# Problem on Algebraic Closure

1. Jun 11, 2010

### logarithmic

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.

2. Jun 11, 2010

### Office_Shredder

Staff Emeritus
$$\mathbb{C}(x)$$ 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 $$y \in \mathbb{C}(x)$$

3. Jun 11, 2010

### logarithmic

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: Jun 11, 2010
4. Jun 11, 2010

### Office_Shredder

Staff Emeritus
What do you have for your definition for $$F(a)$$ if F is a field?

5. Jun 11, 2010

### logarithmic

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].