Abstract Nonsense

The Degree of an Extension

Point of Post: In this post we discuss the notion of the degree of an extension and prove some theorems relating to it, including the multiplicativeness for towers.

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Motivation

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Ok, so now on to our first order of buisness after defining field extensions–the size of an extension. Roughly we want now to measure how “big” the extension of our field is. There is no obvious way to do this in general (thinking just about the definitions, one field contained in another) and even in the more restricted classes where some definitions of “size” more easily present themselves, we can be mislead. For example, one may be apt to say that a subextension of $\mathbb{C}/\mathbb{Q}$ of the form $\mathbb{Q}(\alpha_1,\cdots,\alpha_n)$ is “smaller” than one of the form $\mathbb{Q}(\beta_1,\cdots,\beta_{n+1})$ because it has “less things appended”. While initially appealing this is an absolutely terrible idea. First of all one of the conditions we should desire in our notion of “size” is that, at least in nice cases (the class of extensions of the form $\mathbb{Q}(\gamma_1,\cdots,\gamma_n)$ counting), objects of the same size can’t nontrivially sit inside one another. In less cryptic terms if $L,K$ are two “nice” subextensions of $k/F$ with $L\subseteq K$ then $\text{size}(L)=\text{size}(K)$ should imply that, in fact, $L=K$. Our counting of appended elements fails terribly at this–for example $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt[4]{2})$ yet $\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}(\sqrt{2})$ is an abelian group of infinite rank. More egregiously though, our “appendage counting” if you will, is not even well defined. I’m sure you can convince yourself that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,1,2,3)$ for every $\alpha\in\mathbb{C}$. You may protest saying “it’s stupid to add rational numbers–it should work in other cases”. Ce n’est pas vrai mon ami. Indeed, it may be non-obvious at first, but $\mathbb{Q}(X)=\mathbb{Q}(\sqrt[n]{2},\zeta)$ where $X$ is the $n$ roots of the polynomial $x^n-2$ and $\zeta=e^{\frac{2\pi i}{n}}$. So, we can definitely conclude that “appendage counting” is not the way to go.

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This all said, I made it clear earlier that there are no obvious ways to define size based solely on the definition “an extension is a field which is a subset of another field”. But, we were actually able to think about extensions quite differently–they are just $F$-algebras which happen to be fields. And $F$-algebras, if anything, are $F$-vector spaces, and vectors spaces are things we know very well how to “size up”.

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The Degree of an Extension

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Let $F$ be a field and $k/F$ an extension. We define the index of $k/F$, denoted $[k:F]$, to be the dimension $\dim_F k$ where we are thinking about $k$ as an $F$-algebra.

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Now, we said last time that we are not going to think explicitly about field extensions as being $F$-algebra (i.e. rings $R$ with maps $F\to Z(R)$) but instead just as fields containing $F$. How do we reconcile these two view points? Well, if one goes back and looks at how we go from a map $f:F\to Z(R)$ to defining an $F$-vector space structure on $R$ one sees that we defined $\alpha\cdot r$ for $\alpha\in F$ and $r\in R$ to be $f(\alpha)r$ where this is just normal $R$-multiplication. Thus, we see that given an extension $k/F$ thought of as some kind of embedding $f:F\to k$ we are thinking of $k$ as an $F$-vectors space by the action $\alpha\cdot x=f(\alpha)x$. But, the thing we said is that we might as well identify $F$ with $f(F)\subseteq k$ so that we’d have the identification $\alpha\leftrightarrow f(\alpha)$ and so $\alpha\cdot x$ literally just becomes $\alpha x$. This is the key to our desired reconciliation. To make everything kosher with our conflation of literal extensions and just fields containing $F$ we should define the $F$-vector space structure in the latter by just normal multiplication in the ambient field.

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For example, the implied $\mathbb{Q}$-space structure on $\mathbb{R}$ is just the structure where the scalar multiplication of $\mathbb{Q}$ by $\mathbb{R}$ is just normal $\mathbb{R}$-multiplication.

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Remark: If the above was confusing (doubtlessly, in no small part to my poor writing) just keep in mind that for us an extension of $F$ is a field $k$ containing $F$, and $k$ has an $F$-vector space structure by defining scalar multiplication of $F$ by $k$ to just be normal $k$-multiplication. The above just discusses why this is the “right definition” if we want to conflate our two different ways of thinking about extensions of $F$.

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Let’s see what the degree of the various examples we discussed last time

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It’s easy to see that $[\mathbb{C}:\mathbb{R}]=2$.

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Again, it’s not hard to see that $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ whereas $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4$–this makes concretely how “bad” the size measurement suggested via “appendage counting was.

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Much less obvious than the previous two is the fact that $[\text{Mer}(\mathbb{D}):\mathbb{C}]=\aleph_0$. Here’s a plausibility argument. We note that $[\text{Mer}(\mathbb{D}):\mathbb{C}]\geqslant\aleph_0$ since $\{1,t,t^2,\cdots\}$ is a countably infinite linearly independent set. That said, we claim that the cardinality of $\text{Mer}(\mathbb{D})$ is at most $\mathbb{C}^{\aleph_0}$. Indeed, it’s trivial that $\#\text{Hol}(\mathbb{D})=\#\text{Mer}(\mathbb{D})$, and so it suffices to bound the cardinality of $\text{Hol}(\mathbb{D})$. To do this we note that every holomorphic map $f:\mathbb{D}\to\mathbb{C}$ is continuous, and since $\mathbb{C}$ is Hausdorff each $f$ is determined on a dense subset. That said, $\mathbb{D}$ is a subspace of $\mathbb{C}\approx\mathbb{R}^2$ which is second countable, and thus $\mathbb{D}$ is second countable. Thus, we see that we can inject $\text{Hol}(U)$ into $\mathbb{C}^D$ (where $D$ is a countable dense subset of $\mathbb{D}$) but $\#\mathbb{C}^D=\mathbb{C}^{\aleph_0}$. Thus, by pure cardinality reasons we may conclude that $\aleph_0\leqslant[\text{Mer}(\mathbb{D}):\mathbb{C}]<\mathfrak{c}$ and so if we assume something like the Continuum Hypothesis, we’re golden, if not, we’re not. It shouldn’t be too hard to make everything above more rigorous though, so I’ll leave that to you.

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It’s easy to make large estimates of the degree of a given field over its prime subfield. For example, if $k$ is a field with $k_0=\mathbb{Z}/p\mathbb{Z}$ then evidently $[k:k_0]<\infty$ if and only if $|k|<\infty$, in which case $[k:k_0]=\log_p(|k|)$. If $k_0=\mathbb{Q}$ then it’s easy to see that $k$ is countable implies that $[k:k_0]\leqslant\aleph_0$.

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We note that, right off the bat, our definition of size (i.e. index) has one of the desired qualities: it forbids nice objects of the same size to sit inside one another. Of course, what I mean by this is that if $k/F$ is an extension of finite index and $L/F$ is a subextension (necessarily of finite index) then $[k:F]=[L:F]$ implies $k=F$. We call this nice class of extensions, those with finite index, finite extensions. These type of extensions shall be our bread and butter when talking about field extensions. The reason for this is given any extension, even one of infinite degree, we can often times “specialize” the argument to some finite subextension (e.g. consider $\mathbb{Q}(\sqrt[5]{2})$ opposed to $\mathbb{C}$).

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References:

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;.