# Normal Field Extensions

1. Mar 15, 2006

### gonzo

How do I show that the following field extension is normal?

Q($$\sqrt{2+\sqrt{2}}$$):Q

As far as I could tell from my limited understanding is that I need to show that:

$$\sqrt{2-\sqrt{2}}$$

is also an element of the new field, which is required for the minimum polynomial to split over the field (which is what is required to make it normal as I understand it).

However, I can't figure out any way to show this.

I think there is also a way to do it by starting with the Galois group for the extension and using that to prove normality, but that seemed circular to me and still required me to show the above problem.

Any help would be appreciated, thanks.

Last edited: Mar 15, 2006
2. Mar 15, 2006

### Hurkyl

Staff Emeritus
I can think of several approaches.

The first one that sprung to mind is to simply solve for $\beta := \sqrt{2 - \sqrt{2}}$. You know the minimum polynomial, and you can express it as a function of $\beta$ alone. (Because you're going to treat $\alpha := \sqrt{2 + \sqrt{2}}$ as a known)

The second idea that sprung to mind is to simply play with the numbers. There are common manipulations you can do with radicals to yield their conjugates.

The third idea that sprung to mind is that you know the Galois group of the splitting field of the minimal polynomial, as well as the degree of your field extension. Surely that tells you something.

3. Mar 15, 2006

### mathwonk

i think hurkyl is suggesting you compare the degree of the extension, to the order of the galois group. hint: these numbers are equal if the extension is "biquadratic", i.e. obtained by adjoining two square roots successively.

4. Mar 16, 2006

### gonzo

At first I thought the solve for it idea was good--till I sat down to try and do it and realized I had no idea how to procede.

Could you be more specific in what you meant with this. Same with the second idea ... playing with the numbers is how I've mostly tried to do it, but I don'treally see how to get a negative number under the radical.

The Galois idea seems to run into the same problem ... it assumes the second number (with the negative under the radical) is part of the field extension.

5. Mar 16, 2006

### Hurkyl

Staff Emeritus
Hrm. I'm going to advise trying to do the problem all three ways. When you get one, try and do it the other two ways also!

How far did you get? Where you able to write the minimum polynomial in one way? Both ways?

What would you do if the outer radical wasn't there? Why wouldn't that work in this case?

I didn't say look at the Galois group of this field extension...

6. Mar 16, 2006

### gonzo

I can't do it any of the ways, let alone all three, so that advice isn't really helpful.

The minimum polynomial is easy, as is factoring it, but I still don't see what you mean. So I have some function with an unknown and alpha and beta, which is only a valid factorization in my field if beta really is in the field of course, otherwise I can do $\alpha$ and $\beta^2$. for example:

$( \gamma - \alpha ) ( \gamma + \alpha ) ( \gamma^2 - \beta^2 ) = 0$

Then what?

And I don't understand your comment about the second method. Obviously $Q( \sqrt{2} ) : Q$ is a subfield of the bigger one, since it contains $\sqrt{2}$ and so also contains $- \sqrt{2}$

But so what? I can't just put that under the bigger radical, it just means I can use $\sqrt{2}$ in my calculations.

As for the Galois part, I just haven't learned enough to know what you are talking about I guess. This is a problem working up to Galois Theory, just covering some of the beginning stuff.

It's okay if you don't want to tell me anything specific, but please let me know so I can start asking somewhere else for help with this problem.

Thank you.

Last edited: Mar 16, 2006
7. Mar 16, 2006

### Hurkyl

Staff Emeritus
(let f be the minimum polynomial of $\sqrt{2 + \sqrt{2}}$

I thought I was being rather specific -- I just wasn't being very detailed.

You know the minimum polynomial. (doesn't involve $\beta$)
You know the factorization of the minimum polynomial. (does involve $\beta$)

So you have two expressions for the same thing -- that lets you make an equation. Then you try to solve the equation.

