Quotients of $\mathbb Z[X]$ by irreducible polynomials 
Suppose we have an irreducible polynomial $f\in \mathbb Z[X]$ and say $r$ is a root of $f$ in $\mathbb C$. Is it true that $\mathbb Z[X]/(f(X))$ is isomorphic to $\mathbb Z[r]$ ? 

I ask since it seems to work for cyclotomic polynomials but I do not understand why. 

Can we generalize it to an integral domain $R$ (instead of $\mathbb Z$) and an algebraic closure of $\mathrm{Frac}(R)$ (instead of $\mathbb C$) ?

 A: This adds details to DonAntonio's answer and generalizes it to the case in which $R$ is a UFD. The main point that requires clarification is why the kernel of the morphism he defines consists of the principal ideal generated by $f$.
There is a theorem that says that if $R$ is a UFD and $K$ its field of fractions, then $R[X]$ is also a UFD, and the prime factorization of a polynomial  $f$ in $R[X]$ is $f = ap_1 \cdots p_n$, where the $p_i$'s are irreducible in $K[X]$ and $a \in R$ is a constant. (Technically, I should also write $a = uq_1 \dots q_n$ for the factorization of $a$ in $R$.) Also, if $p \in R[X]$ is nonconstant, then it is irreducible in $R[X]$ if and only if it is irreducible in $K[X]$ and its coefficients have no common factors (Gauss's Lemma). So, except for constant factors, the decomposition of any polynomial is the same in $R[X]$ or in $K[X]$.
In your situation, you have a polynomial $f$ that is assumed irreducible in $R[X]$, and $f(r) = 0$. Of course, $f$ cannot be constant. From the above, $f$ is also irreducible in $K[X]$. Now if $g \in R[X]$ and $g(r) = 0$, then certainly $f$ divides $g$ in $K[X]$. If we write the decomposition $g = a p_1 \dots p_n$ in $R[X]$, this is also a valid decomposition in $K[X]$, hence $f$ must be, except for a constant factor, one of the $p_i$'s. But $f$ and $p_i$ are both irreducible in $R[X]$, so they differ by a factor which is a unit of $R$. Therefore $f$ divides $g$ in $R[X]$.
This proves that the morphism in DonAntonio's answer induces an isomorphism between $Z[x]/(f)$ and $Z[r]$.
A: Just apply the first isomorphism theorem for ring homomorphisms:
$$\phi:
\Bbb Z[x]\to\Bbb Z[r]\;,\;\;\phi(q(x)):=q(r)$$
