Show that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ is not isomorphic to $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ I am given in the question that $\frac {(\mathbb Z/p)[x]}{(x^2)}$ has only one maximal ideal.
So my thinking for this is that   $(\mathbb Z/p)[x]$ is an integral domain.
Now my original thinking was to use Chinese Remainder Theorem as I thought the $\gcd(x,x) = x$ which means they are not coprime so they can't be isomorphic.
I have a secondary thought of showing that   $\frac {(\mathbb Z/p)[x]}{(x)}$ has one maximal ideal, therefore its product $\frac {(\mathbb Z/p)[x]}{(x)} \times \frac {(\mathbb Z/p)[x]}{(x)}$ will have two maximal ideals, but I am not sure whether this is appropriate too.
 A: Hint:
$\mathbf Z/p\mathbf Z[x]/(x^2)$ has a nonzero nilpotent element, whereas
$\mathbf Z/p\mathbf Z[x]/(x)\times\mathbf Z/p\mathbf Z[x]/(x)\simeq\mathbf Z/p\mathbf Z\times\mathbf Z/p\mathbf Z$, a product of fields.
A: There are several ways to show the assert.
The hint is to look at the maximal ideals of the product. Note that a product of two nontrivial rings will always have at least two distinct maximal ideals. Indeed, if $A$ and $B$ are nontrivial rings, choose $I$ and $J$ maximal ideals in $A$ and $B$ respectively. Then $I\times B$ and $A\times J$ are distinct maximal ideals in $A\times B$.
The ring $(\mathbb{Z}/p)[x]/(x)$ is nontrivial, because it is isomorphic to $\mathbb{Z}/p$.
On the other hand, the ideals of $(\mathbb{Z}/p)[x]/(x^2)$ are in bijection with the ideals of $(\mathbb{Z}/p)[x]$ containing $x^2$: these are only $(x^2)$ itself, $(x)$ and $(\mathbb{Z}/p)[x]$.
A: Here is another way to look at this question, maybe longer but that show what an isomorphism would "concretely" look like, and why it cannot work.


*

*The product in the ring $A:=\frac {(Z/p)[x]}{(x^2)}$ has the form $(a+bX)(a'+b'X)=aa'+(ab'+ba')X \pmod{X^2}$.


Thus, this ring is isomorphic to $(Z/p \times Z/p, +, * )$ where 
$$\tag{1}(a,b) * (a',b') = (aa',ab'+ba')$$


*

*whereas $B:=\frac {(Z/p)[x]}{(x)} \times \frac {(Z/p)[x]}{(x)} \equiv Z/p \times Z/p$ has the "classical" multiplication rule:


$$\tag{2}(a,b) \times (a',b') = (aa', bb')$$
Clearly (2) cannot be assimilated to (1). 
Let us be more precise by a reasoning by contradiction. 
Let us assume that a ring isomorphism $\varphi$ exists between $A$ and $B$.
This isomorphism could be transfered, through the following diagram:
$$\require{AMScd}
\alpha \begin{CD}
A @>{\varphi}>> B\\
@VVV  @VVV \\
\mathbb{Z/p}\times\mathbb{Z/p} @>{\psi}>>\mathbb{Z/p}\times\mathbb{Z/p}
\end{CD}\beta$$
into this one:
$$\beta \circ \varphi \circ \alpha^{-1}=:\psi:(aa',ab'+ba')\mapsto(aa',bb')$$
Thus, we would have in particular, taking $a=a'=1$, for all $b, b'$:
$$\tag{3}\psi(1,b+b')=(1,bb').$$
The fact that such an isomorphism exists is impossible for different reasons.
Here is a basic proof that the impossibility is of a serious nature: $\varphi$ is not even well defined. 
Here is why. Due to (3), we could write:
$$\psi(1,0)=\begin{cases}\psi(1,(p-1)+1)=(1,p-1)\\\psi(1,(p-2)+2)=(1,2(p-2))\end{cases}$$
The contradiction comes from the fact that in general $p-1 \neq 2(p-2) \pmod{p}$ i.e., $p \neq 3 \pmod{p}$, thus a same element can have more than one image.
(I leave as an exercice the proof of the contradiction for the case $p=3$).
The impossibility of the existence of such an isomorphism $\psi$ yields the impossibility of the existence of isomorphism $\varphi$.
A: I would simply show you can't create a bijection that satisfies the properties to make the two isomorphic. $\Bbb{Z}_p [x] \big/ \langle x^2 \rangle$ will have polynomials of a greater order, so you cannot construct said bijection.
