# Separable Algebras 3

Fix a field ${k}$ and let ${k_s}$ be a separable closure. Let ${G=Gal(k_s/k)}$. Today we prove the strongest structure theorem so far: The category of separable ${k}$-algebras is anti-equivalent to the category of finite ${G}$-sets (where the action is continuous). Recall that one way to phrase the ${G}$ action being continuous on ${X}$ is to say that ${X}$ is a union of sets on which the action factors through some finite quotient ${Gal(L/k)}$.

To show the theorem let’s construct our functor ${F: Sep_k \rightarrow G}$-Set (I made this notation up just now, so it isn’t standard). Define ${F(A)=Hom_k (A, k_s)}$. Recall from last time that our separable ${k}$-algebra must have the form ${\prod L_i}$ where ${L_i/k}$ are finite separable field extensions. Thus a map ${A\rightarrow k_s}$ kills all factors except one which lands inside a finite separable extension of ${k}$.

The ${G}$-action on ${F(A)}$ is the one given by acting on ${k_s}$. More specifically, given ${\sigma\in G}$ and ${f\in F(A)}$, we need to define a new ${k}$-algebra map ${\sigma\cdot f}$, but we do this by mapping ${x\mapsto \sigma(f(x))}$. If you try this trick in other situations, be careful. It works here because any element ${\sigma\in G}$ fixes ${k}$ and hence preserves the ${k}$-algebra structure. Suppose ${Im(f)\subset E}$, then the action factors through ${Gal(E/k)}$ and hence the action is continuous.

Now we get the rest of ${F}$ being a contravariant functor for free because we defined it to be ${Hom_k(-, k_s)}$, so any ${\phi: A\rightarrow B}$ gives us ${F(B)\rightarrow F(A)}$ by composing ${(B\stackrel{f}{\rightarrow} k_s)\mapsto (A\stackrel{\phi}{\rightarrow} B\stackrel{f}{\rightarrow} k_s)}$. Of course a morphism in ${G-set}$ must respect the ${G}$-action, but this is true by construction.

We must check that we have a bijection on Hom sets. Suppose we have a ${G}$-homomorphism ${F(B)\rightarrow F(A)}$, i.e. ${Hom_k(B, k_s)\rightarrow Hom_k(A, k_s)}$. We’ll show that ${Hom(F(B),k_s)\simeq B\otimes k_s}$ and ${Hom(F(A),k_s)\simeq A\otimes k_s}$. Thus applying ${Hom(-,k_s)}$ to both sides gives us a map ${A\otimes k_s\rightarrow B\otimes k_s}$. Keeping track of the action we can take the invariants to get ${(A\otimes k_s)^G=A\rightarrow (B\otimes k_s)^G=B}$. Thus from knowledge of ${F(B)\rightarrow F(A)}$ we can completely recover our map ${A\rightarrow B}$ which shows the functor is fully faithful.

The above argument requires us to keep careful track of the action to know it works. Let’s check the isomorphism ${G: A\otimes k_s\simeq Hom(F(A), k_s)}$. The map is given by ${a\otimes \lambda\mapsto f}$ where the map ${f(x)=x(a)\cdot \lambda}$ (evaluation on the first factor followed by multiplication). The action on the left is ${\sigma\cdot(a\otimes\lambda)=a\otimes \sigma(\lambda)}$ and the action on the right is conjugation ${\sigma\cdot f=f^\sigma}$. Let’s check equivariance of ${G}$. Consider ${G(\sigma\cdot(a\otimes\lambda))=x\mapsto x(a)\cdot \sigma(\lambda)}$
${=\sigma(\sigma^{-1}(x(a)\cdot\lambda)}$
${=f^\sigma (x)=\sigma\cdot G(a\otimes \lambda)}$. Thus the isomorphism preserves the ${G}$-action and we see the previous paragraph goes through.

Lastly we need to know the functor is essentially surjective. Let ${X}$ be an arbitrary ${G}$-set. Since ${X}$ is a disjoint union of its orbits and if ${X=F(A)\coprod F(B)}$, then ${X=F(A\times B)}$, we may assume without loss of generality assume the action of ${G}$ is transitive. We know that ${G}$ factors through ${Gal(L/k)}$ for some finite extension ${L}$. Let ${x\in X}$ so that the orbit of ${x}$ is all of ${X}$. The stabilizer of ${x}$ is some subgroup ${H}$ of ${G}$ and so we can define the fixed field ${A=L^H}$. Now we’re done, because ${F(A)=Hom_k(A, k_s)=Hom_k(A, L)}$ and the ${G}$-action is transitive by Galois theory. Thus the map ${X\rightarrow F(A)}$ determined by ${x\mapsto (A\hookrightarrow L)}$ is a ${G}$-isomorphism.

