Field that properly contains the field of complex numbers

Question

Problem

Give an example of a field that properly contains the field of complex numbers $\mathbb{C}$

Attempt

Let us consider the polynomial ring over the field of complex numbers ,i.e.,

$\mathbb{C}$[x]= {$c_nx^n+.......+c_o|c_i \in \mathbb{C}$}

Field of quotients of $\mathbb{C}$[x] is

$\mathbb{C}$(x)={$\frac{f(x)}{g(x)}|f(x),g(x) \in \mathbb{C}[x],g(x) \neq 0$}

I am not able to move from here.

Answer 1

For example, take $2 + i \in \mathbb C$. Define the polynomials $f(x) = 2 + i, g(x) = 1$, and note that $f/g \equiv 2+i$ is contained in $\{f/g : f,g \in \mathbb C[x], g \not \equiv 0\}$. Now, you know how to show that $\mathbb C$ is contained in this field : take $x \in \mathbb C$ instead of $2+i$, and the argument is exactly the same.

To show that containment is proper, note that $\frac 1x$ is a quotient of polynomials, which does not correspond to any complex number, because as we observed above, all polynomials corresponding to complex numbers are constants, but this one is not.

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More precisely, given a non-trivial field $F$, the quotient ring of the ring of polynomials with coefficients in $F$, always contains $F$ strictly as an image of a homomorphism from $F$ to this field (injective, as field homomorphisms are always injective).

Answer 2

That's the example that I would give. Note that if $z_0\in\mathbb C$, if you define $f(x)=z_0$ and $g(x)=1$, then $\frac{f(x)}{g(x)}$ belongs to your field. In other words, you can see $z_0$ as an element of $\mathbb{C}(x)$.

Answer 3

You find $\Bbb C$ by making $g(x)=1$ and all the terms in $f(x)\ 0$ except the constant term. $c_0$ ranges over $\Bbb C$ and there is your copy.