The Tower Rule

Yesterday, we noticed that [\mathbb Q(\sqrt 2):\mathbb Q]=2. We can use a similar argument to deduce that [\mathbb Q(\sqrt[4] 2):\mathbb Q(\sqrt2)]=2. What does that tell us about [\mathbb Q(\sqrt[4] 2):\mathbb Q]? You may have guessed that [\mathbb Q(\sqrt[4] 2):\mathbb Q]=4. Now you’re either thinking that 4=2+2 or 4=2\cdot2. Those of you thinking that 4=2\cdot 2 are correct!

In general, if I have fields E, F, and G, with E\subseteq F\subseteq G, I can think of G as a vector space over the field F and F as a vector space over E. What’s more, I can think of G as a vector space over the field E. Then I can look at the dimensions, and get what I declare to be the coolest result in general field theory:

[G:E]=[G:F][F:E]

Remember, if you want, you can think about this as \dim_EG=\dim_FG\cdot\dim_EF, but the former way I think looks nicer.

Proof:

Let \{a_1,a_2,\dots,a_m\} be a basis for the vector space F over the field E. Let \{b_1,b_2,\dots,b_n\} be a basis for the vector space G over the field F. We want to find a basis for the vector space G over the field E, and we want it to have m\cdot n elements. There’s really only one logical choice, so let’s hope it works. We’ll show that \mathcal B=\{a_ib_j\mid 1\leq i\leq m,\ 1\leq j\leq n\} is a basis for G over the field E.

So take v\in G. We know there is a unique way to write it v=f_1b_1+f_2b_2+\cdots+f_nb_n where f_i\in F. But then each of these can be written uniquely as f_j=e_{1,j}a_1+e_{2,j}a_2+\cdots+e_{m,j}a_m. So we have

v=\displaystyle\sum_{i=1}^m\sum_{j=1}^ne_{i,j}a_ib_j.

Nice! So \mathcal B spans G (with coefficients in the field E). Now lets check linear independence. Suppose

\displaystyle\sum_{i=1}^m\sum_{j=1}^n\lambda_{i,j}a_ib_j=0.

We can think of this once again as a vector space over the field F, and let f_j=\lambda_{1,j}a_1+\cdots+\lambda_{m,j}a_m \in F, giving us

\displaystyle\sum_{j=1}^n f_jb_j=0.

But \{b_1,\dots,b_n\} is a basis for the vector space G over the field f, and hence a linearly independent set, so it must be that for every j, 0=f_j=\lambda_{1,j}a_1+\cdots+\lambda_{m,j}a_m. Now we have \{a_1,\dots,a_n\} is a basis and hence linearly independent, so each of the \lambda_{i,j} must be zero, for any 1\le i\le m or 1\le j\le n. This is exactly what it means for \mathcal B to be linearly independent. WOOHOO!

\square

If you didn’t think that was cool, you have no business calling yourself a decent human being.

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One Response to The Tower Rule

  1. Pingback: Galois Theory « Andy Soffer

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