How to prove normality of a field extension with Q(\sqrt{2+\sqrt{2}}):Q?

[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.

 

[Hurkyl]

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.

 

[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.

 

[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.

 

[Hurkyl]

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!

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.

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

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.

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

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.

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

 

[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.

 

[Hurkyl]

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

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.

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

The minimum polynomial is easy, as is factoring it, but I still don't see what you mean.

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 let's you make an equation. Then you try to solve the equation.

And I don't understand your comment about the second method.

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.

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.

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.

 

[gonzo]

(Since we don't just hand out answers to homework problems here)

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:

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

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.

 

[Hurkyl]

The factored equation is in $\beta^2$ and taking roots is not an allowed operation.

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)

 

[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?

 

[mathwonk]

i thought i had completely given it away.

 

[Hurkyl]

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 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.

 

[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?

 

[mathwonk]

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

 

[gonzo]

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

 

[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:

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

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

 

[Hurkyl]

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?

 

[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?

 

[Hurkyl]

The book says it's cyclic,

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

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

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

but $\tau$ and $\tau^3$ do weird things to the coefficients in my basis

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)

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?

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 $\alpha \mapsto \beta$?

 

[gonzo]

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)

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?

 

[Hurkyl]

This is where I get confused. What exactly is meant by fixed?

An element s of any endomorphism f is fixed if and only if f(s)=s.

An automorphism f of Q(a) fixes Q if and only if f(q) = q for all q in Q.

If we "forget" the field structure, and just say that Q(a) is a vector space over Q, then the elements of Gal(Q(a) / Q) are merely linear transformations. In particular, linear transformations can change vectors (elements of Q(a)), but must leave scalars (elements of Q) alone.

So, I think you're simply getting confused by coordinates. You're doing more than just applying an element of the Galois group: you're also rewriting the result according to the basis you've chosen.

But this is where I get all confused. What is allowed with a homomorphism ... how do we check that we really have a homomorphism?

You use ring theory! If it helps, recall that $\mathbb{Q}(\alpha) \cong \mathbb{Q}[x] / \langle f_{\alpha}(x) \rangle$.

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

In Q(a), there are two ways to map $\sqrt{2} \mapsto -\sqrt{2}$. One of them does what you say. The other one maps $\alpha \mapsto -\beta$.

Hey, we can lift this into a statement using Galois theory! The problem is that $\mathop{\text{Gal}}(\mathbb{Q}(\alpha) / \mathbb{Q}(\sqrt{2}))$ is nontrivial (and has order two). So, once you've specified where $\sqrt{2}$ goes, you still have two choices for how the rest of the field transforms.

 

[Hurkyl]

Are you sure? I agree with your previous assessment:

But no longer. The group is cyclic. Stupid intuition failiing me.

 

[gonzo]

Then which map generates it? I still don't see it. Can you just tell me straight out? I'm getting tired of this one problem.

 

[gonzo]

I found another way to prove that it is cyclic easily, but that doesn't help me find the actually elements of the Galois group, the map. If it makes it easier for you to help me that is NOT part of the problem, I just want to know (even though it isn't homework anyway).

To prove it is cyclic you could do this, where a,b,c and d are the roots of the min. poly:

U = (a+b)(c+d)
V= (a+c)(b+d)
W = (a+d)(b+c)

If it is NOT cyclic then these should be fixed, but two of them are irrational and thus not fixed by the galois group, and so it must be C4.

Doesn't help find what they are though.

 

[gonzo]

I even thought of another easy proof for the group being C4. $Q(\sqrt{2}]$ is the only subfield, and it is order 2, which means the Galois group will only have one subgroup of order 2, which is obviously C4 and not C2xC2, and this one is clearly generated by $\alpha\mapsto-\alpha$.

It's the other two that I'm not seeing.

I figure there are 24 permutations of the 4 roots, and only 4 of these form the Galois group. 2 of them are easy, but I have no clue at all on the other two.

 

[Hurkyl]

Well, if it's cyclic, and it's not the identity map, and it's not the map a --> -a, then the other two maps must be generators! You know the maps: you wrote them down in post #20.

I even thought of another easy proof for the group being C4. $Q(\sqrt{2})$ is the only subfield

That's a neat way! How did you prove it's the only subfield (other than Q), though?

