What is \mathbb{C}(x) and why is it not algebraically closed?

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The discussion revolves around the field \mathbb{C}(x), which is the field of quotients of polynomials with complex coefficients where x is treated as an indeterminate. Participants are exploring the nature of this field extension and questioning why it is not considered algebraically closed.

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  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the meaning of \mathbb{C}(x) and its relationship to algebraic and transcendental elements. Questions are raised about the nature of polynomials in this context and whether there exist complex polynomials whose roots are not complex numbers.

Discussion Status

The discussion is active, with participants providing definitions and examples related to field extensions and polynomial forms. There is an exploration of the differences between \mathbb{C}(x) and \mathbb{C}(a), as well as inquiries into the implications of algebraic versus transcendental elements.

Contextual Notes

Some participants express uncertainty about the definitions and properties of field extensions, particularly regarding the nature of polynomials in \mathbb{C}(x) and the implications of being algebraically closed.

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Why is it that \mathbb{C}(x) (\mathbb{C} adjoined with x) is not algebraically closed? Here x is an indeterminate.

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

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

So apparently \mathbb{C}(x) is somehow not algebraically closed. So is there a complex polynomial whose root isn't a complex number? Any help is greatly appreciated.
 
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\mathbb{C}(x) 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 \mathbb{C}(x). (note these will NOT be polynomials in x). For example:

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

is a quadratic polynomial. The question is whether you can find a solution y \in \mathbb{C}(x)
 
So the notation \mathbb{C}(x) isn't related to the notation \mathbb{C}(a), 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:
\frac{2ix^2 -3}{4i-x}.
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:
What do you have for your definition for F(a) if F is a field?
 
Office_Shredder said:
What do you have for your definition for F(a) if F is a field?

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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