## September 1, 2023

### Grothendieck–Galois–Brauer Theory (Part 4)

#### Posted by John Baez

The word ‘separable’ is annoying at first. In Galois theory we learn that ‘separable field extensions’ are the nice ones to work with, though their definition seems dry and technical. There’s also a concept of ‘separable algebra’. This is defined in a different way — and not every separable field extension is a separable algebra! So what’s going on?

Today I’ll remind you of these two concepts and show you why a finite-dimensional extension of a field $k$ is a separable extension of $k$ if and only if it’s a separable algebra over $k$.

I got my arguments for this from Tom Leinster and Qing Liu. Their arguments are nice because they use differential geometry. More precisely, they use an algebraic approach to those ‘$d x$’ thingies you see in calculus and differential geometry. They’re called Kähler differentials, and I introduced them last time.

Suppose $K$ is a field with a subfield $k$. Remember from Part 2 that $x \in K$ is algebraic over $k$ if there’s a nonzero polynomial $P$ with coefficients in $k$ such that $P(x) = 0$. There is then a unique polynomial $P$ with coefficients in $k$ such that

• $P(x) = 0$,
• $P$ has the lowest possible degree for a nonzero polynomial that vanishes on $x$,
• the leading coefficient of $P$ is $1$.

This is called the minimal polynomial of $x$.

If $x \in K$ is algebraic over $k$, we say it’s separable if the derivative of its minimal polynomial is nonzero. And $K$ is a separable extension of $k$ if all its elements are separable.

So much for separable extensions of fields. What about separable algebras over fields? I’ve defined those many times here, but last time I explained the module of Kähler differentials $\Omega^1(A)$ of an algebra $A$ over a field and showed that if $A$ is finite-dimensional, it’s separable iff $\Omega^1(A) \cong 0$. So you can use this fact as our definition for now. Today we’ll put this fact to work!

I got this proof from Tom Leinster during our conversations on this stuff:

Lemma 18. Suppose the field $K$ is a separable extension of the field $k$. Then $\Omega^1(K) \cong 0$.

Proof. It’s enough to show that $d x = 0$ for any $x \in K$.

We know $x$ has a minimal polynomial $p$ over $k$. So $P(x) = 0$, and therefore $d(p(x)) = 0$. Now the usual rules for differentiating sums, products and constants all hold for Kähler differentials, and these imply that $d(P(x)) = P'(x) d x$, where $P'$ is the derivative of $P$ computed using the usual rules. So $P'(x) d x = 0$. By separability, $P'$ is not the zero polynomial. So $P'$ is a nonzero polynomial of smaller degree than $P$, which by minimality implies that $P'(x) \ne 0$. Since $P'(x)$ is an element of the field $K$, it follows that $P'(x)$ is invertible. And since $P'(x) d x = 0$, we have $d x = 0$.    █

Remember that a field $K$ extending $k$ is called a finite extension if it’s finite-dimensional as a vector space over $K$.

Theorem 19. Suppose the field $K$ is a finite separable extension of the field $k$. Then $K$ is a separable algebra over $k$.

Proof. We just saw that $\Omega^1(K) \cong 0$, and in Theorem 14 we saw that any finite-dimensional commutative algebra $A$ over $k$ with $\Omega^1(A) \cong 0$ is separable.    █

We really do need the finiteness condition here! You see, every separable algebra over a field is automatically finite-dimensional, but not every separable field extension is finite. I’ll show you a proof of the first fact someday when I classify separable algebras over a field. For the second fact, we’ll see later today that every extension of $\mathbb{Q}$ is separable — but of course they’re not all finite-dimensional.

Does Theorem 19 have a converse? Yes!

Theorem 20. Suppose the field $K$ is a finite extension of the field $k$. If $K$ is a separable algebra over $k$ then $K$ is a separable extension of $k$.

One way to prove this is by classifying all separable algebras over a field. But there’s also a proof using just Kähler differentials: the idea is to show that if $K$ is a finite extension of $k$ that’s not separable, then $\Omega^1(K) \ncong 0$. After struggling to prove this myself, I gave up and found a proof in a book. I’ll sketch it below.

But first, to build some intuition for this subject, it’s time to look at a field extension that’s not separable!

### Inseparable field extensions

Let’s find a field extension that’s not separable, and work out its Kähler differentials. How can we do this? How can we find a field extension containing an element whose minimal polynomial has derivative zero?

Here we are computing derivatives of polynomials ‘formally’, defining

$\frac{d}{d x} \sum_{i = 0}^n a_i x^i = \sum_{i = 1}^n i a_i x^{i-1}$

The derivative of a constant polynomial is zero — but a constant polynomial can’t be the minimal polynomial of anything. So how can a nonconstant polynomial have derivative zero?

