Global choice and algebraic closures

It was pointed out to me today that I glazed over a set-theoretic point in my proof that every field has an algebraic closure. We appealed to Zorn’s lemma, which says:

Given a partially ordered set $\mathcal P$, if every chain $\mathcal C\subseteq\mathcal P$ has an upper bound, then $\mathcal P$ has a maximal element.

An assumption we made here that seems innocuous is that $\mathcal P$ is a set. If it is a proper class (class and not a set), we may run into problems. So what if the collection of all field extensions of a given field is a proper class? Then we can’t use Zorn’s Lemma.

I think, though haven’t attempted to prove it, that this collection of field extensions is really a set, meaning we’re safe in applying Zorn’s lemma. But it’s an interesting diversion to see what happens if we try to prove Zorn’s lemma for classes.

It’s well known that Zorn’s lemma is equivalent to the axiom of choice, so it would be nice to find an “axiom of choice for classes.” Such a thing exists, and it’s called the “axiom of global choice.” It says the same thing about classes that the axiom of choice says about sets. So if you accept the axiom of global choice, we just need to prove a “global Zorn’s lemma.”

Unfortunately, the standard proof of Zorn’s lemma goes something like this: Assume it’s false. Then use the axiom of choice to construct (via transfinite induction) larger and larger elements. These form a well-ordering. We can keep going and we’ll eventually get to a well-ordering that has too many elements to be contained in the set $\mathcal P$.

The problem is that if we replace the axiom of choice with the axiom of global choice, we’ll be unable to make this well-ordering bigger than our “poclass” $\mathcal P$.

If you want more information on this sort of set theory stuff, you should read Zach Norwood’s posts (this is a great blog to which you should definitely subscribe):

Basic facts about AC, part I

Basic facts about AC, part II

On an unrelated note, coming soon will be a series of posts on some generalized abstract nonsense.

Fixed fields

If I have a field $k$ and a field automorphism $\sigma$, I can ask about the fixed points of $\sigma$. That is, all of the $x\in k$ such that $\sigma(x)=x$. Notice that if $\sigma$ fixes $x$ and $y$, then $\sigma(x+y)=\sigma(x)+\sigma(y)=x+y$ and $\sigma(xy)=\sigma(x)\sigma(y)=xy$, so $\sigma$ fixes $x+y$ and $xy$. In fact, the set of fixed points of $\sigma$ is itself a field! I just checked two of the rules. You can check the rest.

This field is called the fixed field of $\sigma$, and it’s super super important. If we have two automorphisms, $\sigma$ and $\tau$, we can take the intersection of their fixed fields. The intersection of fields is always a field. What field is it? It’s the field consisting of everything fixed by both $\sigma$ and $\tau$.

But why stop at two automorphisms? What if I take a group $G$ of automorphisms of $k$. That has a fixed field. It’s called the fixed field of $G$. I could take the entire automorphism group of $k$, or I could just take a subgroup of it. The bigger the group I take, the more fields I intersect, so the smaller my fixed field is.

That’s an important idea. Big group means small fixed field. Small group means big fixed field. We’ll make this precise very soon.

Automorphisms of Q

We’ve talked a bit about automorphisms, but we haven’t seen very much of them, so I wanted to do an example. What is $\mbox{Aut}(\mathbb Q)$ (the group of automorphisms of the field of rational numbers)?

It definitely contains the identity (a.k.a. lame) automorphism, since every field has that one. Is there anything else? No, and here is why. Let $\phi$ be an automorphism of $\mathbb Q$. That is, let $\phi\in\mbox{Aut}(\mathbb Q)$. Let $p/q$ be a rational number. Then we can write $p/q$ as

$p/q=\frac{\overbrace{1+\cdots+1}^{p\ times}}{\underbrace{1+\cdots+1}_{q\ times}}$

If we apply $\phi$,, we get that

$\phi(p/q)=\frac{\overbrace{\phi(1)+\cdots+\phi(1)}^{p\ times}}{\underbrace{\phi(1)+\cdots+\phi(1)}_{q\ times}}=\frac{p\cdot\phi(1)}{q\cdot\phi(1)}=p/q.$

This is to say, any field automorphism of a field extending $\mathbb Q$ fixes $\mathbb Q$. Another way to say this is that we can reach any number in $\mathbb Q$ just from 1 and operations which any field automorphism respects (namely addition and division).

