Establishing isomorphisms between polynomial quotient rings 
Let $A$ be a commutative ring, $I$ be an ideal of $A$ and:
$$\varphi: A[X] \to \frac{A}{I}[X]$$
$$\sum_{i = 0}^{n} a_i X^i \mapsto \sum_{i = 0}^{n} \overline{a_i} X^i$$
where $\overline{a_i} = a_i +  I$. Prove that:
a) $\varphi$ is a ring homomorphism
b) $I[X] = \{ \sum_{i = 0}^{n} a_i X^{i} \ \vert \ a_i \in I, n \in \mathbb{N}\}$ is an ideal of $A[X]$ and $\frac{A[X]}{I[X]} \cong \left( \frac{A}{I} \right)[X]$.

a) and the first part of b) are easy. I'm having trouble with finding the last isomorphism though. I think it should be:
$$\psi:\frac{A[X]}{I[X]} \to \left( \frac{A}{I} \right)[X] $$
$$ \left(\sum_{i = 0}^{n} a_i X^{i}\right) + I[X] \mapsto \sum_{i = 0}^{n} (a_i + I)X^i = \sum_{i = 0}^{n} \overline{a_i} X^i $$
EDIT: Made the required corrections. Thanks a lot! Now I understand these things better than I did when I first asked the question.
 A: You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $\mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $\left(\sum_{i = 0}^{n} a_i X^{i}\right) + I[X]$.
Similarly, after the $\mapsto$, the left hand side of the equation should be replaced by $\sum_{i = 0}^{n} (a_i + I)X^{i}$.
(You also need to correct the earlier "$\overline{a_i} = a_i I$" to $\overline{a_i} = a_i + I$.)
The right hand side of the equation after $\mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
