Showing that $\Bbb Z[x]/\langle n, p(x)\rangle \cong \Bbb Z_n[x]/\langle p(x)\rangle$. 
Show that $\Bbb Z[x]/\langle n, p(x)\rangle \cong \Bbb Z_n[x]/\langle p(x)\rangle$.

I'm thinking of using the third isomorphism theorem for rings, but don't really now how to put the pieces together. It states that if $R$ is a ring and $I \subset J \subset R$, then $$(R/I)/(J/I) \cong R/J$$
So if I write $\Bbb Z_n[x]$ as $\Bbb Z/n\Bbb Z$, then I think I have that $n \Bbb Z \subset \langle p(x) \rangle \subset \Bbb Z[x]$ so $$(\Bbb Z[x]/ n \Bbb Z)/(\langle p(x)\rangle/n \Bbb Z) \cong \Bbb Z[x]/\langle p(x)\rangle$$ but the lhs of this doesn't seem to be $\Bbb Z_n[x]/\langle p(x)\rangle$?
 A: Consider the homomorphisms $\mathbb{Z}[X]\to\mathbb{Z}_n[X]\to\frac{\mathbb{Z}_n[X]}{<p(X)>}$ given by first applying the canonical projections (first applying the projection $\mathbb{Z}\to\mathbb{Z}_n$ at each coefficient of the polynomial, and then taking remainder dividing by $p(x)$), it is quite clear that $<n>$ and $<p(x)>$ are included in the kernel of the composition $\mathbb{Z}[X]\to\frac{\mathbb{Z}[X]}{<p(x)>}$, prove every element of the kernel is in $<n, p(x)>$ and apply First Isomorphism Theorem for Rings
A: The third isomorphism theorem works fine for this, but that not the right way to use it. $\newcommand{\Z}{\Bbb Z}$
Let’s say we have a commutative ring $R$, and two element $a$ and $b$ of $R$. If we denote the class of $b$ in $R/\langle a \rangle$ as $\bar b$, note that
$$
\langle \bar b \rangle = \langle a,b \rangle/\langle a \rangle \tag1
$$
and then
$$
\frac{R}{\langle a,b \rangle} \cong \frac{R/\langle a \rangle}{\langle a,b \rangle/\langle a \rangle} = \frac{R/\langle a \rangle}{\langle \bar b \rangle}.
$$
Hence, for an integer $n \geq 2$ and a polynomial $p$ in $\Z[x]$ we have that
$$
\frac{\Z[x]}{\langle n,p \rangle} \cong \frac{\Z[x]/\langle n \rangle}{\langle \bar p \rangle} \cong \frac{\Z_n[x]}{\langle \bar p \rangle} \tag2
$$
where the second $\cong$ follows from the well-known isomorphism
$$
\Z[x]/\langle n \rangle \cong \Z_n[x]. \tag3
$$
A small technical remark: The two $\bar p$ that appears in $(2)$ are not the same thing, the first one (as I said before $(1)$) is the class of $p$ on $\Z[x]/\langle n \rangle$, while the second one is the polynomial in $\Z_n[x]$ which corresponds to (the true) $\bar p$ under the isomorphism $(3)$.


*

*Proof of $(1)$: Again, for each $r \in R$, denote the class of $r$ in $R/\langle a \rangle$ as $\bar r$, and note that
$$\begin{align*}
\langle a,b \rangle/\langle a \rangle 
&= (\langle a \rangle + \langle b \rangle)/\langle a \rangle \\
&= \{\bar x + \bar y : x \in \langle a \rangle,\, y \in \langle b \rangle\} \\
&= \{\bar y : y \in \langle b \rangle\} = \langle \bar b \rangle.
\end{align*}$$

*Proof of $(3)$: Let $\pi \colon \Z \to \Z_n$ be the canonical projection, and $\tilde\pi \colon \Z[x] \to \Z_n[x]$ the map that sends a polynomial $a_0+a_1x+\cdots+a_rx^r$ in $\Z[x]$ to the polynomial $\pi(a_0)+\pi(a_1)x+\cdots+\pi(a_r)x^r$ in $\Z_n[x]$. This is a surjective ring homomorphism with kernel $\langle n \rangle$, as is easily verified.

