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Field Extensions and Root Fields

  1. Aug 9, 2012 #1
    I have a quick question. How does the following look?


    Proposition:

    Every extension of degree 2 is a root field.

    Proof:

    Let F be a field. Let p(x) ε F[x] and suppose p(x) has degree n. Then p(x) has n roots, say c1,c2,...,cn. Let E be the extension of F that contains the aforementioned roots, and suppose [E:F]=2. Now, we know F(c1,c2,...,cn) is the root field of p(x) over F. Previously, we have shown that if an extension over a field is of a degree that is prime, then there is no proper field between the field and its extension*. The extension E clearly contains F, and by our hypothesis c1,c2,...,cnε E. Thus, F(c1,c2,...,cn) is subset of E. Then since,
    F(c1,c2,...,cn) ≠ F, F(c1,c2,...,cn) = E. Hence, our extension is indeed a root field. QED


    *I simply used the theorem that is analogous to LaGrange's Theorem for Finite Groups, but for fields!

    Now, this is my "second version" of the proof. I am going back and trying to prove a few exercises different way. I am not sure if I set up my assumptions correctly. Any criticism would be very helpful! Thanks! :)
     
  2. jcsd
  3. Aug 10, 2012 #2

    micromass

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    No, that's not correct.

    Here, you basically assume the existence of a polynomial whose roots are in E. You can't do that.

    What you have to do is assume that [E:F]=2 and actually prove that there's a polynomial whose roots are in E and that make E into a root field.
     
  4. Aug 11, 2012 #3
    I knew something was dodgy about my assumptions. I just took another look at it. Here is an outline of how I want to prove it. (Before I write up the proof nicely, I usually write my solution line by line to make sure I have not made any unjustified statement.)

    Every extension of degree 2 is a root field.

    Proof: Let F be a field, and let E be an extension such that [E:F]=2.

    1. Let {a,b} be our basis for E.

    2. By defintion of a basis, every element of E can be written as a linear combination of {a,b} with coefficients in F.

    3. Then F(a,b) is clearly a subset of E. But, if we let x ε E, then x is some linear combination of {a,b}. Hence, x ε F(a,b). [This statement may not be needed]

    4. Thus, E=F(a,b)

    5. Since E is a finite extention, every element in E is algebraic over F.

    6. Let p(x) be the polynomial who roots are a, b. (i.e. p(x)=(x-a)(x-b))

    7. Then, E is a root field for p(x) over F.

    QED

    Thank you for all the help! :)
     
  5. Aug 11, 2012 #4

    micromass

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    OK, but that's not good enough. You want the coefficient of p(x) to be elements of F. This is not always the case.

    For example, [itex][\mathbb{C},\mathbb{R}]=2[/itex]. We can take a=i and b=1. Then [itex]p(x)=(x-1)(x-i)=x^2-(1+i)x+i[/itex] does not have its coefficients in [itex]\mathbb{R}[/itex].
     
  6. Aug 11, 2012 #5
    Hmmm...That's right.. (x-a)(x-b)ε E[x]. Just how (x+i)(x-i)ε C[x].

    Maybe I said too much in Line 6. Would be suffice to just say that p(x) ε F[x] whose roots are a and b is the polynomial that E is the root field for over F?
     
  7. Aug 11, 2012 #6

    micromass

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    You have to check your definition of root field. I guess that a root field that is spanned by the roots of a polynomial whose coefficients are in F. So I think you do need to find a polynomial with coefficients in F.

    You're right that for [itex][\mathbb{C}:\mathbb{R}][/itex], that you have to take (x+i)(x-i). Can you generalize this?
     
  8. Aug 11, 2012 #7
    I had never heard of a root field until I read this post. When I learned ring theory, we referred to them as splitting fields :approve:

    ...I miss algebra.
     
  9. Aug 11, 2012 #8

    micromass

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    Yeah, we used the term splitting field as well.
     
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