# 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.