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

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The discussion centers on the nature of the field \mathbb{C}(x), which is defined as the field of quotients of polynomials with complex coefficients where x is an indeterminate. Participants clarify that \mathbb{C}(x) is not algebraically closed because it includes polynomials with coefficients in \mathbb{C}(x) rather than just complex numbers. An example polynomial is provided to illustrate that solutions may exist in \mathbb{C}(x) that are not complex numbers. The conversation also touches on the definitions of field extensions and the structure of F(a) when F is a field. Overall, the key point is that \mathbb{C}(x) contains elements that prevent it from 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).
 
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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].
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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