General intuition behind Ring isomorphism when it involves Polynomial Rings and help for proving it. I want to ask in general but I will give an example that I am trying to prove. My question is how you can show one quotient or non quotient ring of polynomials is isomorphic to another ring/field. For me this is not so intuitive, because for example when we deal with non polynomial ring we have just single elements for Identity in the domain and also in the co-domain by single elements I mean they consist of 1 symbol. I don't say this is in every case. But when this is proven in general for example in textbooks when they proved it, they do it in a very general way. Now I have 2 examples that I am trying to solve as follows: $$ \ (1)R[x]/(f) \cong \mathbb{C}, \text{ where } f \in R[X] \\(2)R[X]/(x^2+1)\cong \mathbb{C}$$
Ok, so I will not prove the first part of the theorem for homomorphism because this is not related to my question. To show that a homomorphism exists, we must show that the additive identity in the domain $R[X]/(x^2+1)$ is mapped onto the additive identity in the co-domain $\mathbb{C}$. In general they always give the example for this with simply take $\ \varphi(0) \in R $ and an arbitrary element $a \in R$ and from here we have: $$\varphi(a+0)=\varphi(a)+\varphi(0)$$ but in the domain ring $\varphi(a+0)$ is just $\varphi(a).$ and because the co-domain is a ring also, it is containing the inverse of $\varphi(a) $ we multiply  both sides of the equation by its inverse $\varphi(a)$. Then we got our proof: $$0=\varphi(0)$$ and the multiplication is a bit more sophisticated to show. I want to point out I was using some random $R$ and $\dot R$ for the example. And now I want to ask how this logic applies when it comes to rings of polynomials. In the (2) example we can use directly (*) $\ker(\varphi)=\{\ a\in R \mid \varphi(a)=0 \ \}$. First I want to ask whether this means that if I have a quotient ring, I can directly use this $(*)$ definition and I don't need to bother, like the example above, that I must show the identity for addition and multiplication from the domain is mapped onto the identity in the co-domain. Now I have tried to proof the (2) example like follows $$(**) \ \varphi(f)=f(i) \\ (***) \ f=a_ox^n+a_1x^{n-1}+,....,+a_2x^2a_1x+a_o$$ Now I need a help to understand the problem first I could not interpret $(**)$ what it means, I just used it here without understanding it. And I am stuck here for example to show that $\varphi \in \operatorname{Hom}(R[X], \mathbb{C})$. I will take two polynomials from $R[X]$, I will add them and I will multiply them by some  $i \in \mathbb C$ and I will be left with these two equations: $$1) \ \varphi(f+g)=(f+g)(i)=f(i)+g(i) = \ ? \\ 2) \ \varphi(fg)=(fg)(i)=f(i)g(i)=\varphi(f) \varphi(g)$$ Maybe the question got too big but I wanted to ask some particular things that I don't understand not just to solve my problem also I have found this "Are the Complex Numbers Isomorphic to the Polynomials, mod $x^2+1$?" but I didn't understand how they get to this conclusion. I have tried to understand the concept from literature and from lectures but i didn't. I will be thankful for any help on this subject. 
 A: Purely leading you by intuition, I will construct an isomorphism:
Put $$R:= \mathbb{R}[X]/(X^2+1)$$
A first observation is that in this ring (identifying classes and their representants) we have $$X^2 +1 = 0$$
I.e. 
$$X^2 = -1$$
Now, an element in $R$ is a polynomial in $\mathbb{R}[X]$ with the extra rule $X^2 = -1$.Thus, every element in the quotient ring can be written as $a + bX$ where $X^2 = -1$. Moreover, you can show that this representation is unique! (exercise)
Note that every element in $\mathbb{C}$ can be written as $a + bi$ with $i^2 = -1$. This suggest the following isomorphism 
$$\phi: R \to \mathbb{C}: a + bX \mapsto a + bi$$
(exercise that this is an isomorphism).

An alternative way is to prove that $\mathbb{R}[X] \to \mathbb{C}: P \mapsto P(i)$ is a ring epimorphism with kernel $(X^2 +1)$ and then apply the isomorphism theorem.

