$R[x] / \langle f(x)\rangle $ with $\deg f=1$ This question might look like a silly.
Despite most of the people are interested in $\deg f \geq 2$ for polynomial $f(x) \in 
\mathbb{Z}[x]$ either it is irreducible or not, I'm focused on the case $\deg f =1$.
So I considered the case $\mathbb{Z}_n[x]/\langle f(x)\rangle$ for $f(x) \in \mathbb{Z}[x]$ with $\deg f=1$ and I tried to find the isomorphic ring with that.
Since $f$ always has a root $\alpha$ in $\mathbb{Z}_n$, defining the $\phi :\mathbb{Z}_n[x]  \to \mathbb{Z}_n$ by $f(x) \to  f(\alpha)$
So, by isomorphic thm, my conclusion is 
$\mathbb{Z}_n[x]/\langle f(x)\rangle (= \{\beta + \langle f(x)\rangle\mid \beta \in \mathbb Z_n\})\simeq \mathbb{Z}_n$.
It looks like a clear for me, But I eager to check my conclusion is true or not.
Plus For all ring $R$, Could we say $R[x] / \langle f(x)\rangle  \simeq R$ for $f(x) \in R[x]$ ??
(Here $\deg f= 1$)
p.s.)
What if the $f$ is irreducible over $R$?
We can't say "$R[x] / \langle f(x)\rangle \simeq R$" (because $f$ doesn't have a any root in $R$)
Then what the ring is isomorphic with   $R[x] / \langle f(x)\rangle?? $ 
Thank you.
 A: I cannot give you a full answer, but I think that this already might help you for your intuition. Actually the behaviour of polynomials over finite rings is quite complicated and an own field of study. I have just some remarks:
(1) It is not in general true that a polynomial $ f = ax+b $ with $a, b \in \mathbb{Z}$ must have a zero mod $m$ for every $m$. E.g. consider $f=2x+1$ mod $4$. So your map does not work in general.
(2) Taking again $f = 2x+1$ mod $4$, you see that $f^2 = 4x^2+4x+1 \equiv 1$ mod $4$. So in this case, $f$ is a unit, hence $\mathbb{Z}_4[x]/(f)$ is the zero ring. 
(3) Now some positive news: If $R$ is any ring with unity and $f = ax+b$, where $a$ is a unit in $R$, then $f$ had the zero $\alpha = -ba^{-1}$ and the map you described, gives you an isomorphism from $R[x]/(f)$ to $R$. The reason is that in this case you can devide every polynomial $g \in R[x]$ by $f$  with some residue $r \in R$, and $r = 0$ iff $ g(\alpha) =0$.
(4) In a much more special situation, we can perfectely describe the ring you are interested in: Let $R$ be a factorial domain with quotient field $K$. Let $a,b \in R$ with $\gcd(a,b)=1$ ( which is equivalent to $ax+b \in R[x]$ being irreducible). If $f = ax+b$, then $R[x]/(f)$ is isomorphic to $R[-b/a]$ which is the smallest subring of $K$ containing $R$ and $ c =-b/a$. To see this, consider the map $\phi: R[x] \to R[c]$ with $\phi(g) = g(c)$. It is easy to proof that it is a surjective homomorphism. A polynomial $g$ is in the kernel iff $g(c) = 0$ iff $f$ divides $g$ in $K[x]$ (since it is Euclidean) iff $f$ divides $g$ in $R[x]$ (by the Lemma of Gauß). So the kernel is $(f)$.
I hope this helps!
