# Abstract Nonsense

## Separable Extensions (Pt. I)

Point of Post: In this post we discuss separable extensions and the separable closure of an extension.

$\text{ }$

Motivation

$\text{ }$

Last time we discussed normal extensions which were the nice kind of extensions which (among several other definitions) are just the splitting field for a set of polynomials over the ground field. In this post we’d like to discuss another kind of “nice” extension, namely separable extensions. Roughly a separable extension is one where there are multiple roots which become indistinguishable (inseparable, if you will) to the Galois group of the extension. For example, consider the finite field $\mathbb{F}_p$ and the rational function field $\mathbb{F}_p(t)$. Then, we claim that $\mathbb{F}_p(t)/\mathbb{F}_p(t^p)$ is a simple extension of degree $p$. Indeed, it’s a simple extension since $\mathbb{F}_p(t)=\mathbb{F}_p(t^p)(t)$. To see that it’s degree $p$ it suffices to show that $u^p-t^p\in\mathbb{F}_p(t)[u]$ is irreducible (since it annhilates $t$). To do this we merely note that if $u^p-t^p$ factors in $\mathbb{F}_p(t^p)[u]$ it factors in $\mathbb{F}_p(t)[u]$ but we already know the irreducible factorization there–$u^p-t^p=(u-t)^p$. Clearly then (since everything in sight if a UFD) the factorization in $\mathbb{F}_p(t)[u]$ must look like $(u-t)^m(u-t)^n$ for some $m,n$. But this, in particular, implies that $t^m\in\mathbb{F}(t^p)$ which clearly says that $m=p$ and thus our factorization was trivial.  Ok, so what? The important thing to note is that while $[\mathbb{F}_p(t):\mathbb{F}_p(t^p)]=p$ you can easily verify that $\text{Gal}(\mathbb{F}_p(t)/\mathbb{F}_p(t^p))$ is trivial since they are just permutations of the roots of $u^p-t^p=(u-t)^p\in\mathbb{F}_p(t)[u]$ [which, in case this isn’t obvious, there is only one such root].

$\text{ }$

Similar to the case of $\mathbb{Q}(\sqrt[3]{2})$ we see that the Galois group of our extension has order strictly less than the degree of our extension. The key observation though is that it is for a fundamentally different reason. The reason that $|\text{Gal}(\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q})|<[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ is an issue of normality [in the sense of normal extensions]–the polynomial, for which the Galois group is permuting its roots, doesn’t have all of its root in the field. In our case though the polynomial the Galois group is permuting roots of has all of its roots in the field, the problem is that it has repeated roots. The Galois group’s raison d’etre is to permute the roots the best it can, but when there are repeated roots the Galois group has a difficulty doing this because it can’t distinguish (separate) the repeated roots.

$\text{ }$

Thus, our topic of this thread will be those fields which do not impede the Galois group in doing its job, or at least do not impede it in the way described above.

$\text{ }$

May 4, 2012

## Normal Extensions

Point of Post: In this post we shall discuss normality for extensions.

$\text{ }$

Motivation

$\text{ }$

This post is where we start admitting to a god-given truth–not all field extensions are created equal. Obviously this could be interpreted as saying not all extensions are isomorphic, but this is too strong. Instead of going head-long at the isomorphism problem for extensions I’d rather first discuss some simple properties that extensions have that are isomorphism invariants. Namely, properties that an extension does or does not enjoy, that shall enable us to tell extensions apart.

$\text{ }$

In this post we discuss one of these invariants, the notion of normality. Intuitively normality is the failure for a given extension to be a splitting field of some polynomial. For example, we have discussed before the field $\mathbb{Q}(\sqrt[3]{2})$. This is a perfectly fine extension, but it is NOT a splitting field. Why? Because splitting fields $L$ of some polynomial $f_0(x)\in F[x]$ have the very nice property that if you give me an irreducible $f(x)\in F[x]$ that has a root in $L$ then $f(x)$ has all its roots in $L$. We see then that $\mathbb{Q}(\sqrt[3]{2})$ is definitely not a splitting field since $x^3-2\in\mathbb{Q}[x]$ is irreducible, has the root $\sqrt[3]{2}$ contained in itself, yet does not have the other two roots of $x^3-2$ contained in itself.

$\text{ }$

Of course, there are other descriptions of normal extensions besides the two given above (being a splitting field/having the all-or-none roots policy for irreducible polynomials), but this is the property to keep in mind. Normality is one-half of the equation to defining Galois extensions which, in our very simple look at field theory, are the “perfect” (not in the technical sense) extensions.

$\text{ }$

April 30, 2012

## Maps of Extensions and the Galois Group (Pt. II)

Point of Post: This is a continuation of this post.

$\text{ }$

April 24, 2012

## Maps of Extensions and the Galois Group

Point of Post: In this post we discuss the morphisms in the “category of all extensions”.