Our functor ${F}$ is fully faithful and essentially surjective and hence is an anti-equivalence of categories.

# Finite Groups as Galois Groups

So my old proof isn’t really working on wordpress for some reason, so I’ve taken it as a sign to do it in a different way. This method is far more complicated than the old way (in which I just call upon some theorems and look at orders and then am done), but I think it better gets at what is going on.

Anyways, since we were on the topic of Galois theory, here is a fact I found astonishing the first time I heard it (maybe it is quite obvious to you). Every finite group arises as the Galois group of a field extension, moreover we can choose the two fields to be number fields. Recall that a number field is just a subset of the complex numbers that is algebraic over $\mathbb{Q}$.

Proof: Let G be a finite group. Then by Cayley’s theorem $G\cong H< S_n$ for some n. But then there is a prime p, such that n<p, meaning $S_n. Let’s find a Galois extension $K/\mathbb{Q}$ such that $S_p\cong Gal(K/\mathbb{Q})$.

Our natural choice is the splitting field of $f(x)=x(x^2+1)(x^2-1)(x^2-4)\cdots (x^2-m^2)+1/p$, (note the sarcasm) where we chose our prime to be of the form $p=2m+3$. Thus we have that $deg(f(x))=p$, it is irreducible by Eisenstein’s criterion, and it has exactly 2 roots in $\mathbb{C}\setminus\mathbb{R}$. Thus, $Gal(K/\mathbb{Q})\cong S_p$. The details of this just amounts to playing around with cycles.

Now we can just invoke the Fundamental Theorem of Galois Theory. We have the subgroup $H < S_p$ which corresponds to the fixed field, say L, where $K\supset L\supset \mathbb{Q}$ and $Gal(K/L)\cong H\cong G$, which is what we wanted.

Now that this is typed out, I think the other way is better since I didn’t have to skip over the cycle argument. This method is just as mysterious overall, but in fact the other is probably more believable, since I only use things that are standard.

Lastly, let me just clear up some mystery about $f(x)$ at least. The way we know it has $2m+1$ real roots is that $f(x)$ alternates signs between pairs $-m-1/2, -m+1/2$, then $-m+3/2, -m+5/2$, all the way to $m-1/2, m+1/2$. Thus by the IVT, there is a real root between each of those. Thus there are 2 complex roots left over. We can even locate them to be near $\pm i$ if p is sufficiently large by Rouche’s Theorem. Also, $f(x)$ is irreducible iff $pf(x)$ is irreducible, and $pf(x)=px(x^2+1)(x^2-1)\cdots (x^2-m^2)+1$ and here it is clearly seen that p divides all the coefficients except the constant term and $p^2$ does not divide the highest term (Eisenstein).

Oh, and while I’m at it, I might as well lead you in the right direction if you want to check that $Gal(K/\mathbb{Q})\cong S_p$. By order, we know that $Gal(K/\mathbb{Q})\leq S_p$ (at least there is an isomorphic copy in there). So let the copy of $Gal(K/\mathbb{Q})$ act on $S_p$. The action is transitive, so there is a 2-cycle, and since p divides the order there is a p-cycle. These three things should get you there (that it is inside $S_p$, that there is a 2-cycle, and that there is a p-cycle).

So although they exist, they aren’t always the easiest to find. In fact, if we want our base field to be $\mathbb{Q}$, then this is known as the “Inverse Galois Problem” and is still open. Some cases have been resolved. For instance, it is known that every finite solvable group arises as a Galois group of an algebraic extension of $\mathbb{Q}$.

# Descent Theory

I’m going to do a change in plan.

Galois Theory: Let F be a field. In some sense the “universal” Galois group is $Gal(\overline{F}/F)$ where $\overline{F}$ is the algebraic closure, since given any algebraic extension $K/F$ we have that $Gal(\overline{F}/K) < Gal(\overline{F}/F)$. In fact, there is a bijective correspondence between subgroups of the Galois group and algebraic extensions (this is just loosely speaking to show a connection later on, I’m not being careful about finiteness and things). In this case the we have an inverse corrolation. As the fields get bigger, the groups get smaller.

