Quotient Space of Space of Polynomials Here's the question: Let $V = \mathbb{R}[x]$ be the space of polynomials with real coefficients, and W the space of polynomials divisible by $x^2+1.$ 
Then the quotient space $V/W$ can be identified with the set $\mathbb{C}$ of complex number, and the projection $P\colon \mathbb{R}[x]\rightarrow\mathbb{C}$ with the map $p(x)\mapsto p(i)$ of evaluating a polynomial $p(x)$ at $x = i$.
I understand that two polynomials $p(x)$ and $p'(x)$ are equivalent modulo $W$ if and only if $p(x)-p'(x)$ is divisible by $x^2+1$. This means $p(x)$ and $p'(x)$ are equivalent if and only if $p(i)=p'(i)$ or $p(-i)=p'(-i)$. But why is the conclusion in bold font is true? Thanks!
 A: First, look at the space $\mathbb R[x]/\langle 1+x^2\rangle$ which is the set of equivalence classes of polynomials with real coefficients, modulo $1+x^2$. For example $4+3x+2x^2+x^3 = (2+x)(1+x^2)+(2+2x)$:
$$4+3x+2x^2+x^3 \equiv 2+2x \bmod 1+x^2$$
For any polynomial $\mathrm p(x)$ in $\mathbb R[x]$, we divide by $1+x^2$ and identify it with its remainder. For each $\mathrm p(x)$ in $\mathbb R[x]$ we get a linear polynomial of the form $a+bx$ as the remainder, where $a,b \in \mathbb R$. We can use the notation $[a+bx]$ for the set of all polynomials which leave remainder $a+bx$ when divided by $1+x^2$. For example $4+3x+2x^2+x^3$ belongs to the equivalence class $[2+2x]$.
The space $\mathbb R[x]/\langle 1+x^2\rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $\mathrm{p}(x)$ in the set $[a+bx]$, we know that its members look like $\mathrm{q}(x) \cdot (1+x^2)+a+bx$ for some polynomial $\mathrm{q}(x)$ in $\mathbb R[x]$.
We can see that $\mathbb R[x]/\langle 1+x^2\rangle$ behaves like $\mathbb C$. For example $(1+2\mathrm i)(3+4\mathrm i) =-5+10\mathrm{i}$, while
$$\begin{eqnarray*} \\
[1+2x]\cdot [3+4x] &\equiv& (1+2x)(3+4x) \bmod 1+x^2 \\ \\
&\equiv& 3+10x+8x^2 \bmod 1+x^2 \\ \\
&\equiv& 8(1+x^2) - 5 +10x \bmod 1+x^2 \\ \\
&\equiv& -5+10x \bmod 1+x^2
\end{eqnarray*}$$
On the one hand, in $\mathbb C$, we have $(1+2\mathrm i)(3+4\mathrm i) =-5+10\mathrm{i}$. On the other, in $\mathbb R[x]/\langle 1+x^2\rangle$, we have $[1+2x]\cdot [3+4x] = [-5+10x]$. It seems that $[a+bx] \mapsto a+\mathrm i b$ could be a good candidate for an isomorphism $\mathbb R[x]/\langle 1+x^2\rangle \to \mathbb C$. Let's go back and check our first example: 
$$4+3x+2x^2+x^3 \in [2+2x]$$
$$4 + 3(\mathrm{i}) + 2(\mathrm{i})^2 + (\mathrm{i})^3 = 2+2\mathrm{i}$$
This makes perfect sense. All of the members of $[a+bx]$ look like
$$\mathrm q(x)\cdot(1+x^2)+a+bx$$
for some polynomial $\mathrm{q}(x)$. Applying the rule $x \mapsto \mathrm i$ gives
$$\mathrm q(x)\cdot(1+x^2)+a+bx \longmapsto \mathrm q(\mathrm i)\cdot(1+\mathrm i^2)+a+b\mathrm i = \mathrm{q}(\mathrm i) \cdot 0 + a+b\mathrm i = a+b\mathrm i$$
This show that the rule $\mathrm{p}(x) \mapsto \mathrm{p}(\mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+\mathrm{i}b$. It's not hard to check that this rule is in fact an isomorphism. 
A: One way to think of it is that (the ideal of) $x^2 + 1=0$ in the quotient space.  So $x^2=-1$ in the quotient space.  So $x$ acts as a square root of $-1$ in the quotient space.
A: Let define the following ring homomorphism: $$\varphi:\left\{\begin{array}{ccc}\mathbb{R}[X]&\rightarrow&\mathbb{R}[i]\\P&\mapsto&P(i)\end{array}\right..$$
Using the euclidean division in $\mathbb{R}[X]$, one has $\textrm{im}(\varphi)=\mathbb{R}+i\mathbb{R}$ and $\textrm{ker}(\varphi)=\left(X^2+1\right)$, therefore, one gets: $$\mathbb{R}+i\mathbb{R}\cong\mathbb{R}[X]/\left(X^2+1\right).$$
Hence the result.
