Abstract Nonsense

Field Extensions

Point of Post: In this post we discuss the notion of field extensions and plenty of examples.

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Motivation

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I’ve been a little behind with blogging recently, and since I’ve been doing a lot of Galois theory recently I thought I’d brush up on some of the basics by blogging about it. In this post we are going to discuss the fundamental objects in Galois theory–field extensions. Historically field extensions arose from the desire to extend a given field to obtain the roots of a (some) given polynomial(s). For example, the complex numbers $\mathbb{C}$ could be thought of (algebraically) as mathematician’s answer to the difficulty of the equation $x^2+1=0$ not having any solutions in the field $\mathbb{R}$. Indeed, one can algebraically think about $\mathbb{C}$ (as a ring) as nothing more than $\mathbb{R}[x]/(x^2+1)$ which can basically be described as taking $\mathbb{R}$ and appending a formal variable $\alpha$ to get a ring $\{a+b\alpha:a,b\in\mathbb{R}\}$ subject (besides the obvious ones) to the condition that $\alpha^2=-1$. Of course we have a different name for this $\alpha$, we call it $i$, and we call the result of appending $\mathbb{R}$ with $i$ the field $\mathbb{C}$ of complex numbers. More generally one may ask whether given a field $k$ and some polynomial $f(x)\in k[x]$ where we can find some “larger” field $K$ such that the polynomial $f(x)\in k[x]\subseteq K[x]$ has a root, or more brashly, can be factored into linear terms. The reason for this, besides pure intellectual curiosity, can be seen in the utility of using $\mathbb{C}$ in the study of the algebra of $\mathbb{R}$ and $\mathbb{R}[x]$. Indeed, many questions dealing exclusively with the real numbers or real polynomials can be reformulated into easier to understand, and often times, easier to solve problems involving $\mathbb{C}$. Just a small taste of this can be glimpsed by the notion of divisibility in $\mathbb{R}[x]$. For example It’s hard for me to tell if $x^2+1$ divides $x^5+13x^4+x^3+14x^2+1$ just by looking at it, but luckily I can quickly check that this latter polynomial has $i$ as a root in $\mathbb{C}$ from where (by general nonsense) I know that $x^2+1$ does indeed divide it. Of course, this is a paltry use of extending fields compared to the use that’s touted as one of the most beautiful results in mathematics (at least to mathematicians)–the insolvability of the general quintic. Roughly this “insolvability” says that the notion of a “quadratic equation” (or cubic, or quartic) does not exist for quintic equations. Why is it feasible that this problem should involve field extensions? What the problem is really stating is that we can’t start with the coefficients of our polynomial with (say) integer coefficients and express the roots of the polynomial by performing a finite number of sums, multiplications, and taking of roots (e.g. $\sqrt[n]{x}$). The statement is a non-existence theorem involving rational numbers and $n^{\text{th}}$ roots, but more importantly it’s discussing the roots of a certain polynomial. Thus, if we’d like to make any sort of formal workings with these roots we better have a field that contains them–thus field extensions.

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Field Extensions

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Remark: For all the posts on field theory/Galois theory we shall be working in $\mathbf{CRing}$ which is to say that our rings will have units, and more importantly, our maps need to respect $1$ (i.e. $1\mapsto 1$ under any ring map).

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Let’s begin by defining exactly what we mean by “field extension”. Let $F$ be some given field, then an extension of $F$ or a field extension over $F$ is a field $k$ along with an embedding (of rings) $F\hookrightarrow k$. Note though that the quantifier embedding is really quite unnecessary since any ring map out of field is necessarily injective. Thus, a field extension of $F$ is nothing but a field $k$ along with a distinguished ring map $F\to k$. In this way we see that field extension of $F$ are nothing more than $F$-algebras where we require the algebra to actually be a field.

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Let’s look at some examples:

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The field $\mathbb{C}$ can be thought of as an extension over $\mathbb{R}$ with the inclusion map $\mathbb{R}\hookrightarrow\mathbb{C}$.

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The field $\mathbb{Q}(\sqrt{2})=\left\{a+b\sqrt{2}:a,b\in\mathbb{Q}\right\}$ is readily verified to be a field and can then be thought of as an extension of $\mathbb{Q}$ with the inclusion mapping.