U = (a+b)(c+d)
V= (a+c)(b+d)
W = (a+d)(b+c)

If it is NOT cyclic then these should be fixed, but two of them are irrational and thus not fixed by the galois group, and so it must be C4.

I was curious about this too. I'll buy it if we know the elements of the galois group have cycle structure ( , )( , ) (i.e. the product of two disjoint transpositions), but what if the galois group contained single transpositions? Or is there a cool theorem I'm forgetting? (Which is quite possible)

 

[gonzo]

The whole problem is that those other two maps seem to be order 2 not order 4, which is still the problem. How are they order 4? Swithcing $\alpha$ and $\beta$ twice puts you back where you started in any of the bases I've looked at.

 

[Hurkyl]

Unless I've made a very silly mistake last night while thinking about $\alpha \beta = \sqrt{2}$, the map that sends a-->b does not send b-->a as I had originally thought.

 

[gonzo]

It has to send b to one of the other roots, right? Which other root could it otherwise send it to?

 

[gonzo]

To put it another way:

$\beta=\sqrt{2}/\alpha$
$\alpha\mapsto\beta$
$\beta\mapsto\sqrt{2}/\beta$
but the first equality still has to hold so we get
$\beta\mapsto\sqrt{2}/(\sqrt{2}/\alpha)=\alpha$

 

[shmoe]

To put it another way:

$\beta=\sqrt{2}/\alpha$
$\alpha\mapsto\beta$
$\beta\mapsto\sqrt{2}/\beta$

This last one doesn't follow, you're assuming that this map fixes $$\sqrt{2}$$, but it doesn't. To find what this map does to $$\beta$$, use $$\beta=\alpha^3-3\alpha$$, or work out where sqrt(2) goes using $$\sqrt{2}=\alpha^2-2$$.

 

[mathwonk]

im puzzled as to what you all are doing. doesn't the remark that b = sqrt(2)/a prove that the field Q(a) also contains b? hence is normal?

or am i way behind? certainly my earlier comments were incorrect, since i confused a biquadratic extension, namely one where both square roots adjoined were of elements in Q, with this situation where one adjoins the square root of 2, i.e. of an element of Q, and then a square root of an element of the larger field Q(sqrt(2)).

Indeed this is a standard way to get a non normal extension, generated by the real 4th roots of 2, for instance.


But if you want the galois group elements here, it seems they are:

1) identity

2) sqrt(2) fixed, hence a goes to the other root of its minimal polynomial over Q(sqrt(2)), namely -a. thus b goes to -b.

3) sqrt(2) goes to the other root of its minimal polynomial over Q, namely -sqrt(2). a goes to some root of its transformed minimal polynomial under the previous map on sqrt(2), say a goes to b. then since a field map takes quotients to quotients, b = sqrt(2)/a goes to -sqrt(2)/b = -a.

4) This one also takes sqrt(2) goes to the other root of its minimal polynomial over Q, namely -sqrt(2). Then a must go to the other root of its transformed minimal polynomial under the previous map, namely a goes to -b. then b = sqrt(2)/a goes to -sqrt(2)/-b = a.

the orders of these elements seem to be 1,2,4,4. In particular the group is cyclic of order 4.


the key observation being that of gonzo, that b = sqrt(2)/a.


I have not been able to follow all your discussion but to compute galois groups of an extension like Q(c,a) by hand, one simply sends c to some other root of its minimal polynomial over Q, then follows what happens to the minimal polynomial of a over Q(c) and sends b to some root of that transformed polynomial.


am i in the ball park? i admit i always was weak at algebra. which is why i wrote the book on my webpage. my book is full of thorough details, and shares everything i myself understand, but to understand things well i recommend reading a book by someone who gets it, like artin, or jacobson.

 

[gonzo]

Thanks, that was what I was confusing ... if a goes to b, then b goes to -a, I missed that. That makes it easy.

Thank you.

 

[gonzo]

By the way, Mathwonk, can you give me a link to your web page, I would be interested in looking through it if you have some of this stuff explained. It always helps to get a different perspective.

 

[mathwonk]

goto http://www.math.uga.edu/~roy/

download the algebra course notes 843-2, and read pages 43-46 for a 2 calculations of the galois group of X^4-2.

the whole section 19, pages 34-46 is relevant.