$\text{ }$

Motivation

$\text{ }$

We have spent a fair amount of time talking about the structures that are field extensions, but we have yet to mention what the “structure preserving maps” between such extensions “should” be. Well, of course, it’s not exactly obvious how to define such maps, but if we stop thinking about an extension $k/F$ as being a field $k$ and a field $F$ with $k\supseteq F$ and instead think about it as an $F$-algebra $k$ which also happens to be a field then we know what to do. Namely, we already know what the morphisms in $F-\textbf{CAlg}$ (the category of commutative $F$-algebras) are. Namely, if $k,k'$ are two $F$-algebras then an $F$-algebra map $f:k\to k'$ is nothing more than a ring map $k\to k'$ which also respects the $F$-vector space structure (i.e. $f(\alpha k)=\alpha f(k)$ for all $\alpha\in F$).

$\text{ }$

That said, despite the fact that this is really the reason that we define maps of extensions the way we do, this is rarely said in basic books on the subject. Instead such books like to define a map $f:k\to k'$ of extensions $k/F,k'/F$ to be a ring map such that $f_{\mid F}=\text{id}$.  Of course, it’s easy to see that such definitions are equivalent. If $f$ is a map of extensions as defined in the first paragraph then necessarily $f(\alpha)=\alpha f(1)=\alpha$ for all $\alpha\in F$. Conversely, if $f$ is a map of extensions as just defined then $f(\alpha x)=f(\alpha)f(x)=\alpha f(x)$ for all $\alpha\in F$ and $x\in k$ (since $f(\alpha)=\alpha$ for all $\alpha\in F$ by assumption).

$\text{ }$

Being a map of extensions is a pretty strong condition. For example, we shall see that if $k/F$ is a finite extension then every map of extensions $k\to k$ is necessarily invertible. Moreover, in general mathematics one can construct a lot of information from an object by examining its automorphism group. This is (hyperbolically) nowhere more true than in the study of field extensions. We shall see that examining the automorphism group of an extension shall enable us to (more so in certain nice cases) read off certain extension-theoretic properties of our extension–this is the so-called field of Galois theory.

April 24, 2012

## Finite Fields

Point of Post: In this post we discuss the classification of finite fields, and discuss some other incidental properties relating to such.

$\text{ }$

Motivation

$\text{ }$

There are very few quick, clean, and satisfying classifications in mathematics. Among these though is the classification of finite fields. Now, while this may not seem like a very important class of fields, finite fields have stupendous importance in many areas of mathematics, perhaps most prominently in number theory. That said, for now we shall view them mostly as curiosities to test our knowledge against.

$\text{ }$

Ok, so why are finite fields so “curious”? What’s so interesting about them is that even though we know entirely what their additive and multiplicative structures look like up to isomorphism it is very difficult to construct them explicitly. For example, we shall show that there is a field $\mathbb{F}_4$ with four elements. Now, clearly $\mathbb{F}_4\cong\mathbb{Z}_2^2$ as groups (if this isn’t obvious, we shall prove this below) and also that $\mathbb{F}_4^\times\cong\mathbb{Z}_3$.  That said, try to sit down and construct a field with four elements just by writing out a multiplication table–I bet you will have a hard time.

$\text{ }$

In fact, there is only one surefire technique for constructing such a field. Indeed, suppose that we wanted a field of order $p^n$ where $p$ is some prime (as we shall see these are the only orders of fields), then to construct a field $k$ of order $p^n$ it seems sensible to construct a degree $n$ extension of the field $\mathbb{Z}_p$. Moreover, we have a good way to do this. Namely, if we were able to find a polynomial $f(x)\in\mathbb{Z}_p[x]$ which is irreducible and of degree $n$ then we know that $\mathbb{Z}_p[x]/(f)$ is a field extension of $\mathbb{Z}_p$ of degree $n$. That said, it’s difficult to actually find such a polynomial. For example, the problem of finding fields of order $4$ is simple because $f(x)=1+x+x^2$ is an irreducible (it is identically $1$ as a function) polynomial in $\mathbb{Z}_2[x]$ and so $\mathbb{Z}_2/(f)$ is such a field–but, these are really small numbers.

$\text{ }$

No, instead of trying to construct such finite fields explicitly we shall construct them as splitting fields of certain polynomials in $\mathbb{Z}_p[x]$.

$\text{ }$

April 12, 2012

## Splitting Fields and Algebraic Closures (Pt. II)

Point of Post: This is a continuation of this post.

April 5, 2012

## Splitting Fields and Algebraic Closures (Pt. I)

Point of Post: In this post we discuss the notion of splitting fields for sets of polynomials as well as the algebraic closure of a given field.

