logarithmic
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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.
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.