Isomorphism of Polynomial Rings I'm having some difficulty with the following problem. 

Let $p$ be a prime integer, and if $a\in\mathbb{Z}$, write $\overline{a}$ for the coset of $a$ in $\mathbb{Z}_{p}$. Show that there is an isomorphism of rings $$\frac{\mathbb{Z}[X]}{\left\langle p, X^{2}+1\right\rangle}\cong \frac{\mathbb{Z}_{p}[X]}{\left\langle X^{2}+\overline{1}\right\rangle}\cong \frac{\mathbb{Z}[i]}{\left\langle p\right\rangle}.$$

Seeing any isomorphism of rings in which we mod out by ideals immediately makes me think that I should be using the First Isomorphism Theorem for rings. However, I'm not sure exactly what maps I should be using between the rings. I think that we could take an evaluation morphism between the first and second rings, but I'm not positive. I'm also confused slightly as to what the ideal $\left\langle p, X^{2}+1\right\rangle$ looks like (that is, the form of a typical element in the ideal is). Thanks in advance for any help!
 A: Let $I,J$ be two ideals of the (commutative) ring $R$. Consider the quotient map $R\rightarrow R/I$. Since this map is surjective, the image of $J$ under this map is again an ideal of $R/I$, denoted by $(I+J)/I$.

By the third isomorphism theorem we have $R/(I+J)\cong (R/I)/((I+J)/I)$.

Now use this isomorphism with $R=\mathbb{Z}[x], I=(p), J=(x^2+1)$. Then $$\mathbb{Z}[x]/(p,x^2+1)=R/(I+J)\cong (R/I)/((I+J)/I)=(\mathbb{Z}[x]/(p))/(p,x^2+1)/(p)\cong \mathbb{Z}_p[x]/(x^2+1)$$
By changing $I$ and $J$ you also get the other isomorphism.
Edit : The elements of $(p,x^2+1)$ is by definition are of the form $f\cdot p + g\cdot (x^2+1)$ for polynomials $f,g\in\mathbb{Z}[x]$.
A: Lets denote $R=\mathbb{Z}\left[x\right]$, $I=\left<p\right>$ and $J=\left<x^2+1\right>$. The first statement is just an application of the third isomorphism theorem
$$\mathbb{Z}\left[x\right]/\left<p,x^2+1\right>=R/\left(I+J\right)\cong\left(R/I\right)/\left(\left(I+J\right)/I\right)=\mathbb{Z}_{p}\left[x\right]/\left<x^2+1\right>$$
The second statement is very similar, can you proceed?