This is to say, the automorphism group of the field of rationals is trivial!

Field automorphisms

It’s worth, I suppose, talking about automorphisms in general. An automorphism is a “renaming” of stuff so that it keeps the same structure. The “auto” means self. The “morph” means shape. The “same” is suppressed, because it really comes from the word “isomorphism,” which means same shape. So this is like a self-similar shape… but not in a fractally sense.

What is required for an automorphism? Well, any special constants need to stay the same. And the operations need to be “preserved,” whatever that means. For a field, it means that 0 and 1 are fixed, and that you can move $+$ and $\cdot$ outsied the function. More specifically, $\phi$ is an automorphism of a field $F$ if it is a function $\phi:F\to F$ with

• $\phi(0)=0$
• $\phi(1)=1$
• $\phi(x+y)=\phi(x)+\phi(y)$
• $\phi(xy)=\phi(x)\phi(y)$
• $\phi$ is a bijection. That is, $\phi$ doesn’t send two different points to the same point, and it sends something to every point. (Bijection is a fancy word meaning “1-1 and onto.”)
Can you think of any field automorphisms? In any field, you always have the trivial automorphism which sends a point $x$ to $x$. It’s lame, but it counts.
What about something a bit more complicated? How about complex conjugation? What if $\phi:\mathbb C\to \mathbb C$ by $\phi(a+bi)=a-bi$ (for $a,b\in \mathbb C$)? Yes, this works. It’s not hard to check any of these rules, but I’m too lazy to do so.
How about something really crazy? Take the field $\mathbb Q(\sqrt2)$, and the automorphism $\psi$ defined by $\psi(a+b\sqrt 2)=a-b\sqrt 2$ (where $a,b\in \mathbb Q$). What? But that sends negative things to positive things? It doesn’t preserve the shape at all!
I assure you, it is a field automorphism. If you don’t believe me, check the rules yourself. The issue you’re potentially having, is that you’re thinking of $\mathbb Q(\sqrt 2)$ as sitting inside $\mathbb R$, and thinking of the ordering on $\mathbb R$. But alas, I said nothing about order. It’s damn near impossible, but try to forget everything you know about the structure of $\mathbb R$, because it will only misguide you.

The intermediate field lattice

If I have two fields $k$ and $K$, and $k\subseteq K$, we call $k$ a subfield of $K$, but it turns out the proper way to think of it is the reverse. We don’t think of $K$ as some starting field with $k$ as a smaller one sitting inside. Instead, we think of $k$ as the starting field, and $K$ as a field extension of $k$. We could do this for any object, but for whatever reason, it works well in Galois theory.

So one thing we could ask is “what are all the fields between $k$ and $K$? We can draw nice pictures called Hasse diagrams. The idea is, we put the bigger fields towards the top, and make it so that fields that contain each other are connected by a chain of lines.

Let’s do an example. Suppose we start with $\mathbb Q$ and extend it to the field $\mathbb Q(\sqrt 2,i)$. Here’s the picture:

You forgot about $\mathbb Q(i\sqrt 2)$, didn’t you? Yeah, it’s hard. Man, if only there was a way to make sure we got all of them. Like, putting them in a one-to-one correspondence with, I don’t know, groups or something like that.

But seriously, how do we know we got everything? Well, degree 2 extensions can always be written as adjoining the square root of something (fun exercise I alluded to previously), and we can check that these are the only possibilities that lie inside $\mathbb Q(\sqrt2,i)$. Since $[\mathbb Q(\sqrt2,i),\mathbb Q]=4$, all the extensions need to be of degree 1,2, or 4. Since 1 and 4 are stupid, we need only consider the degree 2 extensions, and we just checked that we got them all.

