Show $\mathbb Z[x]/(x^2-cx) \ncong \mathbb Z \times \mathbb Z$. 
For integers $c \ge 2$, prove $\mathbb Z[x]/(x^2 - cx) \ncong \mathbb Z \times \mathbb Z$. (Hint: for a ring $A$, consider $A/pA$ for a suitable prime $p$.)

I'm not entirely sure what the hint means, and I don't really have an idea for an approach.  For context, this is part (c) of a question; part (a) was to show that $\mathbb Z[x]/(x^2) \ncong \mathbb Z \times \mathbb Z$, and part (b) was to show $\mathbb Z[x]/(x^2 - x) \cong \mathbb Z \times \mathbb Z$.  I was able to do both, though my approaches for those questions don't seem to apply for this one.  Any help is greatly appreciated.
 A: Since I'm the one who wrote this past prelim problem that you're asking about (I recognized it immediately since you used the notation $c$ in the coefficient of $x$ from the original prelim problem), perhaps I'm "most" suited to answering it. :)
A general thing to keep in mind is that you can show two rings are not isomorphic by showing they don't share some ring-theoretic property preserved by isomorphisms: one has a finite unit group and the other doesn't, one is a field and the other isn't, one is a PID and the other has a nonprincipal ideal, and so on. How did you handle (a)? 
For (c), the point of the hint is to look at the ring structure on both sides after you reduce them mod $p$ for a prime $p$. On the right side you get $(\mathbf Z/p\mathbf Z) \times (\mathbf Z/p\mathbf Z)$, a product of two fields. As Bill Dubuque indirectly hinted at in his comment, there is a big difference on the left side (that is, for $\mathbf Z[x]/(x^2-cx)$) if you reduce it mod $p$ for a prime $p$ where $p \mid c$ or where $p \nmid c$. For $c \geq 2$ there is going to be a prime of the first kind and that's what makes $c \geq 2$ different from $c = 1$.
By the way, the source of this prelim problem is that in a paper I read before the prelim was being prepared, the author used the "fact" that $\mathbf Z[x]/(f(x)g(x)) \cong \mathbf Z[x]/(f(x)) \times \mathbf Z[x]/(g(x))$ for two relatively prime polynomials $f(x)$ and $g(x)$ in $\mathbf Z[x]$, and that is generally incorrect. If the coefficient ring were a field, like $\mathbf Q$, then the isomorphism would be valid. But $\mathbf Z$ is not a field and the proof of the Chinese remainder theorem for polynomials over a field does not go through for polynomials over $\mathbf Z$ all the time. This prelim problem has the simplest example I could think of (simplest nonconstant $f(x)$ and $g(x)$) where there is no ring isomorphism: not just that the "obvious" homomorphism $h(x) \bmod x^2-cx \mapsto (h(0),h(c))$ from $\mathbf Z[x]/(x^2-cx)$ to $\mathbf Z \times \mathbf Z$ is not a ring isomorphism, but that there is no ring isomorphism at all.
A: It is instructive to bring to the fore the conceptual key idea that is implicit in Mike's answer.
Hint $ $  by the Lemma $\,\Bbb Z[x]/(x(c\!-\!x))\cong\:\! \Bbb Z\!\times\! \Bbb Z\,$ $\Rightarrow\, \overbrace{x\,a(x)\!+\!(c\!-\!x)\,b(x) = 1}^{\large x,\ c-x\ \ \rm are\ \color{#0a0}{comaximal}^{\phantom{}}}\ \ \smash{\overset{x\,\to\, 0}\Longrightarrow}\ \  \bbox[4px,border:1px solid #0a0]{c\mid 1} $
Idea $ $ proper factorizations of $\,R/f\,$ $\rm\small\color{#f60}{induce}$ coprime $\small\rm\color{#0a0}{(i.e. comaximal)}$ proper
factorizations of $f\,$ by
Lemma $\,  $ If $\,\begin{align}f\in R\,\ \rm  a\ UFD,\\ {\rm rings}\ G,H\!\neq 0\end{align}\ $ then $\, \begin{align} R/f\,\ &\smash{\overset{\pi_{\phantom{|}}\!}\cong}\,\ G\!\times\! H\\ \color{#f60}{\bf\large  \Rightarrow} f\ \  &\!\!=\ g\ h\end{align} $ with $\ \begin{align} &\color{#0a0}{(g)\!+\!(h)=(1)^{\phantom{|^|}}\!\!}\\ &\,(g),\, (h)\neq\:\! (1)\end{align}\,$ for some $\,g,h\in R^{\phantom{|^|}}\!\!\!$
Proof $\ $ Isomorphism $\,\pi\,$ maps the idempotent $\,\varepsilon =(1,0)\in G\!\times\! H\,$ to one  $\, \bar e =  e\!+\!(f)^{\phantom{|^|}}\!\!\!\,$  in $\,R/f\,$ by  $\pi$ maps $(1,0)^{\phantom{|^|}}\!\!\!=\!(1,0)^2$ to $\, \bar e = \bar e^2.\,$ $\,\bar e\,$ is $\,\small\rm\color{#c00}{nontrivial}$: $\,H\!\neq 0\,\Rightarrow\,\varepsilon \neq (0,0),(1,1)^{\phantom{|^|}}\!\!\!\Rightarrow \color{#c00}{\bar e\neq 0,1}$.