This can only happen if the coefficients $a_i$ with $i \gt 0$ don’t all vanish, yet still $i a_i$ vanishes for all $i \gt 0$. This can never happen if our field has characteristic zero! In characteristic $p$, it happens when $a_i$ is nonzero only for $i$ that are multiples of $p$.

So, a field of characteristic zero can’t have inseparable extensions. And we only get them in characteristic $p$ when we have an element whose minimal polynomial is of the form

$P(x) = \sum_{i = 0}^n a_i x^{p i}$

We could also write this as

$P(x) = \sum_{i = 0}^n a_i \left(x^p\right)^i$

And that’s suggestive, because in characteristic $p$ the function $x^p$ is not only multiplicative:

$(a b)^p = a^p b^p$

It’s also linear:

$(a + b)^p = a^p + b^p$

using the binomial theorem. This weird fact makes algebra in characteristic $p$ really different than in characteristic zero. So this map $x \mapsto x^p$ gets an impressive name: the Frobenius endomorphism. For a field of characteristic $p$ this map clearly has kernel zero, so it’s one-to-one. Thus, for a finite field of characteristic $p$ the Frobenius endomorphism is a bijection.

This implies that we can’t have an inseparable extension of a finite field. Why not? Well, suppose our extension has an element whose minimal polynomial $P$ has $P' = 0$. In characteristic $p$ we’ve seen this means

$P(x) = \sum_{i = 0}^n a_i x^{p i }$

But for a finite field we can write $a_i = b_i^p$ since the Frobenius endomorphism is a bijection! So

$P(x) = \sum_{i = 0}^n \left(b_i x^i\right)^p = \left( \sum_{i=0}^n b_i x^i \right)^p$

In the second step I used the insane fact that the Frobenius endomorphism is linear. But this means $P$ factors into linear terms, so it can’t have been minimal unless it already were linear, which is also impossible.

We’ve just seen some stuff worth recording:

Theorem 21. Every extension of a field of characteristic zero, or a finite field, is separable.

Since we’re in search of inseparable extensions, this just says where not to look. But that’s still very helpful.

The simplest infinite field of characteristic $p$ is

$k = \mathbb{F}_p(x)$

the field of rational functions in $x$ with coefficients in $\mathbb{F}_p$. Let’s try to get an inseparable extension of this!

We need a nonzero polynomial whose derivative vanishes. So use the simplest one:

$P(x) = x^p$

Then we need a field extension with an element whose minimal polynomial is $P$. So use the simplest one:

$K = k[u]/\langle u^p - x \rangle$

That is, we take $k$ and throw in a new element $u$ whose $p$th power is $x$. This does the job!

What are the Kähler differentials $\Omega^1(K)$ like, thinking of $K$ as an algebra over $k$? Since $d$ of everything in $k$ is automatically zero — these act like ‘constants’ — we mainly need to think about $d u$. The relation

$u^p = x$

implies

$p u^{p-1} d u = d x = 0$

since $x \in k$. If we weren’t in characteristic $p$, this would force $d u = 0$, as in the proof of Theorem 18. But since we’re in characteristic $p$ this equation is vacuous. So $d u$ obeys no nontrivial relations. Thus $\Omega^1(K)$ is the free $K$-module on one generator, namely $d u$.

Notice: I didn’t prove the last two sentences, but that’s indeed how it works!

### Kähler differentials for inseparable field extensions

Now I want to sketch the proof of this:

Theorem 20. Suppose the field $K$ is a finite extension of the field $k$. If $K$ is a separable algebra over $k$ then $K$ is a separable extension of $k$.

This is basically Lemma 1.13(c) in Section 6.1.1 of this book:

• Qing Liu, Algebraic Geometry and Arithmetic Curves, Oxford U. Press, Oxford, 2002.

But let’s see how it goes.

Proof. We’ll prove the contrapositive. We’ll assume $K$ is a finite inseparable extension of $k$, and show $K$ is not a separable algebra over $k$. To do this it’s enough to show $\Omega^1(K) \ncong 0$, since in Theorem 14 we saw that a finite-dimensional commutative algebra $A$ over $k$ is separable iff $\Omega^1(A) \cong 0$.

The simplest case is when $K$ is gotten from $k$ by freely throwing in a single inseparable element $u$ — that is, one whose minimal polynomial has $P' = 0$. Then, as in the example we just looked at, $d u \ne 0$. So $\Omega^1(K) \ncong 0$ in this case.

The only other possibility is that $K$ is bigger. In this case it turns out we can always get $K$ from some field $F$ extending $k$ by throwing in one more inseparable element. So, we have inclusions

$k \hookrightarrow F \hookrightarrow K$

Now we have various kinds of Kähler differentials to think about, so we need better notation:

• We have the Kähler differentials of $K$ as an algebra over $k$, which we’ll call $\Omega^1_k(K)$.

