Field Extensions

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  • #1
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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
 

Answers and Replies

  • #2
Office_Shredder
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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
Firepanda
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So would the min poly look like

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

for a in K

Awesome fast reply thanks
 
  • #4
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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
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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
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That's not an equation. You need to use that when you plug x into the minimal polynomial you get zero
 
  • #7
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So i missed a step?

I'm not sure how to find the equation though
 
  • #8
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If the minimal polynomial for x over K is [tex]m(t)=\sum_{i=0}^{2n+1}a_i t^i[/tex] then you get the equation:

[tex]m(x)=\sum_{i=0}^{2n+1}a_i x^i = 0[/tex]

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:
  • #9
Firepanda
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are you asking me to show that every element of the summation is an element of K(x2)?
 
  • #10
Office_Shredder
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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:

[tex][K(x):K] = [K(x):K(x^2)][K(x^2):K][/tex]. What can you conclude about [tex][K(x):K(x^2)][/tex] using the fact that [tex][K(x):K][/tex] is odd?
 
  • #11
Firepanda
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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
Firepanda
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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
Office_Shredder
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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
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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?

not particularly, not sure how it can be 2
 
  • #15
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[tex] Q(\sqrt{2})[/tex] is a degree two extension at most because [tex] \sqrt{2}[/tex] satisfies the polynomial [tex] t^2-2=0[/tex].

Can you find some element a of [tex]K(x^2)[/tex] so that x satisfies [tex] t^2-a=0[/tex]?
 
  • #16
Firepanda
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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
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It's really easier than that.

You want to find a value of a so that [tex] x^2-a=0[/tex].
 
  • #18
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not root(a) then? otherwise im stumped
 
  • #19
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x is fixed, x is what you're extending the field by. You want to find a
 
  • #20
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oh so x^2?
 
  • #21
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Yeah. The polynomial m(t)=t2-x2 is a degree two polynomial (t is the variable) with coefficients in K(x2), and x is a root. So the extension to K(x) can be of degree two at most.

Of course, we also decided the degree has to be odd.

For completeness, to give an example of the previous proof techinque we abandonded, suppose x had as a minimal polynomial [tex] m(t)=t^5-4t^4+t-1[/tex]

Then [tex] x^5-4x^4+x-1=0[/tex] so [tex] x^5+x=x^4-1[/tex] Factoring out an x gives [tex] x=\frac{x^4-1}{x^4+1}[/tex]. The right hand side is an element of K(x2) so x is an element of K(x2) and the two fields are the same
 
  • #22
Firepanda
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So we know it can't be 2 because it needs to be odd or 1.

How do I deduce it must be 1?

i'm assuming that's what we're after.

edit: oh I reread what you wrote, it must be 1 then
 
  • #23
Firepanda
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so i need to go back to this

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

Any tips?' :P

So if I take an element of K, say z, and take an element of K(x^2), say a+b.x^2, a,b in K

and i have

z = a + b.x^2

then i need to manipulate this somehow to show some contradiction?

am i on the right track?
 
  • #24
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K(x2) can be equal to K, that just means K(x) is K also
 
  • #25
Firepanda
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K(x2) can be equal to K, that just means K(x) is K also

I just saw your edit above, should I go back to trying the original method, unless this new method ends soon I don't feel I'm very close to the solution, there seems to be a lot of things I need to check or I'm really complication things for myself.

I need to write the previous method out into some summation form don't I? Could you give me a little tip on that please so I can get working on it tommorow since it's 5am here now and my brain should have been sleeping 4 ours ago!

Thanks for the help! I'll write back tommorow
 
  • #26
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You're overcomplicating things. We already finished the proof.

[K(x):K]=[K(x):K(x2)][K(x2):K]

We know [K(x):K(x2)] has to be odd, because [K(x):K] is. But [K(x):K(x2)] cannot be more than 2 (because we found x is a root of a quadratic polynomial). So it's either 2 or 1, and can't be 2, so must be one. But a degree one extension means that the two fields are the same
 
  • #27
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Oh sorry, I'll put it down to my lack of understanding of this topic so far, didn't really know where I was in the proof. That or it is because it's 5am ;)

Thanks you were a great help
 

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