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More exotically, let $\mathbb{D}\subseteq\mathbb{C}$ denote the (open) unit disc then the ring $\text{Mer}(\mathbb{D})$ of all meromorphic functions on $\mathbb{D}$ is readily verified to be a field, which can be thought of as an extension of $\mathbb{C}$ by the map $\mathbb{C}\to\text{Mer}(\mathbb{D})$ sending a complex number to its associated constant function.

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Let $R\subseteq S$ be integral domains then $\text{Frac}(R)\subseteq\text{Frac}(S)$  (where $\text{Frac}$ denotes field of fractions) and $\text{Frac}(S)$ can be thought of as extension of $\text{Frac}(R)$ by inclusion. For example, applying this to $\mathbb{Z}\subseteq\mathbb{Z}[\sqrt{2}]$ gives our second example. Also, applying this to $\mathbb{C}\subseteq\text{Hol}(\mathbb{D})$ (the ring of holomorphic functions on $\mathbb{D}$) gives our third example.  Another example of this is thinking of $k\subseteq k[x]$, where $k$ is some field, which tells us that $k\subseteq k(x)$ is an extension where $k(x)$ is the rational function field over $k$ (i.e. the quotient of polynomials in $k[x]$ with nonzero denominator).

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Let $k$ be a field and let $f(x)\in k[x]$ be irreducible. We know then that $(f(x))$ is a maximal ideal and thus $k[x]/(f(x))$ is a field, which can be thought of as an extension of $k$ via the map $k\to k[x]/(f(x))$ sending $k\ni x\mapsto x+(f(x))\in k[x]/(f(x))$. In fact, this also generalizes our second example since, as one can check, $\mathbb{Q}(\sqrt{2})\cong\mathbb{Q}[x]/(x^2-2)$.

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Because the faithful way in which a field $F$ is put into an extension $k$ we often times indulge a (very sweet) logical inaccuracy by identifying $F$ with its image $F\to k$ so that we literally think of $F$ as sitting inside of $k$. Thus, one would not lose much by defining a field extension of $F$ to be a field $k$ with $k\supseteq F$. With this map-free point of view we use the notation $k/F$ to denote that $k$ is an extension of $F$ (note there is not quotienting here–it’s just a convenient notation reminding us that $k$ is over $F$).

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Let’s discuss one more important example of field extensions. Namely, let $k$ be any field. Recall that $\mathbb{Z}$ is an initial object in $\mathbf{Ring}$, or in much less obnoxious language, the fact that $k$ is a ring tells us that there is a unique ring map $\mathbb{Z}\to k$. By studying the image and kernel of this map we were led to the notion of prime subring and characteristic of a ring. While characteristic still works well with fields the notion of prime subring needs some improvement. Indeed, for this map $\mathbb{Z}\to k$ we have two choices–either the map has kernel $p\mathbb{Z}$ for some prime $p$ or it’s injective corresponding to the possibilities $\text{char}(k)=p$ and $\text{char}(k)=0$ repsectively (this follows from the more general fact that for an integral domain, the characteristic is either prime or zero). Now if $\ker(\mathbb{Z}\to k)=p\mathbb{Z}$ then by the first isomorphism theorem we know that we get an embedding $\mathbb{Z}/p\mathbb{Z}\to k$ and thus we can think of $k$ as an extension of $\mathbb{Z}/p\mathbb{Z}$. Now, if $\mathbb{Z}\to k$ is injective we know by the universal characterization of quotient fields that our ring map lifts to a ring map $\mathbb{Q}\to k$ which (by necessity) is an embedding, and thus we can think of $k$ as an extension of $\mathbb{Q}$.

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Summarizing the above we see that given a field $k$ we can think of $k$ as being an extension of $\mathbb{Z}/p\mathbb{Z}$ or $\mathbb{Q}$ according to the two cases $\text{char}(k)=p$ and $\text{char}(k)=0$ respectively. In this context we call the $\mathbb{Z}/p\mathbb{Z}$ or $\mathbb{Q}$ the prime subfield of $k$Intuitively it is the smallest subfield of $k$. Formally, it’s the smallest subfield of $k$. You can get it by just intersecting all the subfields of $k$, or more constructively you can take the smallest unital subring of $R$, i.e. $\mathbb{Z}1\subseteq k$ and look at it’s field of fractions inside $k$. We often times denote the prime subfield of $k$ by $k_0$.

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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;.

March 7, 2012 -

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