Covering spaces: For suggestive notation, let’s denote $Y/X$ to mean Y is a covering of X. Then if X has sufficiently nice conditions, we have that there is a universal cover $\overline{X}/X$ with covering map $q: \overline{X}\to X$. Then we have that $Aut_q(\overline{X})\cong \pi_1(X)$ where $Aut_q(\overline{X})$ is the group of “deck transformations,” i.e. the automorphisms $\phi: \overline{X}\to\overline{X}$ such that $q\circ \phi= q$. Now any other cover will “sit below” the universal one, in that the covering $p: Y\to X$ will have a factoring $\overline{X}\to Y\to X$. Moreover $Aut_p(Y)\cong H<\pi_1(X)$. Just as in the Galois case, there is a bijective correspondence between conjugacy classes of subgroups of $\pi_1(X)$ and isomorphism classes of coverings. This time in a sense it is not reversed, though it depends on how you want to look at it.

I found the similarities of these two situations very strange. There must be something deeper. All field extensions are in bijective correspondence to subgroups of the Galois group of the “largest one,” and all (iso classes of) coverings are in bijective correspondence with (up to conjugacy) subgroups of the fundamental group.

It turns out that after some hunting, there is a huge deep field called “the theory of descent” or something similar. It all looks so fascinating, but it is just too far astray from what I’m studying for me to actually learn right now. I thought I could dip a toe in or something and report back my findings, but there doesn’t seem to be any good introductions to the subject or any hope for quickly seeing some of the ideas. So, after a few days of hunting, I’m changing my plans and am going to look for something new to go on about (possibly back to the prime and localization that I built up, then left for dead?).

Actually, if anyone knows of a place to learn some of this stuff, it would be greatly appreciated if you let me know!

# Generalized HT90

I officially promise this is my last post on Hilbert’s Theorem 90, but because of that it is going to go really fast for those who have not seen group cohomology (it is really cool, so I couldn’t pass it up).

An abelian group is a G-module (G a group) if for all $\sigma\in G$ and $a\in A$ there is a unique element $\sigma(a)\in A$ satisfying two conditions: $\sigma(a+b)=\sigma(a)+\sigma(b)$, and $(\sigma\tau)(a)=\sigma(\tau(a))$ for all $\sigma, \tau\in G$ and $a,b\in A$.

Just check any algebra text or here for more information on modules.

Now define an n-cochain of G over A to be a a function of n variables from G into A. If $n=0$ it is just an element of A. $C^n(G, A)$ is the set of all n-cochains, and can be made into a group by the operation $(f+g)(\sigma_1, \ldots, \sigma_n)=f(\sigma_1, \ldots, \sigma_n)+g(\sigma_1, \ldots, \sigma_n)$.

We can also get from $C^n(G, A)$ to $C^{n+1}(G, A)$ (which is what people who know about cohomology were hoping for), by the function $\delta$:

$(\delta f)(\sigma_1, \ldots, \sigma_{n+1})=\sigma_1(f(\sigma_2, \ldots, \sigma_{n+1}))+$

$\sum_{i=1}^n(-1)^i f(\sigma_1, \ldots, \sigma_i\sigma_{i+1}, \ldots, \sigma_{n+1})+(-1)^{n+1}f(\sigma_1, \ldots, \sigma_n)$.

OK. That looks bad, but really in some sense it is the natural choice. I’ll leave it to you to check that this is both a homomorphism and that $\delta\delta f=0$ (i.e. we have a chain complex).

Now if we label them $\delta_0: A\to C^1(G, A)$

$\delta_1: C^1(G, A)\to C^2(G, A)$ and

$\delta_n: C^n(G, A)\to C^{n+1}(G, A)$. Then we form $\displaystyle H^n(G, A)=\frac{\ker\delta_n}{im\delta_{n-1}}=\frac{Z^n(G, A)}{B^n(G, A)}$. We call the elements of $Z^n(G, A)$ the n-cocycles and the elements of $B^n(G, A)$ the n-coboundaries.

So if you don’t like that, we can scratch it now, since in HT90 we only care about $H^1(G, A)$, so let’s take a closer look at that. We can completely classify what the elements of $Z^1(G, A)$ look like. For any $f\in Z^1(G, A)$ and any $\sigma, \tau\in G$ we need $(\delta f)(\sigma, \tau)=\sigma(f(\tau))-f(\sigma\tau)+f(\sigma)=0$. Which is to say that $f(\sigma\tau)=\sigma(f(\tau))+f(\sigma)$.

Now let’s classify what $B^1(G,A)$ looks like. Well, if $g\in B^1(G, A)$, then $g=\delta h$ for some $h\in A=C^0(G,A)$. So $g(\sigma)=(\delta h)(\sigma)=\sigma(a)-a$ for some $a\in A$. Well, I think you might be able to see the previous formulation of the theorem coming from unravelling these definitions.

