are separable algebras! I want to show you that argument, and maybe some more.

### Proofs of Lemmas

**Lemma 15.** For any commutative algebra $A$ over a commutative ring $k$,
$\Omega^1(A) \cong I / I^2$ as $A$-modules.

**Proof.** We’ll use a cool fact: there’s an explicit description of the kernel $I$ of the multiplication map $m \colon A \otimes A \to A$. Namely, $I$ consists precisely of $k$-linear combinations of elements of the form

$a b \otimes 1 - a \otimes b$

To check this, first note that $m(a b \otimes 1 - a \otimes b) = a b - a b = 0$, so these elements really are in $I$. On the other hand, if $\sum_i a_i \otimes b_i \in I$ then $\sum_i a_i b_i = 0$, so $\sum_i a_i \otimes b_i = -\sum_i \left( a_i b_i \otimes 1 - a_i \otimes b_i\right)$ is a linear combination of elements of the above form.

Using this cool fact, we can define

$d \colon A \to I/I^2$

by

$d a = [a \otimes 1 - 1 \otimes a]$

Here the stuff in the brackets is clearly an element of $I$, while the brackets mean we take an equivalence class to get an element of the quotient $I/I^2$.

To check that $d$ is a derivation, we just check each of the four relations. I’ll do the hardest one and leave the rest for you: I’ll show $d( a b ) = a \, d b + b \, d a$.

Now, this identity would be nicer we didn’t switch the letters $a$ and $b$ in the second term at right: $d (a b) = a \, d b + (d a) b$. But for this to make sense we need to know how to multiply a guy in $I/I^2$ *on the right* by a guy in $A$. And this calls for a little thought. Of course we could just define multiplication on the right to be the same as multiplication on the left. But that wouldn’t achieve anything. In fact we have made $A \otimes A$ into an $A,A$-bimodule where the left and right actions of $A$ are different:

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

while

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

The ideal $I \subseteq A \otimes A$ is a sub-bimodule, and the left and right actions of $A$ are still different on $I$. But they agree up to an element of $I^2$!

**Lemma 16.** If $x \in I$, then $a x = x a \mathrm{mod} I^2$ for all $a \in A$, where the left and right actions of $A$ on $I$ are defined as above.

I’ll prove this later. So, the left and right actions of $A$ on $I/I^2$ agree, and $d (a b) = a \, d b + (d a) b$ means the same thing as $d (a b) = a \, d b + b (d a)$. Yet the first one is easier to show:

$\begin{array}{ccl} a \, d b + (d a) b &=& a [b \otimes 1 - 1 \otimes b] + [a \otimes 1 - 1 \otimes a] b \\ &=& [a b \otimes 1 - a \otimes b + a \otimes b - 1 \otimes a b] \\ &=& [a b \otimes 1 - 1 \otimes a b] \\ &=& d(a b) \end{array}$

So, if you check the other less interesting laws of calculus you’ll see $d \colon A \to I/I^2$ is a derivation. To complete our work it suffices to show $d$ is a *universal* derivation.

Suppose $v \colon A \to M$ is any derivation on $A$, where $M$ is any $A$-module. We want to show there exists a unique $A$-module homomorphism $f : I/I^2 \to M$ making the obvious triangle commute:

$v = f \circ d$

For this to hold we need

$f(d b) = v(b)$

for all $b \in A$. For $f$ to be an $A$-module homomorphism we must also have

$f (a \, d b) = a v(b) \qquad \qquad (\star)$

By our “cool fact”, everything in $I/I^2$ is a $k$-linear combination of guys like $[a b \otimes 1 - a \otimes b]$. But by our definition of $d$, we know

$[a b \otimes 1 - a \otimes b] = a \, d b$

So in fact, everything in $I/I^2$ is a linear combination of guys like $a \, d b$, so $(\star)$ determines $f$ uniquely. To finish the job, please check that $f$ is well-defined — that is, independent of the choice of representatives for the expressions in the brackets. █

Now we pay the price for our slick proof. But it’s always good to stick an annoying calculation into a lemma, even when you’re already proving a lemma.

**Lemma 16.** If $x \in I$, then $a x = x a \mathrm{mod} I^2$ for all $a \in A$, where the left and right actions of $A$ on $I$ are defined as above.

**Proof.** It suffices to show this for $x = b c \otimes 1 - b \otimes c$, since our “cool fact” says every element of $I$ is a $k$-linear combination of elements like this.

$\begin{array}{ccl} a x &=& a(b c \otimes 1 - b \otimes c) = a b c \otimes 1 - a b \otimes c \\ x a &=& (b c \otimes 1 - b \otimes c) a = b c \otimes a - b \otimes a c \\ \\ a x - x a &=& a b c \otimes 1 - a b \otimes c - b c \otimes a + b \otimes a c \\ &=& (a b \otimes 1 - b \otimes a)(c \otimes 1 - 1 \otimes c) \end{array}$

The last expression is the product of two elements of $I$, so we’re done! █

Believe it or not, all these calculations have a geometrical meaning. I hope to clarify that sometime. But for now I’ll conclude with a lemma of a very different sort.

**Lemma 17.** If a finitely generated ideal $I$ in a commutative ring $R$ has $I^2 = I$ then $I = p R$ for some idempotent $p \in R$.

**Proof.** We’ll use Nakayama’s Lemma. There are lots of ways to state this lemma, but the most memorable might be

$M = I M \implies m = i m$

This means: if a finitely generated module $M$ of some commutative ring $R$ has $M = I M$ for some ideal $I \subseteq R$, then there exists $i \in I$ such that $i m = m$ for all $m \in M$.

We can apply this taking $M = I$ to be a finitely generated ideal with $I = I^2$. We conclude that there exists $i \in I$ such that $i m = m$ for all $m \in I$. It follows that $i^2 = i$ and $I = i R$. Now just let $p = i$. █

## Re: Grothendieck–Galois–Brauer Theory (Part 3)

Side remark (apologies if you knew this and were just omitting it to keep the exposition shorter).

One can set up derivations from $A$ into $A$-bimodules, and then it turns out that $D:A\to I, a \mapsto a \otimes 1 - 1\otimes a$ is the universal such derivation. To get the universal derivation into symmetric bimodules, one composes $D$ with the abelianization map $I \to I_{ab}$; then the calculations at the end of your post can be viewed as showing that $I_{ab}$ is naturally identifiable with $I/I^2$.

I think that the arguments in your post show, without any extra work, that for a not-necessarily finite-dimensional and not-necessarily-commutative $k$-algebra $A$, separability of $A$ is still equivalent to the $I$ having an identity element, provided one views it as an ideal in the ring $A \otimes A^{op}$ — and this is equivalent to $D:A\to I$ being an

inner derivation. This is the characterization of the separability of $A$ by the vanishing of its 1st Hochschild cohomology.