• We have the Kähler differentials of $K$ as an algebra over $F$, which we’ll call $\Omega^1_F(K)$.

We’re trying to show $\Omega^1_k(K) \ncong 0$. Since $K$ is gotten from $F$ by throwing in a single inseparable element we have $\Omega^1_F(K) \ncong 0$. So, we’d be done if we could find a surjection $\Omega^1_k(K) \to \Omega^1_F(K)$.

Luckily, this is easy to get. For any commutative algebra $A$ over any commutative ring $R$, $\Omega^1_R(A)$ is the abelian group with generators $d a$ for $a \in A$ and relations

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

for $a, b \in A$ and $\alpha \in R$. So, in our situation $\Omega^1_F(K)$ has the same generators as $\Omega^1_k(K)$ but more relations, namely those of the form $d(\alpha a) = \alpha d a$. This gives our surjection $\Omega^1_k(K) \to \Omega^1_F(K)$, as desired.    █

In fact, for any homomorphisms of commutative rings

$k \to F \to K$

there is an exact sequence

$\Omega^1_k(F) \otimes_F K \to \Omega^1_k(K) \to \Omega^1_F(K) \to 0$

so $\Omega^1_k(K)$ maps surjectively onto $\Omega^1_F(K)$. I won’t explain this exact sequence here, but you can find it in many treatments of Kähler differentials. In Qing Liu’s book it’s Proposition 1.8 of Section 6.1.1. In the Stacks Project it’s Lemma 10.131.7. In Matsumura’s Commutative Algebra it’s Theorem 57. In my favorite book on separable algebras:

• Timothy J. Ford, Separable Algebras, American Mathematical Society, Providence, 2017.

it’s called the First Fundamental Exact Sequence, and it’s constructed in Theorem 8.3.3. So, everyone who knows their Kähler differentials knows and loves this exact sequence!

Posted at September 1, 2023 12:41 PM UTC

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### Re: Grothendieck–Galois–Brauer Theory (Part 4)

I think that u^p-1 should be u^p-x.

Posted by: Allen Knutson on September 2, 2023 2:51 PM | Permalink | Reply to this

### Re: Grothendieck–Galois–Brauer Theory (Part 4)

Whoops! Fixed.

Posted by: John Baez on September 3, 2023 12:32 PM | Permalink | Reply to this

### Re: Grothendieck–Galois–Brauer Theory (Part 4)

I’d really like to solve the puzzle and prove that an inseparable extension $K$ of a field $k$ is not a separable algebra over $k$ by proving $\Omega^1(K) \ncong 0$.

I know (for other reasons) that if $K$ is inseparable, there is some element $u \in K$ with $d u \ne 0$. And I know from Theorem 18 that any separable element $u \in K$ has $d u = 0$. So I know that if $K$ is inseparable, it contains some inseparable element $u$ — that is, one whose minimal polynomial $P$ has $P' = 0$ — with $d u \ne 0$.

But I don’t know how to prove this using just Kähler differentials. That’s the challenge.

This result would follow if every inseparable element $u \in K$ has $d u \ne 0$. But I don’t know if that’s true. How could I show it?

I’ve read an inseparable element of $K$ is the same as one whose minimal polynomial $P$ has repeated roots in the algebraic closure $\overline{k}$. But I don’t see how that helps here.

Polya said: if you can’t see how to prove something, there must be something simpler you don’t know how to prove. Try to prove that!

Okay, so let’s try an example. Suppose we’re in characteristic 2 and $u \in K$ has some quadratic minimal polynomial $P$. Can I show $d u \ne 0$?

Since $P$ is monic, we can assume

$P(x) = x^2 + b x + c$

for some $b,c \in k$. Since $P' = 0$ we must have $b = 0$, and indeed any polynomial

$P(x) = x^2 + c$

has $P' = 0$, so this is what $P$ looks like.

Supposedly $P$ must have repeated roots in $\overline{k}$, so (thanks to the magic of characteristic 2) in $\overline{k}$ we can factor it as

$P(x) = (x + q)(x + q)$

where $q \in \overline{k}$ has

$q^2 = c$

We’ve got $P(u) = 0$ so (thanks to the magic of characteristic 2) we have

$u^2 = c$

so $u = \pm q$, so actually (thanks to the magic of characteristic 2)

$u = q$

But how does any of this help me show $d u \ne 0$?

I think it’s enough to show $d u \ne 0$ in $\overline{k}$ since going up to this bigger field induces an injection $\Omega^1(K) \to \Omega^1(\overline{k})$ — I think I read this somewhere.

But I don’t see how that helps. I’ll stop here for now and think some more.