I can't really think of a way to give any more hints without actually giving you the answer, so I'm going to repeat myself, and abandon this approach until you get the answer some other way. (Since we don't just hand out answers to homework problems here)

You know useful tricks for dealing with expressions that have square roots, and you've used them in the past precisely because they allowed you to change a quantity involving a radical into one involving its conjugate. Although you may not have done one where it was all underneath a square root, the particular technique that will be useful doesn't care about that.

The problem, as you've suggested, is that you don't know if $\mathbb{Q}(\sqrt{2 + \sqrt{2}})$ is the splitting field of f.

So, to me, it seems the logical thing to do is to start studying the splitting field of f. (call it E) Maybe you'll find some quality about E that would allow you to prove it equal to the field of interest.

8. Mar 17, 2006

### gonzo

Though this was a problem from a textbook on Galois Theory, it wasn't a "homework" problem. It was a problem I didn't understand in my attempt to learn the material.

Thank you for your time, sorry for the confusion. I have the answer now.

Oh, and one last thing:

Though I can solve it two other ways, this still doesn't make any sense to me. If you are saying I should equate coefficients on both sides (before or after dividing by the factored term with $\alpha$ (which is the only thing I can think of to do), that doesn't help.

The factored equation is in $\beta^2$ and taking roots is not an allowed operation. All that allows me to do as far as I can see is show that $\beta^2$ is an element of the field, not that $\beta$ itself is.

Last edited: Mar 17, 2006
9. Mar 17, 2006

### Hurkyl

Staff Emeritus
That's not entirely accurate. You can take square roots of things that have square roots.

Or, to put it differently, in any field, you know how to find all solutions to x²=a².

When the characteristic of the field is not two, you can even use the quadratic formula to solve quadratic equations -- the discriminant will have a square root if and only if the quadratic has a solution.

(Similarly, I think you can use the cubic formula in any field of characteristic other than 3, although the characteristic 2 case is a little tricky, since it involves solving a quadratic equation)

Last edited: Mar 17, 2006
10. Mar 17, 2006

### gonzo

But that is circular ... that assumes you already know the square root, which is what we are trying to show exists in this field.

I can get $\beta^2$ in terms of known elements of the field. How does that help me show that $\beta$ is an element of the field?

11. Mar 17, 2006

### mathwonk

i thought i had completely given it away.

12. Mar 17, 2006

### Hurkyl

Staff Emeritus
I do know the square roots of $\beta^2$: they are $\beta$ and $-\beta$. And I know the square roots of $\alpha^2$: they're $\alpha$ and $-\alpha$. And I know the square roots of 2: they're $\alpha^2 - 2$ and $2 - \alpha^2$, etc.

When I worked the problem, I got something involving only things whose square roots I did know. Maybe you got a different expression than I did: it might help if you posted what you got.

13. Mar 17, 2006

### gonzo

So, as a general question, is this method always going to work? Will you always get an answer in terms of obvious perfect squares, or are we just lucky with this one?

14. Mar 17, 2006

### mathwonk

in galois theory any answer is always lucky. or as galois put it, "en un mot, les calculs sont impracticables".

15. Mar 18, 2006

### gonzo

What would a basis be for this field? I can't find one with only 4 elements.

16. Mar 18, 2006

### gonzo

Never mind on that one, I was being blind to something obvious. A harder question ... the galois group for this field is supposed to be cyclic, but it looks to me like it is the Klein 4-group instead.

As far as I can tell the galois group is generated by the following two (not one) maps:

$\sigma : \alpha \mapsto -\alpha ; \sqrt{2} \mapsto \sqrt{2}$

$\tau : \alpha \mapsto \alpha ; \sqrt{2} \mapsto -\sqrt{2}$

17. Mar 18, 2006

### Hurkyl

Staff Emeritus
The second one isn't a map. Remember that you have the equation:

$$\alpha^2 - 2 = \sqrt{2}$$

If you're mapping $\sqrt{2} \mapsto -\sqrt{2}$, then you have to map $\sqrt{(2 + \sqrt{2})} \mapsto \sqrt{(2 - \sqrt{2})}$.

Anyways, why do you think it's supposed to be cyclic?

