Galois theory. field extension, constructible/algebraic numbers

[annawells]

Homework Statement

Let K be the subfield of all constructible numbers in C
Let A be the subfield of all algebraic numbers in C
Is the field extension A:K finite?

The Attempt at a Solution

I don't know where to start! I have read and understood proofs that all constructible numbers are algebraic. I can think of some algebraic numbers which are not constructible, such as cube root of 2, coming from the equation (x^3)-2.
But I can't think how to go about seeing if there are a finite amount of such numbers?

please point me in the right direction! :)

 

[micromass]

It's a long time ago since I did Galois theory. But I think this is it.

I conjeccture that [A:K] is infinite.

I see that because of the following inclusions

$$K\subset K[\sqrt[3]{2}] \subset K[\sqrt[3]{2},\sqrt[5]{2}]\subset K[\sqrt[3]{2},\sqrt[5]{2},\sqrt[7]{2}]\subset K[\sqrt[3]{2},\sqrt[5]{2},\sqrt[7]{2},\sqrt[11]{2}]\subset ...$$

So you got an infinite number of proper field extensions. So this means that [A:K] must be infinite.

I'm sorry if all of this doesn't make sense, it's probably to long ago then (:

 

[annawells]

It's a long time ago since I did Galois theory. But I think this is it.

I conjeccture that [A:K] is infinite.

I see that because of the following inclusions

$$K\subset K[\sqrt[3]{2}] \subset K[\sqrt[3]{2},\sqrt[5]{2}]\subset K[\sqrt[3]{2},\sqrt[5]{2},\sqrt[7]{2}]\subset K[\sqrt[3]{2},\sqrt[5]{2},\sqrt[7]{2},\sqrt[11]{2}]\subset ...$$

So you got an infinite number of proper field extensions. So this means that [A:K] must be infinite.

I'm sorry if all of this doesn't make sense, it's probably to long ago then (:

Thanks, that makes sense! So you are looking at all the pth roots for p prime yeh? and each extension has a degree of 2? so with some application of towers law (which says that for M<N<L, [L:M] = [L:N]*[N:M] ) we get that its infinite.

Instead of adding all the elements pth root of 2, would it be okay to make extensions using the elements cube root of p, eg cube root of 3, cube root of 5, cube root of 7, cube root of 11, ...etc?
(I don't have a preference though!)
thanks! :)

 

[micromass]

Doing it with the cube roots of the primes is what I thought up first. But I wasn't sure if all the field extensions would be proper. But I think it should work...

 

[annawells]

okay thanks! is there some lemma/theorem that states that these numbers are not constructable?? for either pth root of 2, or cube root of p?!

 

[micromass]

okay thanks! is there some lemma/theorem that states that these numbers are not constructable?? for either pth root of 2, or cube root of p?!

Yeah, you have the following. If a number is constructible, then the minimal polynomial of that number is a power of 2.

 

[annawells]

okay. sooo, the argument is that:

If z is constructible then its minimal polynomial has a degree which is a power of 2.
For z = 2^(1/p), z has minimal polynomial i(z^p)-2, which has degree p (which is obviously not a power of 2). So z is not constructible.

Is it obvious to say that clearly 2^(1/3) > 2^(1/5) > 2^(1/7) > 2^(1/11) >... so each extension is proper?

 

[micromass]

Hmm, why would 2^(1/3) > 2^(1/5) > 2^(1/7) > 2^(1/11) >... imply that every extension is proper? I don't really see that...

On the other hand, we have that $$\sqrt[5]{2}=\sqrt{2}\sqrt[3]{2}$$. So the extension

$$K[ \sqrt[3]{2} ] \subseteq K [ \sqrt[3]{2}, \sqrt[5]{2}]$$

is not proper, ****

But maybe the other method, with the cube roots of the primes will work...

 

[micromass]

So obviously the extension $$K\subseteq K[\sqrt[3]{2}]$$ is proper. An element in $$K[\sqrt[3]{2}]$$ has the form $$a+b\sqrt[3]{2}$$, were a and b are in K (and thus have degree a power of two).

To prove that the extension
$$K[\sqrt[3]{2}]\subseteq K[\sqrt[3]{2},\sqrt[3]{3}]$$

is proper. You need to show that $$\sqrt[3]{3}$$ cannot be written as $$a+b\sqrt[3]{2}$$...

 

[micromass]

OK, I've got it now. And it's not that hard. Forget everything about proper extensions.

The field extension $$K[\sqrt[p]{2}]:K$$ is proper. Moreover, the minimal polynomial of $$\sqrt[p](2)$$ is $$X^p-2$$. Thus we have that $$[K[\sqrt[p]{2}]:K]=p$$. Thus we have that

$$[K:A] \geq [K[\sqrt[p]{2}]:K] = p$$

for every prime p. so [K:A] must be infinite.

 

[annawells]

OK, I've got it now. And it's not that hard. Forget everything about proper extensions.

The field extension $$K[\sqrt[p]{2}]:K$$ is proper. Moreover, the minimal polynomial of $$\sqrt[p](2)$$ is $$X^p-2$$. Thus we have that $$[K[\sqrt[p]{2}]:K]=p$$. Thus we have that

$$[K:A] \geq [K[\sqrt[p]{2}]:K] = p$$

for every prime p. so [K:A] must be infinite.

yessssss! :) :D thanks a million!

 

[micromass]

To be honest, you still have to do a bit of work sorry...

I mean, you still need to show that $$X^p-2$$ is irreducible over K. I don't find that completely obvious...

 

[annawells]

To be honest, you still have to do a bit of work sorry...

I mean, you still need to show that $$X^p-2$$ is irreducible over K. I don't find that completely obvious...

ohhhh :( okay I will think about it! does anything make a difference that K contains complex numbers no?

 

[micromass]

No, I don't suppose that makes much of a difference...