Thus by $R^{\phantom{|^{|^{|^|}}}}\!\!\!\!$ a UFD, $\:f^{\phantom{|^|}}\!\!\!\mid (1\!-\!e)e\,\Rightarrow\, f=gh,\,\ g\mid 1\!-\!e,\,\ h\mid e$ $\,\Rightarrow\, (g)^{\phantom{|^|}}\!\!\!+\!(h)\supseteq(1\!-\!e)\!+\!(e)=(1),\:$ and the factorization is proper: $\, g\mid 1^{\phantom{|^|}}\!\! \Rightarrow f\mid h\mid e\,\Rightarrow \color{#c00}{\bar e=0}\,$ and $\,h\mid 1^{\phantom{|^|}}\!\!\Rightarrow f\mid g\mid 1\!-\!e\Rightarrow \color{#c00}{\bar e=1}$.
Remark $\, $ Generally $\small\rm\color{#c00}{nontrivial}$ idempotents (i.e. elements satisfying $\,e^2 = e\,$ and $\,\color{#c00}{e\neq 0,1})\,$ are intimately connected to coprime factorizations (of both elements and rings). In fact some integer factorization algorithms work by searching for nontrivial idempotents mod $\,n,\,$ which immediately yield a factorization of $\,n\,$ (generally we can quickly factor $\,n\,$ given any polynomial which has more roots mod $\,n\,$ than its degree, so any nontrivial idempotent or nontrivial square-root will split $\,n,\,$ since it yields a quadratic with $\,\color{c00}3\,$ roots).
Also closely related: the natural map $\, R\, \to R/I \times R/J\,$ is surjective $\!\iff\! I+J = (1),\,$ e.g. see Prop. $1.10$ in Atiyah and Macdonald's Introduction to Commutative Algebra. This is well known in elementary number theory as the case $\,(a,b)=(1,0)\,$ of the CRT solvability criterion, i.e. $\,x\equiv a\pmod{\!i},\, x\equiv b\pmod{\!j}\,$ is solvable $\!\iff$ $\,\gcd(i,j)\mid a\!-\!b,\,$ i.e. $\,a\!-\!b\in(i)+(j)$.
A: Suppose that $\mathbb{Z}[x]/(x^2 - cx) \cong \mathbb{Z} \times \mathbb{Z}$. Let $\varphi$ denote an isomorphism between these rings. I will denote a coset with an overline to simplify notation. Then, $\varphi(\overline{x}(\overline{x-c}))=\varphi(\overline{x})\varphi(\overline{x-c}) = (0,0)$, and $\varphi(\overline{x})$ and $\varphi(\overline{x-c})$ are not $0$ since $\varphi$ is injective. Therefore, $\varphi(\overline{x}) = (m,0)$ and $\varphi(\overline{x - c}) = (0,n)$ or vice versa for some nonzero $m, n \in \mathbb{Z}$ (WLOG, assume the former).
From
\begin{align*}
(0,n) & = \varphi(\overline{x - c}) \\ 
& = \varphi(\overline{x} - \overline{c}) \\
& = \varphi(\overline{x}) - \varphi(\overline{c}) \\
& = (m,0) - (c,c) \\
& = (m-c,-c)
\end{align*}
we get $m = c$ and $n = -c$. Hence, $\varphi(\overline{x}) = (c,0)$ and $\varphi(\overline{x-c}) = (0,-c)$. Pick $\overline{f(x)} \in \mathbb{Z}[x]/(x^2 -cx)$ such that $\varphi(\overline{f(x)}) = (1,0)$. Since every polynomial in the quotient can be represented by a polynomial of degree $0$ or $1$ in $\mathbb{Z}[x]$, and integers in $\mathbb Z[x]/(x^2-cx)$ 
are images of integers, by injectivity of $\varphi$ we can choose $f$ to have degree $1$ in $\mathbb{Z}[x]$. Then, since $c\varphi(\overline{f(x)}) = \varphi(\overline{x})$ we have $c\overline{f(x)} = \overline{x}$ in the quotient. Hence in $\mathbb{Z}[x]$, 
$$
cf(x) = x + g(x)(x^2 -cx)
$$
for some $g \in \mathbb Z[x]$. If $g(x)$ is nonzero then the left side has degree $1$ in $\mathbb{Z}[x]$ and the right side has degree larger than $1$. This means that $g(x) = 0$ and thus we have $x = cf(x)$ in $\mathbb{Z}[x]$ which is a contradiction since $x$ is irreducible in $\mathbb{Z}[x]$ and $c$ is a non-unit in $\mathbb{Z}[x]$ because $c \geq 2$.
