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Here's the question: Let $V = mathbbR[x]$ 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 $mathbbC$ of complex number, and the projection $Pcolon mathbbR[x]rightarrowmathbbC$ 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!

First, look at the space $mathbb R[x]/langle 1+x^2rangle$ 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 $mathbb R[x]/langle 1+x^2rangle$ is given by all of these equivalence classes $[a+bx]$. Given a polynomial $mathrmp(x)$ in the set $[a+bx]$, we know that its members look like $mathrmq(x) cdot (1+x^2)+a+bx$ for some polynomial $mathrmq(x)$ in $mathbb R[x]$.

We can see that $mathbb R[x]/langle 1+x^2rangle$ behaves like $mathbb C$. For example $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$, while

$$begineqnarray* \
[1+2x]cdot [3+4x] &equiv& (1+2x)(3+4x) bmod 1+x^2 \ \
&equiv& 3+10x+8x^2 bmod 1+x^2 \ \
&equiv& 8(1+x^2) - 5 +10x bmod 1+x^2 \ \
&equiv& -5+10x bmod 1+x^2
endeqnarray*$$

On the one hand, in $mathbb C$, we have $(1+2mathrm i)(3+4mathrm i) =-5+10mathrmi$. On the other, in $mathbb R[x]/langle 1+x^2rangle$, 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 $mathbb R[x]/langle 1+x^2rangle to mathbb C$. Let's go back and check our first example:
$$4+3x+2x^2+x^3 in [2+2x]$$
$$4 + 3(mathrmi) + 2(mathrmi)^2 + (mathrmi)^3 = 2+2mathrmi$$
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 $mathrmq(x)$. Applying the rule $x mapsto mathrm i$ gives
$$mathrm q(x)cdot(1+x^2)+a+bx longmapsto mathrm q(mathrm i)cdot(1+mathrm i^2)+a+bmathrm i = mathrmq(mathrm i) cdot 0 + a+bmathrm i = a+bmathrm i$$
This show that the rule $mathrmp(x) mapsto mathrmp(mathrm i)$ takes all of the elements of $[a+bx]$ to the complex number $a+mathrmib$. It's not hard to check that this rule is in fact an isomorphism.

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.

Let define the following ring homomorphism: $$varphi:left{beginarraycccmathbbR[X]&rightarrow&mathbbR[i]\P&mapsto&P(i)endarrayright..$$
Using the euclidean division in $mathbbR[X]$, one has $textrmim(varphi)=mathbbR+imathbbR$ and $textrmker(varphi)=left(X^2+1right)$, therefore, one gets: $$mathbbR+imathbbRcongmathbbR[X]/left(X^2+1right).$$
Hence the result.