Posted by: John Baez on September 3, 2023 1:51 PM | Permalink | Reply to this

### Re: Grothendieck–Galois–Brauer Theory (Part 4)

I think I was getting lost in elementary algebraic manipulations and forgetting an idea from geometry: if you’ve got a polynomial $P$ with a ‘doubled root’, like $P(x) = (x - q)^2$, then $P(x) = 0$ defines a scheme with a double point, which is like an first-order infinitesimal arrow — so it has not just a point but also a nonvanishing tangent vector. It should thus have nonzero Kähler differential!

And I think the way to prove this is to use the universal property of Kähler differentials. This implies that if you can find any nonzero derivation on a commutative ring, it must have nonzero Kähler differentials.

So I just need to dream up some nonzero derivation on the inseparable extension $K$ I was playing around with above. And this should be something like ‘differentiation in the direction of the infinitesimal arrow corresponding to the double point’.

Unfortunately I don’t see how to define this derivation on all of $K$, since I don’t know much about it.

So in my example above, maybe I should instead take a field $F$ which is just an inseparable extension of $k$ by some element $u$, like

$F = k[u]/\langle u^2 + c \rangle$

where $c$ is as above. Then maybe I can show this has $\Omega^1(F) \ncong 0$. Indeed I pretended to show something very similar near the end of my blog post! — though I skipped some important details.

Then, if we have an even larger inseparable extension of $k$ containing $F$ — call it $K$ — we can try to show $\Omega^1(F) \to \Omega^1(K)$ is an injection (as I seem to have read somewhere) so $\Omega^1(K) \ncong 0$.

This is looking good.

Posted by: John Baez on September 3, 2023 2:08 PM | Permalink | Reply to this

### Re: Grothendieck–Galois–Brauer Theory (Part 4)

This is looking good.

Well, not quite — I don’t think there’s an injection here, but something vaguely similar works. I got stuck, but I found the right argument in Qing Liu’s book Algebraic Geometry and Arithmetic Curves. And I’ve added it to my blog article, at the end.

Posted by: John Baez on September 3, 2023 6:21 PM | Permalink | Reply to this

### another approach to the puzzle

Here’s how I was thinking about the puzzle. I’m not sure if it counts as “just using Kahler differentials”?

1. It suffices (exercise!) to prove that the $\bar k$-algebra $B = K\otimes_k \overline k$ is reduced.

2. The formation of Kahler differentials commutes with base change. (See Wiki page for the formal statement of what this means. I believe this is straightforward to prove from the universal property.) So we have $\Omega_{B / \overline k} = 0$.

3. $B$ is a finite dimensional $\bar k$-algebra, so an easy exercise shows it’s a (finite) direct product (in the category of $\overline k$-algebras) of $\overline k$-finite-dimensional local (= having a unique maximal ideal) $\overline k$-algebras $A$ with $\Omega_{A/\overline k} = 0$. It suffices (easy exercise) to show that each $A$ is reduced. Now I’ll rename $\overline k$ to $k$ and assume it algebraically closed.

4. Let $m$ be the maximal ideal of $A$. We have $A = (A/ m)\oplus m$ as $k$-vector spaces. Since $A/ m$ is a field extension of $k$ that is finite dimensional as a $k$-vector space, it must be $k$ since $k$ is algebraically closed. That is, the image of $k$ in $A$ is a $k$-linear direct complement for $m$. We can therefore define a $k$-linear map $d:A\to m/ m^2$ by extending the natural projection $m \to m/ m^2$ by zero on $k$. It’s easy to see that $d$ is a derivation. (Proof of Liebniz: for constants $c_1, c_2 \in k\subset A$ and nonunits $m_1, m_2\in m$, we have $(c_1 + m_1)(c_2 + m_2)= (const) + (c_1 m_2 + c_2 m_1)$ mod $m^2$.)

5. By assumption $\Omega_{A/k}=0$, so the derivation $d:A\to m/m^2$ is the zero map. In particular, the projection $m\to m/ m^2$ is the zero map. This implies $m=0$ so $A=k$ is reduced.

Posted by: sam on September 3, 2023 7:32 PM | Permalink | Reply to this

### Re: another approach to the puzzle

Thanks! I feel bad having yanked the Puzzle out from under you and anyone else who was trying it.

For those who never saw it, it was to prove this using just Kähler differentials: if a finite extension of a field $k$ is a separable algebra over $k$, then it’s a separable extension of $k$. I got so frustrated with this puzzle that I started trying to solved it myself in the comments, then ‘cheated’ and took a proof in a book and added it to at the end of my blog article!

After Sam wrote his comment I simplified the proof in my article some more. In the end I was pretty happy, but I still want to think about his approach… and any other ideas that people have.

Posted by: John Baez on September 3, 2023 11:23 PM | Permalink | Reply to this

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