Posted by Jason Polak on 22. December 2015 · Write a comment · Categories: algebraic-geometry, commutative-algebra · Tags: ,

Let $F$ be a field and $E/F$ be a nontrivial Galois extensions with Galois group $\Gamma$. If $V$ is an $F$-scheme then the points $V(E)$ carry a natural action of $\Gamma$ via the action on $\mathrm{Spec}(E)$. Sometimes, however, $V$ might have two Galois actions. How does this arise?

Perhaps the most natural setting is when $V$ is defined using the restriction of scalars functor. For example, if $X$ is an $E$-scheme, then $V = \mathrm{Res}_{E/F}(X)$ is by definition the scheme whose points in an $F$-algebra $R$ are given by
$$V(R) = X(R\otimes_F E).$$
Then, for any such $R$, the set $V(R)$ has a natural action of $\Gamma$ acting on $E$ in the tensor product $R\otimes_F E$. On the other hand, if $R$ is an $E$-algebra, then $V(R)$ will also have a $\Gamma$-action, via the action of $\Gamma$ on $R$. And, these two actions won't be the same!
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Posted by Jason Polak on 17. September 2015 · Write a comment · Categories: algebraic-geometry · Tags: , ,

One policy of Aleph Zero Categorical is that any lecture notes posted on the arXiv that I manage to see will be announced and advertised here. Today I saw that Lars Kindler and Kay Rülling have posted their notes entited:

I quote from the introduction for a summary of these notes:

These are the notes accompanying 13 lectures given by the authors at the
Clay Mathematics Institute Summer School 2014 in Madrid. The goal of
this lecture series is to introduce the audience to the theory of $\ell$-adic sheaves
with emphasis on their ramification theory. Ideally, the lectures and these
notes will equip the audience with the necessary background knowledge to
read current literature on the subject, particularly [16], which is the focus of
a second series of lectures at the same summer school. We do not attempt
to give a panoramic exposition of recent research in the subject.

Here, reference 16 refers to the paper "A finiteness theorem for Galois representations of function fields over finite fields (after Deligne)" by Esnault and Kerz. Here is one last quotation that gives the prerequisites for these notes:

The text can roughly be divided into two parts: Sections 2 to 4 deal
with the local theory and only assume a basic knowledge of commutative
algebra, while the following sections are more global in nature and require
some familiarity with algebraic geometry.

The algebraic geometry in these notes is actually quite gentle: they define etale morphism and state basic properties, and the define the etale fundamental group and even prove things about it. Not only that, but these notes have an index!

Posted by Jason Polak on 25. August 2015 · Write a comment · Categories: algebraic-geometry, number-theory · Tags: ,

In the Arthur-Selberg trace formula and other formulas, one encounters so-called 'orbital integrals'. These integrals might appear forbidding and abstract at first, but actually they are quite concrete objects. In this post we'll look at an example that should make orbital integrals seem more friendly and approachable. Let $k = \mathbb{F}_q$ be a finite field and let $F = k( (t))$ be the Laurent series field over $k$. We will denote the ring of integers of $F$ by $\mathfrak{o} := k[ [t]]$ and the valuation $v:F^\times\to \mathbb{Z}$ is normalised so that $v(t) = 1$.

Let $G$ be a reductive algebraic group over $\mathfrak{o}$. Orbital integrals are defined with respect to some $\gamma\in G(F)$. Often, $\gamma$ is semisimple, and regular in the sense that the orbit $G\cdot\gamma$ has maximal dimension. One then defines for a compactly supported smooth function $f:G(F)\to \mathbb{C}$ the orbital integral
\Ocl_\gamma(f) = \int_{I_\gamma(F)\backslash G(F)} f(g^{-1}\gamma g) \frac{dg}{dg_\gamma}.
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Suppose we have a $2\times 2$ matrix
M = \begin{pmatrix}
x_{11} & x_{12}\\
x_{21} & x_{22}
with entries in a field $F$. The characteristic polynomial of this matrix is $p(t) := {\rm det}(tI_2 – M) = t^2 – (x_{11} + x_{22})t + x_{11}x_{22} – x_{21}x_{12}$. One might ask: how can we produce a matrix with a given characteristic polynomial? This can be accomplished using the rational canonical form:
t^2 + at + b\mapsto
0 & -b\\
1 & -a
We can calculate that the characteristic polynomial of this matrix to be $t^2 + at + b$. This map gives a bijection between quadratic monic polynomials in $F[t]$ and matrices of the above form. One way to understand this phenomenon is through algebraic groups. To explain, let's stick with $F$ having characteristic zero, though much of what we do can be done in characteristic $p$ for $p$ sufficiently large as well using very different techniques.
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Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: algebraic-geometry, group-theory · Tags:

Suppose one day you run into the following algebraic group, defined on $\mathbb{Z}$-algebras $R$ by
$$G(R) = \left\{ \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{12} & a_{11} & -a_{14} & -a_{13} \\
a_{13} & -a_{14} & a_{11} & -a_{12} \\
a_{14} & -a_{13} & -a_{12} & a_{11}
\end{pmatrix} \in\mathrm{GL}_4(R) \right\}$$
Can you figure out what this group is? I actually did run into this group one day, but luckily I discovered its true identity. Today we'll see one way to do so.