One more thing: the fact that $\mathbb Q(\sqrt 2)$ is on the left and $\mathbb Q(i)$ is on the right is arbitrary. I could order those three fields in any of the 6 different ways, and it would still be correct. My choice was arbitrary. All that matters is the containment relationships.

Constructing regular polygons

Yesterday, as usual, we saw some cool stuff. The gist of it was that for polynomials $P$ that can’t be factored over $\mathbb Q$, if $P(\alpha)=0$ and $P$ is a polynomial of degree $n$, then $[\mathbb Q(\alpha):\mathbb Q]=n$. Here we can replace $\mathbb Q$ with any field $k$, but for constructibility, we only care about $\mathbb Q$.

What about constructing regular polygons? If I want to construct a regular $N$-gon, I need to construct $\cos\left(\frac{2\pi}N\right)$. So for which $N$ can I do this? I’m going to do something a little crazy. I’m going to consider the $\zeta_N=\cos\left(\frac{2\pi}N\right)+i\sin\left(\frac{2\pi}N\right)$. This is a complex number, which you might recognize as an $N$th root of unity. That is $\left(\zeta\right)^N=1$. What I want to know is the smallest polynomial which has $\zeta_N$ as root. I’m not going to justify this, but if you take all the terms $(x-\zeta_N^k)$ where $k$ is relatively prime to $N$, and multiply them together, you get exactly this polynomial.

But the real question is the degree of this polynomial. It has one factor of $x$ for each $k$ relatively prime to $N$. We have a name for that. It’s the Euler totient function $\phi(N)$. I have a prize for anyone who can tell me what the word “totient” means.

It turns out that by working over the complex numbers, I haven’t actually cheated (though I won’t justify this either), and the degree of $[\mathbb Q\left(\cos\left(\frac{2\pi}N\right)\right):\mathbb Q]=[\mathbb Q(\zeta_N):\mathbb Q]=\phi(N)$. Well, we need $\phi(N)$ to be a power of two if we have any chance of constructing it. Here are some facts about $\phi(N)$ you might prove in an introductory number theory course:

• If $p$ is a prime dividing $N$, then $p-1$ divides $\phi(N)$.
• If $n^2$ divides $N$, then $n$ divides $\phi(N)$.
• If $N$ is odd, then $\phi(2N)=\phi(N)$.
• If $N$ is even, then $\phi(2N)=2\phi(N)$.
So the first thing we should do is break up $N$ into its even and odd parts. Write $N=2^k\cdot m$ where $m$ is odd. Repeated applications of the previous rules tells us that $\phi(N)=2^{k-1}\phi(m)$. So $\phi(N)$ is a power of two if and only if $\phi(k)$ is as well. Take some prime $p$ which divides $k$. Then $p-1$ divides $\phi(k)$, so we need $p-1$ to be a power of two. That is, $p$ is a prime which is one more than a power of two.
We have a name for these types of primes. They’re called Fermat primes, and they seem to be exceedingly rare. We only know of 5 (namely 3, 5, 17, 257, and 65537). What’s more, we can only have each of these primes dividing $N$ once (by the second rule).
So we do know of infinitely many constructible $N$-gons. Namely, the square, octagon, 16-gon, 32-gon, etc. But if you want to know how many constructible $N$-gons there were with $N$ odd, you would be asking an extremely difficult question that has been open for nearly 200 years!
Okay. At long last, we’re ready to jump into Galois Theory. Hold on to your socks, because they’re about to get knocked off!

The classical Greek questions

1. Can we build a cube with twice the volume of a given cube?
2. Can we trisect an arbitrary angle?
3. Can we construct a square with the same area as the unit circle?
Here’s the first idea we will need. Let $\alpha$ be the root of a cubic polynomial with coefficients in $\mathbb Q$. When I say a cubic polynomial, I want one that can’t be factored. So, while $\sqrt{17}$ is a root of $P(x)=x^3-x^2-17x+17$, it’s silly, because $P(x)=(x^2-17)(x-1)$. We call polynomials like $P$ reducible because we can reduce it to smaller polynomials. We want an irreducible polynomial. In some sense, you should think of irreducible polynomials as “prime polynomials,” but don’t ever say this to an algebraist, because strictly speaking this is a different concept (but in our case they are the same).
I want to show that $[\mathbb Q(\alpha):\mathbb Q]=3$. In fact, if $\alpha$ is the root of an irreducible polynomial with coefficients in $\mathbb Q$ of degree $n$, then $[\mathbb Q(\alpha):\mathbb Q]=n$.

