# Abstract Nonsense

## The Degree of an Extension (Pt. II)

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

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One of the interesting things about extensions is that they transfer well. Namely, suppose that we have some extension $k/F$ and some subextension $L/F$. A natural question is if there is some relation between the three degrees $[k:F]$, $[k:L]$, and $[L:F]$ that are floating around. Somewhat surprisingly, not only is there some connection between them, but it’s about as nice as one can hope:

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Theorem: Let $k/F$ be an extension and $L/F$ a subextension. Then,

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$[k:F]=[k:L][L:F]$

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In fact, let $\left\{\alpha_i\right\}_{i\in\mathcal{I}}$ and $\{\beta_j\}_{j\in\mathcal{J}}$ be bases for $L/F$ and $k/L$ respectively. Then, $\mathcal{B}=\left\{\alpha_i\beta_j:(i,j)\in \mathcal{I}\times\mathcal{J}\right\}$ is a basis for $k/F$ which has cardinality $\#(\mathcal{I})\#(\mathcal{J})$.

Proof: Let’s first prove that $\mathcal{B}$ really is a basis for $k/F$. To prove that $\mathcal{B}$ is linearly dependent we merely suppose that

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$\displaystyle \sum_{(i,j)\in\mathcal{I}\times\mathcal{J}}\lambda_{(i,j)}\alpha_i\beta_j=0$

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with $\lambda_{(i,j)}\in F$, all but finitely many zero. We see then that we may rewrite this as

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$\displaystyle \sum_{j\in\mathcal{J}}\left(\sum_{i\in\mathcal{I}}\lambda_{(i,j)}\alpha_i\right)\beta_j=0$

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Now, note that each inner sum lies in $L$ and since the $\{\beta_j\}$ are independent over $L$ we may conclude that the inner sum is zero for each $\in\mathcal{J}$. Now, fix a $j$, the fact that

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$\displaystyle \sum_{i\in\mathcal{I}}\lambda_{(i,j)}\alpha_i=0$

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and that the $\{\alpha_i\}$ are independent over $F$ tells us that each $\lambda_{(i,j)}$ is zero, and since this was for an arbitrary $j$ we may conclude that all the $\lambda_{(i,j)}$ are zero. To see that $\mathcal{B}$ spans $k$ over $F$ we merely note that if $\alpha\in k$ is arbitrary then (by virtue of the fact that $\{\beta_j\}$ is a basis for $k/L$) there exists $\mu\in L$ for each $j\in\mathcal{J}$ such that

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$\displaystyle \sum_{j\in\mathcal{J}}\mu_j\beta_j=x$

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But, since $\{\alpha_i\}$ is a basis for $L/F$ there exists $\lambda_{(i,j)}\in F$ such that

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$\displaystyle \mu_j=\sum_{i\in\mathcal{I}}\lambda_{(i,j)}\alpha_i=\mu_j$

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and so

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$\displaystyle x=\sum_{j\in\mathcal{J}}\mu_j\beta_j=\sum_{j\in\mathcal{J}}\left(\sum_{i\in\mathcal{I}}\lambda_{(i,j)}\alpha_i\right)\beta_j=\sum_{(i,j)\in\mathcal{I}\times\mathcal{J}}\lambda_{(i,j)}\alpha_i\beta_j$

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The only order of business left to take care of is to show that $\#(\mathcal{B})=\#(\mathcal{I})\#(\mathcal{J})$ which, if there is any justice in the world, should be true because the natural surjection $\mathcal{I}\times\mathcal{J}\to\mathcal{B}:(i,j)\mapsto \alpha_i\beta_j$ is an injection. To see this suppose that $\alpha_{i_0}\beta_{j_0}=\alpha_{i_1}\beta_{j_1}$. Since the $\{\beta_j\}$‘s are a basis for $k/L$ this implies that either $\alpha_{i_0}=\alpha_{i_1}=0$ or $j_1=j_0$ and $\alpha_{i_0}=\alpha_{i_1}$. Clearly the first can’t happen since $\{\alpha_i\}$ is a basis for $L/F$ is a basis and thus can’t contain zero, and so the second must be true. Now, we know that $j_1=j_0$ and since the distinctly indexed $\alpha$‘s are distinct we may conclude that $i_0=i_1$ as well. The conclusion follows. $\blacksquare$

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From this we get two immediate, important corollaries:

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Corollary: Let $k/F$ be an extension and $L/F$ a subextension. Then, $k/F$ is finite if and only if $k/L$ and $L/F$ are.

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Corollary: Let $k/F$ be a finite extension. Then, if $L/F$ is a subextension then $[k:L]$ and $[L:F]$ divide $[k:F]$

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The first corollary is often times useful for allowing us to prove a certain extension is finite by breaking it into a sequence of subextensions and treating each piece separately.

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The second corollary is surprisingly powerful. For example, it seems kind of obvious that $\sqrt{2}$ can’t be written as a rational polynomial in $\sqrt[5]{2}$, right? Go ahead and try to prove this. In fact, it’s not extremely easy from standard (non-clever) means, but the second corollary allows us to discount this possibility immediately. Indeed, if this were true then $\sqrt{2}\in\mathbb{Q}(\sqrt[5]{2})$ which will tell us that $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt[5]{2})$, but this can’t be true since (as we shall soon verify) $[\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt[5]{2}):\mathbb{Q}]=5$ and $2\nmid 5$.

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Perhaps even cooler is the sort of amazing fact that if $k/F$ is a finite extension of degree $n$ and $m\nmid n$ then there can’t be a field $k\supseteq L\supseteq F$ with $[L:F]=m$–in particular, if $[k:F]$ is prime then the only fields $L$ with $k\supseteq L\supseteq k$ is $L=F$ or $L=k$.

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Our last note in this post shall be one on notation. We have been having to say “Let $k/F$ be an extension and $L/F$ a subextension”, which is a lot of words. A more preferable notation attached to the notion of a tower of fields. In particular, we shall call a descending chain of fields $k_n\supseteq k_{n-1}\supseteq\cdots\supseteq k_1$tower and denote it $k_n/k_{n-1}/\cdots/k_1$. The tower term comes from the other way to visually represent a tower which is by the Hasse diagram of the fields, with respect to inclusion. For example, $k_n/k_{n-1}/\cdots/k_1$ would be written

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$\begin{array}{c}k_n\\ \vert \\ k_{n-1}\\ \vert\\ \vdots \\ \vert\\ k_1\end{array}$

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Often times the degree of the extension is written beside the line, so that if $m_i=[k_{i+1}:k_i]$ then we’d write

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$\begin{array}{cc} k_n & \\ \vert & _{m_{n-1}}\\ \vdots & \\ \vert & _{m_2}\\ k_2 & \\ \vert & _{m_1}\\ k_1 & \end{array}$

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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;.