Theorem statement: If $K/F$ is a finite Galois extension and $G=Gal(K/F)$, then $H^1(G, K^\times)$ is trivial.

Proof: Let a be a cocycle. Then let $\alpha: K\to K$ by $c\mapsto \sum_{\sigma\in G}a(\sigma)\sigma(c)$. As in the last post, $\alpha$ is not 0 by linear independence. So let $c\in K$ such that $\alpha(c)\neq 0$ and set $b=\alpha(c)$.

Then for any $\tau\in G$ we have

$\displaystyle\tau(b)=\sum_{\sigma\in G}\tau(a(\sigma)\sigma(c))$

$\displaystyle =\sum_{\sigma\in G}\tau(a\sigma)(\tau\sigma)(c)$

$\displaystyle =\sum_{\sigma\in G}a(\tau)^{-1}a(\tau\sigma)(\tau\sigma)(c)$. Now we use that $a$ is a cocycle (in the kernel) to continue the equality as

$\displaystyle = a(\tau)^{-1}\sum_{\sigma\in G}a(\tau\sigma)(\tau\sigma)(c)$

$=a(\tau)^{-1}b$.

Aha, so $a(\tau)=b\tau(b)^{-1}$ is a coboundary! Thus every cocycle is a coboundary, so the quotient is trivial.

Test your understanding by now trying to prove the other formulation as a corollary to this (remember you assume that G is cyclic in that version and have to relate it back to the norm).

# Hilbert’s Theorem 90…the math

So now that I’ve presented the origins of Hilbert’s Theorem 90, I thought it might be good to actually present it mathematically with a proof.

Statement: Suppose K is a finite Galois extension of F with $G=Gal(K/F)$ cyclic, say $G=<\sigma >$ of order n. If $a\in K$, then $N_{K/F}(a)=1$ if and only if $a=b/\sigma(b)$ for some $b\in K$.

Let’s parse this a little before the proof. First off let’s define the notation $N_{K/F}(a)$. Notice that we can view K as a vector sapce over F, so $t_a: K\to K$ by $u\mapsto au$ is a linear transformation. So we want a norm, so we define $N_{K/F}(a)=\det (t_a)$. i.e. the determinant of the matrix that represents that linear transformation. Thus, we get all our nice results from linear algebra like independence of basis choice. If we fight with definitions and linear algebra for a bit we also get that if the extension is Galois, $\displaystyle N_{K/F}(a)=\prod_{\phi\in Gal(K/F)} \phi (a)$. This is left as an (not required) exercise to familiarize yourself with the terms before moving on. Note that we can determine $Tr_{K/F}(a)$ similarly to be the trace of $t_a$.

Proof of the theorem: Let’s do the backwards way first (I’m going to suppress the norm notation since we know where it is happening). Suppose $a=b/\sigma(b)$ for some $b\in K$. Then let’s simply use the result (that you proved!) to see that $\displaystyle N(a)=\left(\frac{b}{\sigma(b)}\right)\left(\frac{\sigma(b)}{\sigma^2(b)}\right)\cdots \left(\frac{\sigma^{n-1}(b)}{\sigma^n(b)}\right)=1$ since $\sigma$ has order n.

Now for the forwards direction. Suppose $N(a)=1$. Again using a standard linear algebra argument we know that $1, \sigma, \ldots, \sigma^{n-1}$ are K-linearly independent (in fact, any set of field automorphisms of K are K-linearly independent over the vector space of all functions from K to K).

Thus, $\displaystyle \phi = a\cdot 1+(a\sigma(a))\sigma + (a\sigma(a)\sigma^2(a))\sigma^2+\cdots + \left(\prod_{i=0}^{n-1}\sigma^i (a)\right)\sigma^{n-1}$ from K to K is not 0. By the thing you proved the coefficient on $\sigma^{n-1}$ is $N(a)$ which is 1. More importantly, there is some $c\in K$ such that $\phi(c)\neq 0$, so let’s define $b=\phi(c)$.

But $\sigma(b)=\frac{1}{a}(b-ac)+\sigma^n(c)=\frac{1}{a}(b-ac)+c=\frac{b}{a}$. Thus $a=b/\sigma(b)$. And we are done.

Does this proof remind anyone of Lagrange resolvents? Eh, whatever, I won’t dig for a connection now.

I also mentioned that there is a generalization of this.

Statement: If K/F is a finite Galois extension and $G=Gal(K/F)$, then $H^1(G, K^\times)$ is trivial. I’m debating whether or not to parse and prove this version next time, or to just drop the Hilbert Theorem 90 posts. Suggestions?