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Problem on Algebraic Closure

  1. Jun 11, 2010 #1
    Why is it that [tex]\mathbb{C}(x)[/tex] ([tex]\mathbb{C}[/tex] adjoined with x) is not algebraically closed? Here x is an indeterminate.

    My first question is what does the field extension [tex]\mathbb{C}(x)[/tex] even mean? If E is a field extension of F, and a is an transcendental element of E over F, then the notation [tex]\mathbb{C}(a)[/tex] is defined to mean the field of quotients of [tex]\mathbb{C}[a][/tex] (set of polynomials with complex coefficients). If a is algebraic then [tex]\mathbb{C}(a)[/tex] is defined to be [tex]\mathbb{C}[a][/tex] (which is the same thing as the field of quotients of [tex]\mathbb{C}[a][/tex], since in this case [tex]\mathbb{C}[a][/tex] is a field).

    Now x is an indeterminate, is that the algebraic or transcendental case? i.e. does [tex]\mathbb{C}(x)[/tex] mean field of quotients of polynomials with complex coefficients, or does it mean just polynomials with complex coefficients.

    So apparently [tex]\mathbb{C}(x)[/tex] is somehow not algebraically closed. So is there a complex polynomial whose root isn't a complex number? Any help is greatly appreciated.
     
  2. jcsd
  3. Jun 11, 2010 #2

    Office_Shredder

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    [tex]\mathbb{C}(x)[/tex] has to be field of quotients, because it has to be a field.

    It's not a complex polynomial. It's a polynomial with coefficients in [tex]\mathbb{C}(x)[/tex]. (note these will NOT be polynomials in x). For example:

    [tex]p(y)=x^4y^2 + (3x^2-\frac{2}{x})y + \frac{3}{x}[/tex]

    is a quadratic polynomial. The question is whether you can find a solution [tex] y \in \mathbb{C}(x)[/tex]
     
  4. Jun 11, 2010 #3
    So the notation [tex]\mathbb{C}(x)[/tex] isn't related to the notation [tex]\mathbb{C}(a)[/tex], where a is algebraic or transcendental?

    Hmm I thought field of quotients meant something in the form p(x)/q(x) i.e. has elements like:
    [tex]\frac{2ix^2 -3}{4i-x}[/tex].
    So this isn't right either?

    Also, while on the subject of field extensions, why is it that: if E is a field extension of F, and a in E is algebraic over F, if b in F(a) then F(b) is a subfield of F(a).
     
    Last edited: Jun 11, 2010
  5. Jun 11, 2010 #4

    Office_Shredder

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    What do you have for your definition for [tex]F(a)[/tex] if F is a field?
     
  6. Jun 11, 2010 #5
    If F is a field, then F[a] would be the set of all polynomials in a, (i.e. a typical element would be b_n * a^n + ... + b_1 * a + b_0 where all the b_ i in F), then F(a) is defined to be the field of quotients of F[a].
     
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