# This Week’s Finds in Arithmetic Geometry

It has been going around the math blogosphere that in honor of John Baez’s 20 year anniversary of doing This Week’s Finds we all do one in our area. So here’s a brief This Week’s Finds in Arithmetic Geometry. Hopefully this will raise awareness of the blog that essentially pioneered math/physics blogging (and if you’re into arithmetic geometry some papers you might not have caught yet).

Since the content and style of Baez’s This Week’s Finds vary so much, I’ll just copy what Jordan Ellenberg did here and give some papers posted the last week that caught my attention.

The fields of definition of branched Galois covers of the projective line by Hilaf Hasson caught my attention because I just met him and saw him speak on this exact topic a few weeks ago at the Joint Meetings. Although the results are certainly interesting in themselves, the part that a blog audience might appreciate is the set of corollaries to the results.

Recall that a major open problem in number theory is the “Inverse Galois Problem” which asks which groups arise as Galois groups. I even posted an elementary proof that if you don’t care what your fields are, then any finite group arises as ${Gal(L/K)}$. In general (for example if you force ${K=\mathbb{Q}}$), then the problem is extremely hard and wide open.

If you haven’t seen this type of thing before, then it might be surprising, but you can actually use geometry to study this question. This is exactly the type of result that Hilaf gets.

Next is New derived autoequivalences of Hilbert schemes and generalised Kummer varieties by Andreas Krug. This topic is near and dear to me because I study derived categories in the arithmetic setting. I haven’t taken a look at this paper in any depth, but I’ll just point out why these types of things are important.

In the classification of varieties one often tries to study the problem up to some type of birational equivalence otherwise it would be too difficult. Often times birational varieties are derived equivalent, but not the other way around. So one could think of studying varieties up to derived equivalence as a slightly looser classification.

When trying to figure out what two varieties that are derived equivalent have in common, a typical sticking point is that you need to know certain automorphisms of the derived category (i.e. autoequivalence) exist to get nice cohomological properties or something. When papers constructing new autoequivalences come out it always catches my attention because I want to know if the method used transfers to situations I work in.

Lastly, we’ll do La conjecture de Tate entière pour les cubiques de dimension quatre sur un corps fini by François Charles and Alena Pirutka. If you don’t know what the Tate conjecture is, then lots of people refer to it as the Hodge conjecture in positive characteristic.

If you’ve ever seen any cohomology theory, then you should be at least passingly familiar with the idea that certain sub-objects (subvarieties or submanifolds etc) can be realized as classes in the cohomology. Sometimes this is due to construction and sometimes it is a major theorem.

The particular case of the Tate conjecture says the following. Consider the relatively easy to prove fact. If you take a cycle on your variety ${X/k}$, then the cohomology class it maps to (in ${\ell}$-adic cohomology) will be invariant under the natural Galois action ${Gal(\overline{k}/k)}$ (because it is defined over ${k}$!). The Tate conjecture is that any Galois invariant cohomology class actually comes from one of these cycles.

The fact that mathematicians can have honest arguments over whether or not the Tate conjecture or the Hodge conjecture (a million dollar problem!) is harder just gives credence to the fact that it is darned hard. If you weren’t convinced, then just consider that this paper is proving the Tate conjecture in the particular case of smooth hypersurfaces of degree ${3}$ in ${\mathbb{P}^5}$ just for the cohomology classes of degree ${4}$. People consider this progress, and they should.

# 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 $\mathbb{Q}$.