$\text{ }$

Splitting Fields and Algebraic Closures

$\text{ }$

Up until this point we have committed a bit of an atrocity–well, sort of. Continually we have been considering some field $F$ and some extension $k$ and saying “well, if $f(x)\in F[x]$ is some polynomial and $f(\alpha)=0$ with $\alpha\in k$ then we can consider the field $F(\alpha)$ generated by $F$ and $\alpha$ in $k$“. My point being that to consider things like $F(\alpha)$, which fits with our original goal to consider minimal fields over which we can discuss the roots of a polynomial, we actually have to start with/find a field containing a root. Unfortunately, up until this point we have failed to do this. We have done a lot of math assuming we already have such a field, but alas, none are to be found. In this post we remedy this by actually showing that for any polynomial $f\in F[x]$ we can find a field extension $k/F$ such that $f(\alpha)=0$ for some $\alpha\in k$ (i.e. such that it has a root in $k$) but that we can factor $f=(x-\alpha_1)\cdots(x-\alpha_n)$ (i.e. that it has all of its roots in $k$). It turns out that this is not too bad, but lurking quietly in the background is a much more interesting, and deep, question. Namely, the way in which we shall construct such a field for a given $f$ shall be simple-minded, but at the price of this simplicity comes a lack of “cohesion”. Namely, we shall have a difficult time trying to compare results about a polynomial $f$ to a polynomial $g$ by looking at the respective fields we constructed since, in a sense, they don’t related well together. What would be nice is if we didn’t have to look for these fields containing the roots of $f,g,h,\cdots$ separately, if instead we could think about all of these fields living inside some bigger (realistically huge) field. In other words, can we find a an extension $k/F$ such that $k$ contains all of the roots of all $f\in F[x]$? This is a somewhat sensitive question construction wise. In a perfect world we’d just take the “union” of all the fields we get by examining the polynomials individually. Alas, being the semi-rigorous minded students that we are, this just won’t do. We shall construct the algebraic closure by a sneaky trick first thought up by Artin.

$\text{ }$

April 5, 2012

## Algebraic Extensions (Pt. III)

Point of Post: This is a continuation of this post.

March 25, 2012

## Algebraic Extensions (Pt. II)

Point of Post: This is a continuation of this post.

March 25, 2012

## Algebraic Extensions (Pt. I)

Point of Post: In this post we discuss the notion of algebraic extensions, including minimal polynomials and the fact that the algebraic elements of an extension form a field.

$\text{ }$

Motivation

$\text{ }$

We have talked extensively about making new fields by appending new elements to the field. In this post we make an important distinction between what kind of elements can get appended to a field–roughly one being infinite in nature and the other being finite. Roughly the idea is when the new element we have appended to our field is a root of some polynomial over our original field. For example, $\sqrt{2}$ can be appended to $\mathbb{Q}$ and it satisfies the polynomial $x^2-2\in\mathbb{Q}[x]$. These objects shall be very important to us for several reasons.

$\text{ }$

First off they are the torsion analogues in the theory of fields. How exactly is this? In a literal way, every extension $k/F$ carries the structure of a $F[x]$ module via $f(x)\cdot \alpha=f(\alpha)$ and obviously the elements that satisfy some polynomial are exactly the torsion elements of this module. Secondly, these act as the torions elements when we try to describe our extension in terms of generators and relations (i.e. the quotient of some free algebra). To make this last part clear, let’s think about how we get a “presentation” for a cyclic group $\langle g\rangle=G$. We begin by recalling that we have a group homomorphism $\mathbb{Z}\to G$ which takes $1$ to the generator $g$. We see then that $G$ is free of relations if and only if this map is injective–if the kernel is trivial. But, often times the group $G$ is not free, in the sense that the generator $g$ satisfies some non-trivial relation (e.g. $5g=0$) in which case the kernel of $\mathbb{Z}\to G$ shall be a “relation set” (i.e. in the case $5g=0$ we’d have the kernel is $(5)$).

$\text{ }$

Let’s now perform this same procedure with some ‘simple’ extension of $F$, in other words an extension of the form $F(\alpha)$ for some $\alpha$. We take the free object in our case, which is just $F[t]$ (which is just the polynomial ring). Just as before we get an $F$-algebra map $F[t]\to F(\alpha)$ with $t\mapsto \alpha$. In general we see that $p(t)\mapsto p(\alpha)$ so that this map will have a trivial kernel precisely when $\alpha$ satisfies no polynomial in $F$, and it does have a trivial kernel if $\alpha$ does. Thus, comparing it to the case of groups we see that the “non free” extensions are those whose generators satisfy some kind of polynomial equation over the ground field.

$\text{ }$

The second reason that objects which satisfy polynomial equations are so important to us is that they are basically what finite extensions are comprised of. Indeed, we shall eventually see that finite extensions are nothing more than the field appended finitely many of these elements which satisfy polynomial equations.

$\text{ }$

Lastly, these objects are important to us because they were the main motivation for consider field extensions. Namely, we wanted to solve some problem in our original field, or in the polynomial ring of that field, and decided that we could make everything simpler if we just made the roots appear–if we just added them in. Of course, by literal definition, these appended things will satisfy polynomial equations. Thus, if we are able to get more information about field extensions which are just the ground fielded appended these “roots” we will be all the better off since this is our chief concern (as of now).

$\text{ }$

March 25, 2012