# Homework Help: Field Extensions

1. Mar 28, 2010

### Firepanda

I think these are related:

Definition. Let F be an extension field of K and let u be in F. If there exists a nonzero polynomial f(x) in K[x] such that f(u)=0, then u is said to be algebraic over K. If there does not exist such a polynomial, then u is said to be transcendental over K.

Proposition. Let F be an extension field of K and let u in F be an element algebraic over K. If the minimal polynomial of u over K has degree n, then K(u) is an n-dimensional vector space over K.

Definition. Let F be an extension field of K. If the dimension of F as a vector space over K is finite, then F is said to be a finite extension of K.
The dimension of F as a vector space over K is called the degree of F over K, and is denoted by [F:K].

Proposition. Let F be an extension field of K and let u be in F. The following conditions are equivalent:

(1) u is algebraic over K;

(2) K(u) is a finite extension of K;

(3) u belongs to a finite extension of K.

What I know:

I've been staring at [K(x):K] = 2n+1 , n>=0, for a while now.

So this means the degree of the minimal polynomial is odd

A basis of K(x) over K is 1, x, x2, ..., x2n

Not sure what to do with all this information though.

Any help would be appreciated. Thanks

2. Mar 28, 2010

### Office_Shredder

Staff Emeritus
To show that K(x)=K(x2), you just need to show that x is in K(x2).

Write down the minimal polynomial for x, and see if you can use that to write x as an element in K(x2)

3. Mar 28, 2010

### Firepanda

So would the min poly look like

a2n+1x2n+1+a2nx2n+...+a1x+a0?

for a in K

4. Mar 28, 2010

### Office_Shredder

Staff Emeritus
Minor point of clarification, x is a root of an equation not a variable. It might be better to right down the polynomial as a function of t, say m(t), and then notice m(x)=0. Can you solve for x in terms of the a's and some x2s then?

5. Mar 28, 2010

### Firepanda

So i can write something like

a2n+1(x2)n.x + a2n(x2)n+...+a1(x2).x-1+a0(x2).x-2

is that enough, or is it wrong? Seems wrong, I don't really see how it uses the 2n+1 property, would have had the same result with any degree.

ah i just noticed your edit, hang on a sec for an edit

6. Mar 28, 2010

### Office_Shredder

Staff Emeritus
That's not an equation. You need to use that when you plug x into the minimal polynomial you get zero

7. Mar 28, 2010

### Firepanda

So i missed a step?

I'm not sure how to find the equation though

8. Mar 28, 2010

### Office_Shredder

Staff Emeritus
If the minimal polynomial for x over K is $$m(t)=\sum_{i=0}^{2n+1}a_i t^i$$ then you get the equation:

$$m(x)=\sum_{i=0}^{2n+1}a_i x^i = 0$$

Notice that xk is in K(x2) if k is even. If k is odd, you can write it as xk-1*x and you have an element of K(x2) times x. Can you use this to solve for x in terms of things from K(x2)?

Last edited: Mar 28, 2010
9. Mar 28, 2010

### Firepanda

are you asking me to show that every element of the summation is an element of K(x2)?

10. Mar 28, 2010

### Office_Shredder

Staff Emeritus
No, because that's not true. Suppose I gave you this formula:
am + b + cm + d=0

Can you solve for m?

m in this case is going to be the x that's factored out of every odd degree monomial.

Do you understand what the ultimate goal of doing this is? It sounds like you're a bit lost as to why I'm making this suggestion

Also, I thought of an alternative method if you like this better:

$$[K(x):K] = [K(x):K(x^2)][K(x^2):K]$$. What can you conclude about $$[K(x):K(x^2)]$$ using the fact that $$[K(x):K]$$ is odd?

11. Mar 28, 2010

### Firepanda

Ok, so I believe its a summation divided by another summation in the same way as your simple factoring example above.

Where its the summation of all the even a terms divided by the summation of all the odd a terms.

edit: actually i just wrote out all the factoring, is it something to do with taking each element of the summation pairwise?

this is deadly confusing for me

Edit: saw your edit, im not too concrete with all the theroy yet, it's new stuff so I'm not 100% sure on how and why and where to manuipulate things.

12. Mar 28, 2010

### Firepanda

i like the new method, it seems cleaner, with [K(x):K] being odd then that implies either [K(x):K(x^2)] is 1 or [K(x^2):K] = 1

So either K(x)=K(x^2) or K(x^2)=K

so i need to show that K(x^2) is not equal to K

Any tips?

I don't knwo any elements of K(x^2) or K I don't think.

13. Mar 28, 2010

### Office_Shredder

Staff Emeritus
Umm, they could both just be odd numbers.

The key is that [K(x):K(x2)] is odd. But also it can only be either one or two. Do you see why?

14. Mar 28, 2010

### Firepanda

not particularly, not sure how it can be 2

15. Mar 28, 2010

### Office_Shredder

Staff Emeritus
$$Q(\sqrt{2})$$ is a degree two extension at most because $$\sqrt{2}$$ satisfies the polynomial $$t^2-2=0$$.

Can you find some element a of $$K(x^2)$$ so that x satisfies $$t^2-a=0$$?

16. Mar 28, 2010

### Firepanda

a part of me wants to say root a but ill go out on a limb here

b+cx2 is in K(x2), b, c in K

so (b+cx2)2 - a = 0

(b2+2bcx2+c2x4) -a = 0

unless (b2+2bcx2+c2x4) is in K(x2) then I don't see how i can find an a in K(x2) that will cancel it out

hmm..

17. Mar 28, 2010

### Office_Shredder

Staff Emeritus
It's really easier than that.

You want to find a value of a so that $$x^2-a=0$$.

18. Mar 28, 2010

### Firepanda

not root(a) then? otherwise im stumped

19. Mar 28, 2010

### Office_Shredder

Staff Emeritus
x is fixed, x is what you're extending the field by. You want to find a

20. Mar 28, 2010

oh so x^2?