Note that in general it won't be true that $\mathbb{R}[X]/(f) \cong \mathbb{C}$. Indeed, if $f \in \mathbb{R}[X]$ is a reducible polynomial the quotient ring $\mathbb{R}[X]/(f)$ isn't even a field.
A: The intuition should be that $R[X]/(f)$ is the “minimal extension” of $R$ where $f$ has a root.
Indeed, if you call $\xi$ the image of $X$ in the quotient ring and $f(X)=a_0+a_1X+\dots+a_nX^n$, then, in the quotient ring,
$$
f(\xi)=a_0+a_1\xi+\dots+a_n\xi^n=(a_0+a_1X+\dots+a_nX^n)+(f)=0+(f)
$$
Conversely, suppose that $S$ is a (commutative) ring, with a ring homomorphism $\varphi\colon R\to S$. Suppose there exists $\alpha\in S$ such that
$$
\varphi(a_0)+\varphi(a_1)\alpha+\dots+\varphi(a_n)\alpha^n=0
$$
Then you can define uniquely a ring homomorphism $\varphi_\alpha\colon R[X]\to S$ by mapping $r\in R$ to $\varphi(r)$ and $X$ to $\alpha$. By assumption, the kernel of $\varphi_\alpha$ contains $(f)$, so $\varphi_\alpha$ induces a unique ring homomorphism $\psi\colon R[X]/(f)\to S$ such that $\psi(\xi)=\alpha$.
A common case is when $S$ is an overring of $R$ and $\varphi$ is just the inclusion map. In this case, the image of $\psi$ is $R[\alpha]$, the minimum subring of $S$ containing $R$ and $\alpha$.
In your case, if $R=\mathbb{R}$ and $f$ is irreducible, then $\mathbb{R}[X]/(f)$ is a field extension of $\mathbb{R}$, the minimal one where $f$ has a root, because it embeds in every field extension where $f$ has a root, via the homomorphism $\psi$ described above (which is injective because its domain is a field).
This “minimality” is, of course, up to isomorphisms.
Suppose $f(X)=X^2+1$. Then we're essentially adding an element $\xi$ such that $\xi^2=-1$. Hey! We know of a field extension of $\mathbb{R}$ where the polynomial has a root: it's precisely $\mathbb{C}$. So, how do we prove the isomorphism?
Consider the ring homomorphism $\varphi_i$ (where $\varphi$ is the inclusion map). Then $\varphi_i$ is surjective, because every element of $\mathbb{C}$ can be written as $a+bi=\varphi_i(a+bX)$.
Thus $\psi$ is both injective (because its domain is a field) and surjective (because its image is $\mathbb{R}[i]=\mathbb{C}$).
What if $f(X)=X^2-X+1$ or, more generally, $f(X)=X^2+aX+b$ with $a^2-b<0$? Well, we need to find a root of it in $\mathbb{C}$. This is $\alpha=(-a+i\sqrt{b-a^2})/2$. The map $\varphi_\alpha$ is again surjective (prove it) and the isomorphism follows as before.
Alternatively, you need to find in $\mathbb{R}[X]/(f)$ an element whose square is $-1$. You know that $\xi^2+a\xi+b=0$ in the quotient ring. Therefore
$$
4\xi^2+4a\xi+a^2=a^2-b
$$
that becomes $(2\xi+a)^2=-p^2$, where $p=\sqrt{b-a^2}$. Hence
$$
\left(\frac{2\xi+a}{p}\right)^{\!2}=-1
$$
as required.
What if $f$ is reducible? Well, in this case the quotient ring is no longer a field, but it still is an overring of $\mathbb{R}$ where $f$ has a root. If you split $f=f_1^{m_1}f_2^{m_2}\dots f_k^{m_k}$ as a product of (distinct) irreducible polynomials, then
$$
\mathbb{R}[X]/(f)\cong \mathbb{R}[X]/(f_1^{m_1})\times\mathbb{R}[X]/(f_2^{m_2})\times \dots \times \mathbb{R}[X]/(f_k^{m_k})
$$
Exercise: what's $\mathbb{R}[X]/(f^m)$, where $f$ is irreducible?