Proof:

Remember that $[\mathbb Q(\alpha):\mathbb Q]=\dim_{\mathbb Q}\mathbb Q(\alpha)$. We’re considering $\mathbb Q(\alpha)$ as a vector space over the field $\mathbb Q$, and want to know its dimension. Let $P(x)=c_0+c_1x+\dots+c_nx^n$ be an irreducible polynomial with $c_i\in\mathbb Q$ and $P(\alpha)=0$. The vector space $\mathbb Q(\alpha)$ is definitely spanned by $1, \alpha, \alpha^2,\alpha^3,\dots$. I claim that $1,\alpha,\dots, \alpha^{n-1}$ form a basis. This would be a basis with $n$ elements, making $[\mathbb Q(\alpha):\mathbb Q]=n$.

We have a linear dependence $c_0+c_1\alpha+\dots+c_n\alpha^n=0$, so any basis must have fewer than $n+1$ elements. Now suppose we have a linear dependence among $1,\alpha,\alpha^2,\dots,\alpha^{n-1}$. Say,

$\lambda_0+\lambda_1\alpha+\lambda_2\alpha^2+\dots+\lambda_{n-1}\alpha^{n-1}$

Then we have a polynomial $Q(x)=\lambda_0+\lambda_1x+\lambda_2x^2+\dots+\lambda_{n-1}x^{n-1}$ of smaller degree with $Q(\alpha)=0$.

Why is this a problem? We can take combinations of $P(x)$ and $Q(x)$, and try to make the polynomial of smallest degree in this way. Let’s call it $R(x)=A(x)P(x)+B(x)Q(x)$ for some $A,B$. This is the “greatest common divisor” of $P$ and $Q$. You thought GCDs only existed for integers? Nope. There’s a general structure (of which the integers are an example) which always has a concept of a GCD.

The important thing is that $R(x)$ is a divisor of $P(x)$. But we assumed $P(x)$ was irreducible, so the only possible divisors are $P$ itself and $1$. Since $Q(x)$ has smaller degree than $P(x)$, so does $R(x)$, so it must be that $R(x)=1$. But then $R(\alpha)\neq 0$. What? Contradiction!

This means we have $\{1,\alpha,\alpha^2,\dots,\alpha^{n-1}\}$  is a basis, so $[\mathbb Q(\alpha):\mathbb Q]=n$.

$\square$

This “back-and-forth” between bases for vector spaces and polynomials is a really cool trick, and is a fun part of Galois theory.

Now we can answer the questions.

1. NO! If we could construct a cube with twice the volume of a given cube, we could in particular construct a cube of volume 2, so we could construct the side length $\sqrt[3]2$. But $x^3-2$ is an irreducible polynomial (check that yourself) with $\sqrt[3]2$ as a root, so this would mean we have a degree 3 extension $\mathbb Q(\sqrt[3]2)\subseteq \mathbb K$. But every extension of $\mathbb Q$ in $\mathbb K$ has degree that is a power of two. Three is certainly not a power of two. Contradiction!
2. NO! If we could construct arbitrary angles $\theta$ given $3\theta$, we could in general construct the length $\cos\theta$ from $\cos3\theta$. If $3\theta=60^\circ$, then $\theta=20^\circ$, which is a root of the polynomial $4x^3-3x-\frac12$ (this comes from the triple-angle formula for cosine).
3. NO! If we could, we could construct the side length $\sqrt\pi$. This is the most absurd, because then we could construct $\pi$ which isn’t the root of any polynomial. Some might want to even write $[\mathbb Q(\pi):\mathbb Q]=\infty$, which is accurate. We can’t get to infinity by only finite constructions.
So there you have it. Field theory rocks.