18. Mar 18, 2006

### gonzo

My post got cut off, I didn't get to type it all out. The book says it's cyclic, and I found out that it is, sort of, but I'm having trouble coinciding the basis which I am unsure about with my choice of maps, which I am also unsure about.

Turns out $\tau : \sqrt{2} \mapsto -\sqrt{2}$ is enough to generate a cyclic mapping group, but my coeffiecients get a little wonky.

Here is the basis I think is closest (with a,b,c,d rational of course):
with $\alpha=\sqrt{(2 + \sqrt{2})}$

$a + b\sqrt{2} + c\alpha + d\alpha\sqrt{2}$

I thought at first this wouldn't work since we missed $\beta = \sqrt{(2 - \sqrt{2})}$ but this turned out not to be a problem since $\beta = \sqrt{2}\alpha - \alpha$.

My $\tau$ is order 4, and $\tau^2$ is even nice and maps $\alpha$ to $-\alpha$ which I wanted, but $\tau$ and $\tau^3$ do weird things to the coefficients in my basis, and I don't really understand what I'm doing enough with this since I am just starting in on Galois stuff, so I'm not sure.

I thought any maps in the Glaois group had to be all combinations of mapping $\sqrt{2} \mapsto -\sqrt{2}$ (yes, I realize that also interchanges alpha and beta, which it is supposed to do) and $\alpha \mapsto -\alpha$.

I mean, I can't "just" change alpha into beta and -beta, can I? I need to map all occurances of root two in the field to minus root two, right?

19. Mar 18, 2006

### Hurkyl

Staff Emeritus
Are you sure? I agree with your previous assessment: I thought it was "obvious" that the Galois group was the Klein 4-group -- intuitively, the only things you can do are to pick the "other" square root. Two square roots were taken in $\alpha$, so $\mathop{\text{Gal}}(\mathbb{Q}(\alpha) / \mathbb{Q}) \cong \mathbb{Z}_2 \times \mathbb{Z}_2$.

There are some other choices of bases:
The obvious one $a + b\alpha + c\alpha^2 + d\alpha^3$
as well as $a + b\alpha + c\beta + d\alpha\beta$
and $a + b\alpha + c\beta + d\sqrt{2}$.

$\sqrt{2}$ is not a generator of $\mathbb{Q}(\alpha)$: you haven't supplied enough information to define a homomorphism.

Then that's a big problem: the coefficients are rational, and thus must be fixed by any element of the galois group! (But maybe that's not what you meant to say)

Well, work it out. You know that $\sqrt{2} = \alpha^2 - 2$ correct? What happens when you hit this equation with a homomorphism that maps $\alpha \mapsto \beta$?

20. Mar 18, 2006

### gonzo

This is where I get confused. What exactly is meant by fixed? I mean, look at the simple extension Q(i). Any element equals a+bi, and there is only one Galois map, i to -i, which transforms a+bi to a-bi, so can't you say that the coefficient b didn't stay unchanged here? In my case, I got something more like a+(b+3)i which is what I meant by wonky and why it felt wrong, or in the bases I used something like:
$a-b\sqrt{2}-(c+2d)\alpha+(c+d)\alpha\sqrt{2}$

I also am sure that the Galois group is cyclic unless the book is out to lunch. And yes, obviously from your example, mapping alpha to beta _also_ maps root two to minus root two.

Also, I don't see what it matters that root two doesn't generate the whole field, mapping root two to negative root two changes alpha into beta and beta into alpha, which I guess ends up being the same in some sense as just saying alpha to beta, though it was easier to work out as root two to negative root two.

By the way, your last two bases are the same since alpha beta is root two.

Should't the four maps be:
$\alpha\mapsto\alpha$
$\alpha\mapsto-\alpha$
$\alpha\mapsto\beta$
$\alpha\mapsto-\beta$

since these are the four roots of the minimum polynomial. But then one of these would have to generate the whole Glaois Group (and the last one is the only one that seems like it can do this).

But this is where I get all confused. What is allowed with a homomorphism ... how do we check that we really have a homomorphism? Do we need to just go through the grunge and make sure it preserves products and sums, or can we see that directly somehow?