# Abstract Nonsense

## 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.

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Motivation

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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.

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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 $(5)$).

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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.

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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.

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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).

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Algebraic Extension

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Let $k/F$ be an extension, we call $\alpha\in k$ algebraic if there exists some $f(t)\in F[t]$ such that $f(\alpha)=0$. Equivalently, we say that $\alpha$ is algebraic if the guarnateed $F$-algebra map $\text{ev}_\alpha:F[t]\to k$ sending $t\mapsto \alpha$ (named as such because it really is just evaluation, $f(t)\mapsto f(\alpha)$)  has a non-trivial kernel. If $\alpha\in k$ is not algebraic we call it transcendental.

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Let’s take a look at some examples of algebraic and transcendental elements of some extensions.

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Clearly $\sqrt{2}$ is algebraic in the extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ since $f(\sqrt{2})=0$ where $f(t)=t^2-2$.

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Consider the extension $F(t)/F(t^2)$ where $F(t)$ is the rational function field and $F$ is some field. We note then that $t$ is algebraic since it satisfies $x^2-t^2\in F(t^2)[x]$.

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The element $t$ in the extension $F(t)/F$ is transcendental since the map $F[x]\to F(t)$ with $x\mapsto t$ is nothing more than the identity map on $F[t]$ and thus injective.

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It takes a fair amount of work but one can show that in the extension $\mathbb{C}/\mathbb{Q}$ one has that $e$ and $\pi$ are transcendental. That said, it’s still an open question as to whether $e$ is a transcendental element of the extension $\mathbb{C}/\mathbb{Q}(\pi)$.

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The first fundamental results about algebraic elements in some field extension is that they satisfy some monic polynomial of minimal degree. For example, $i$ is an algebraic element of the extension $\mathbb{C}/\mathbb{R}$ and while $i$ satisfies many polynomials in $\mathbb{R}[t]$ it satisfies one of minimal degre–namely $t^2+1$. Moreover, we know that if $f(t)\in\mathbb{R}[t]$ annihilates $i$ then $f(t)$ is divisible in $\mathbb{R}[t]$ by $t^2+1$. What we alluded to in the first sentence of this paragraph is that, in general, an algebraic element has such a unique monic polynomial of minimal degree.

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To see why this is true suppose that $k/F$ is an extension and $\alpha\in k$ is algebraic. Then, we know that the $F$-map $\text{ev}_\alpha:F[t]\to k$ with $t\mapsto\alpha$ has some non-trivial kernel sitting inside $F[t]$. But, since $F$ is a field we know that $F[t]$ is a PID and so $\ker\text{ev}_\alpha=(m(t))$  for some monic polynomial $m(t)$. Moreover, this polynomial has the property that if $f(t)\in F[t]$ annihilates $\alpha$ then $m(t)\mid f(t)$, since we know that $f(t)\in\ker \text{ev}_\alpha=(m(t))$ so that $f(t)$ is a multiple of $m(t)$. Of course, this is the unique such polynomial satisfying these properties and so we can unambiguously define it to be the minimal polynomial of $\alpha$, which we shall denote as $m_{\alpha,F}(t)$ or $m(\alpha,F)(t)$. If it’s clear the ground field we are consider then we shall often times just shorten this to $m_\alpha$.

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References:

[1] Morandi, Patrick. Field and Galois Theory. New York: Springer, 1996. Print.

[2] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[3] Lang, Serge. Algebra. New York: Springer, 2002. Print.

[4] Conrad, Keith. Collected Notes on Field and Galois Theory. Web. <http://www.math.uconn.edu/~kconrad/blurbs/&gt;.

[5] Clark, Pete. Field Theory. Web. <http://math.uga.edu/~pete/FieldTheory.pdf&gt;.