One thing to try to identify unknown algebraic groups in practice is to check out their matrix multiplication. If $A = (a_{ij}), B = (b_{ij})$ have the above form then $C = AB$ has first row
c_{11} = a_{11} b_{11} + a_{12} b_{12} + a_{13} b_{13} + a_{14} b_{14} \\c_{12}= a_{12} b_{11} + a_{11} b_{12} – a_{14} b_{13} – a_{13} b_{14} \\c_{13}= a_{13} b_{11} – a_{14} b_{12} + a_{11} b_{13} – a_{12} b_{14} \\c_{14}= a_{14} b_{11} – a_{13} b_{12} – a_{12} b_{13} + a_{11} b_{14}
(We only need to specify the first row to get the whole matrix in this group). Can you tell what the group is yet? At first, I couldn't, so let's try something else. One can check now that this group is actually commutative, and this would be enough to determine what it is given the context of where it came from, but let's assume we don't know that. Instead, let's take the determinant:
\mathrm{det}(A) = (a_{11} + a_{12} + a_{13} – a_{14})(a_{11} + a_{12} – a_{13} + a_{14})\\\times(a_{11} – a_{12} + a_{13} + a_{14})(a_{11} – a_{12} – a_{13} – a_{14})$$
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Let $ R$ be an integral domain and $ K$ is fraction field. If $ K$ is finitely generated over $ R$ then we say that $ R$ is a $ G$-domain, named after Oscar Goldman. This innocuous-looking definition is actually an extremely useful device in commutative algebra that pops up all over the place. In fact, the $ G$-domain usually comes up in the context of quotienting by a prime ideal, so let's call a prime $ P$ in $ R$ a $ G$-ideal if $ R/P$ is a $ G$-domain. In this post, we shall see a few applications of this concept, following Kaplansky's book Commutative Rings" for the theory and some standard examples for illustrations. At the end, we shall see a short paragraph proof of the Nullstellentsatz assuming the theory of $ G$-ideals.

Why is this concept so useful? Perhaps because of the following result: an ideal $ I$ in $ R$ is a $ G$-ideal if and only if it is the contraction of a maximal ideal in $ R[x]$ (although we won't use it, it's interesting to note that the nilradical is actually the intersection of all the $ G$-ideals). It's worth looking at one direction of a proof of this result since it's so short. First, the reader should prove that $ K$ can be generated by one element over $ R$ if and only if it is finitely generated, as an exercise.
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Posted by Jason Polak on 03. August 2013 · Write a comment · Categories: algebraic-geometry, group-theory · Tags: ,

Conventions: $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.

Consider the algebraic groups $ \mathbb{G}_a$ and $ \mathbb{G}_m$. They are the only one-dimensional connected groups and they are both solvable. What about two-dimensional connected groups? It turns out that if $ \mathrm{dim} G\leq 2$ then $ G$ is solvable.

For $ \mathrm{dim} G= 3$, this is no longer true, for instance $ \mathrm{SL}_2$ is $ 3$-dimensional but not solvable since it is perfect, i.e. equal to its commutator subgroup. So let's prove our theorem:

Theorem. Let $ G$ be a connected algebraic group over $ k$. If $ \mathrm{dim} G = 2$ then $ G$ is solvable.

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Continuing our previous series, $ G$ is an algebraic group over an algebraically closed field $ k$ and we identify $ G$ with $ G(k)$.

Here is an interesting fact:

Theorem. In a connected solvable group the unipotent part $ G_u$ is a closed connected normal subgroup of $ G$ and contains the commutator subgroup $ [G,G]$.

Why is this interesting? It will allow us to prove that if $ T\subseteq G$ is a normal torus in a connected group $ G$, and $ G/T$ is also a torus then $ G$ itself is a torus!
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The previous series on algebraic groups is over. Actually, I barely got to the root system and root datum of a reductive group, but I found that the whole slew of material was getting too complex to organise on this blog, which I feel is better for more self-contained posts. Instead, I have begun to write a set of notes in real LaTeX describing the root datum of a reductive group and root systems in general, most of which I have completed, and I will release a draft soon.

In the mean time, I will continue posts on algebraic groups by reviewing some theorem and then illustrating it with an application such an an exercises from some textbook. Let's start by reviewing a simple one: let $ k$ be an algebraically closed field. Then an irreducible finite set in the Zariski topology on $ k^n$ (as a classical variety) necessarily has one element. (A topological space is irreducible if it cannot be written as the union of two proper closed subsets.)

We added the disclaimer "as a classical variety" since over an algebraically closed field $ k$ it is sufficient to consider the closed points of $ \mathrm{Spec}(k[x_i])$—otherwise the statement would be false.
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Posted by Jason Polak on 07. June 2013 · Write a comment · Categories: algebraic-geometry, group-theory · Tags: , ,

IMG861From Highlights 12 and Highlights 13, we have gained quite a bit of information on connected reductive groups $ G$ of semisimple rank 1. Recall, this means that $ G/R(G)$ has rank 1 where $ R(G)$ is the radical of $ G$, which is in turn the connected component of the unique maximal normal solvable subgroup of $ G$.

But wait, why have we been looking at groups of semisimple rank 1 at all? Let's take a quick look at how we can get a good source of this groups inside a general group $ G$.


Let $ G$ be a connected algebraic group over $ k=\overline{k}$. We are not assuming that $ G$ is reductive or anything else besides this. We start by dividing up the tori of $ G$ into two kinds: the regular tori and the singular tori. Both of these species will be important in our study of algebraic groups.
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