More awesomeness of the tower rule

A few days ago we noticed that as we build up \mathbb K, by placing down intersection points, we never need to solve anything more complicated than the quadratic equation. We may need to do it a gazillion times, but every constructible number can be build by using nothing fancier than the quadratic equation.

Let’s do some magic. Suppose we construct some number \alpha. Maybe we had to put down points at a bazillion (which, in the metric system, can be more than a gazillion) intersections. Each time we place a point, it gives us a new constructible length. Let’s write down these lengths in the order they are constructed. Call them a_1,a_2,\dots, a_n=\alpha (here n takes the place of a gazillion). Let E_0=\mathbb Q, E_1=E_0(a_1)=\mathbb Q(a_1), E_2=E_1(a_2)=\mathbb Q(a_1,a_2), and so on, up until E_n=E_{n-1}(a_n)=\mathbb Q(a_1,\dots,a_n). Another way to describe this is to say that E_k is the smallest field containing \mathbb Q and a_i for 1\le i\le k.

The first thing to notice is that E_0\subseteq E_1\subseteq\dots\subseteq E_n. So what can we say about E_n? It’s a field? But wait! It’s actually a vector space too! It’s a vector space with scalar field E_m for any m! So let’s consider [E_n:E_0]=[E_n:\mathbb Q].

The magical tower rule tells us that [E_n:E_0]=[E_n:E_{n-1}][E_{n-1}:E_0]. But we can actually continue expanding this product a lot more:


So it suffices to figure stuff out about [E_{k+1}:E_k]. This is the extension we get when we add one new point.

But we already know something about what we get when we put down a new point. It is either a quadratic extesnion (if a circle is involved), or a linear extension (if it is just two lines intersecting). Linear extensions are lame, because it’s not an extension at all. It’s just the same field back. But quadratic extensions are not so lame.

In any event, [E_{k+1}:E_k] is either 1 or 2, so [E_n:E_0] must be a power of two. This is particularly handy, and an awesomely beautiful result. We are now really ready to snatch some cool results.

Tomorrow we’ll solve all the classical greek questions. Wednesday we’ll figure out which regular polygons we can construct. Then on Thursday, as promised, we’ll really start Galois Theory. Sorry to delay it so long, but there’s a lot that needs to be said before we can really attack the problem we want to see (the unsolvability of the general quintic).


One Response to More awesomeness of the tower rule

  1. Pingback: The classical Greek questions « Andy Soffer

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