Abstract Nonsense

Crushing one theorem at a time

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_1tower 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;.

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March 8, 2012 - Posted by | Uncategorized | , , , , ,

2 Comments »

  1. […] put this to use. In our last post we pointed a cool consequence of the multiplicativity of the degrees in a tower. We said that […]

    Pingback by Algebraic Extensions (Pt. II) « Abstract Nonsense | March 25, 2012 | Reply

  2. […] this point have concerned us appending a single element to a given field. Second, thanks to our product formula we know that if we can cut the extension into a tower where each piece is finite then the top […]

    Pingback by Algebraic Extensions (Pt. III) « Abstract Nonsense | March 25, 2012 | Reply


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