Complex analysis - existence of field $\mathbb{C}$ In the following: $\mathbb{F}$ is defined to be a field containing $\mathbb{R}$ and in which the equation $x^{2}+1=0$ can be solved. 
Then define a set $\mathbb{C}$ to be subset of $\mathbb{F}$ whose elements are of the form $x+iy$ where $x,y\in\mathbb{R}$ where $+$ and $i$ have only symbollical meaning. Defining addition and multiplication in usual way, then $\mathbb{C}$ is indeed a field: but here is what I do not understand;
How does the fact that we can write
$x+iy=(x+i0)+y(0+i1)$
shows that in fact $\mathbb{F}$ and $\mathbb{C}$ are identical? I presume it somehow shows that $\mathbb{F}$ $\subseteq$ $\mathbb{C}$ and $\mathbb{C}$ $\subseteq$ $\mathbb{F}$ but I don't have any clear idea why this is.
Also, when we say field of all polynomials modulo polynomial $x^{2}+1$, what does this exactly mean? 
And some basic algebraic manipulation question: is there any nice way of putting $(1-i)^{n}+(1+i)^{n}$ in a closed form, except for expressing them in modulus-argument form then using de'Moivre? I could try using binomial theorem but after a few steps it is redueced to sums of binomial coefficients which I could not put in a closed form.
Thanks
 A: The smallest field extension of $\Bbb R$ in which the irreducible polynomial $x^2 + 1$ has a root is $\Bbb R[x] / (x^2 + 1)$. We want to find an isomorphism between $\Bbb R[x] / (x^2 + 1)$ and $\Bbb C$. To do this, define a homomorphism $\varphi : \Bbb R[x] \to \Bbb C$, $\varphi(x) = i$. By direct computation, the ideal $(x^2 + 1)$ is in the kernel. Since $x^2 + 1$ is irreducible and $\Bbb R[x]$ is a PID, $(x^2 + 1)$ is maximal. Thus $\ker \varphi = (x^2 + 1)$. Now the first isomorphism gives the desired result.
A: $\mathbb C$ can be defined as $\mathbb R[X]/(x^2+1)$. This is a quotient of a ring by a maximal ideal*, so that it is a field. Now, by properties of quotients, we have two corollaries:
i ) Every polynomial in the quotient has degree less than that of $x^2+1$, so that every element of the quotient is of the form $a+bx$
ii)The element $x^2+1$ is the kernel of the map , so that $x^2+1=0 \rightarrow x^2=-1$
So we have a field with elements of the form $ax+b$ , with the property that $x^2=-1$. Then the map $ax+b \rightarrow ai+b$ is a field isomorphism.


*

*Because it is an ideal generated by an irreducible polynomial.

