Why does the argument that $\mathbb{Q}[x] / \langle x^{2}\rangle \simeq \mathbb{Q}[x] / \langle x^{2} - 1\rangle $ not apply in $\mathbb{Z_2}$? So I have an exercise asking if $\mathbb{Q}{[x]} / \langle x^{2}\rangle  \simeq \mathbb{Q}{[x]} / \langle x^{2} - 1\rangle $ and if $\mathbb{Z_2}{[x]} / \langle x^{2}\rangle  \simeq \mathbb{Z_2}{[x]} / \langle x^{2} - 1\rangle $. My opinion is yes for $\mathbb{Q}$ and no for $\mathbb{Z_2}$, using contradiction:
Suppose we have an isomorphism $\phi : \mathbb{Q}{[x]} / \langle x^{2}\rangle  \rightarrow \mathbb{Q}{[x]} / \langle x^{2} - 1\rangle $, that implies $\phi(x^{2}) = \phi(0) = 0 \implies \phi^{-1}(0) = \phi(x^{2}) = 0$, which is not possible since $\phi^{-1}(x^{2}) = 1$, it will work in $\mathbb{Q}$ but we have $0 = 1$ in $\mathbb{Z_2}$.
Could someone please explain to me how to do this?
 A: (Thanks to Vercassivelaunos for the correction. My previous answer was quite far off.)
Define a homomorphism of rings (check it's well-defined)
$$
\varphi \colon k[x] \to k[y]/(y^2-1),  
x \mapsto y-1, 1 \mapsto 1.
$$
In characteristic 2, $$y^2 -1 = 0 \iff (y-1)^2 = 0 \iff y = 1,$$
so the kernel is $(x^2)$, and apply the fundamental homomorphism theorem to conclude.
In characteristic zero this map $\varphi$ is not an isomorphism. In fact, you should convince yourself that however we write down a map it cannot possibly be an isomorphism (see Showing that $\frac{\mathbb{R}[x]}{\langle x \rangle}$ and $\frac{\mathbb{R}[x]}{\langle x-1 \rangle}$ are not isomorphic as $\mathbb{R}[x]$ modules.).
A: $\text{Nil}(\frac{\Bbb Q[x]}{x^2-1})$,$\text{Nil}(\frac{\Bbb Z_2[x]}{x^2-1})$ are trivial nilradicals because where $\Bbb F$ is a field we must have  $$f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x^2-1\rangle}\big)\iff f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x-1\rangle}\big)\;\land\;f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x+1\rangle}\big)$$ $$\therefore\;f(x)\in\text{Nil}\big(\frac{\Bbb F[x]}{\langle x^2-1\rangle}\big)\iff f(x)=0+\langle x^2-1\rangle$$ $x\pm 1$ are irreducibles, $\langle x\pm 1\rangle$ are maximal and thus $\frac{\Bbb F[x]}{\langle x\pm 1\rangle}$ are fields but $$X_1:=x+\langle x^2\rangle_{\Bbb Q[x]}\in\frac{\Bbb Q[x]}{\langle x^2\rangle}\;\;\text{and}\;\;X_2:=x+\langle x^2\rangle_{\Bbb Z_2[x]}\in\frac{\Bbb Z_2[x]}{\langle x^2\rangle}$$ are non-trivial nilpotents as $$X_1^2=x^2+\langle x^2\rangle_{\Bbb Q[x]}=\langle x^2\rangle_{\Bbb Q[x]}\;\;\;\;\;X_2^2=x^2+\langle x^2\rangle_{\Bbb Z_2[x]}=\langle x^2\rangle_{\Bbb Z_2[x]}$$
Therefore $\text{Nil}(\frac{\Bbb Q[x]}{x^2-1})$,$\text{Nil}(\frac{\Bbb Z_2[x]}{x^2-1})$ are trivial but $\text{Nil}(\frac{\Bbb Q[x]}{\langle x^2\rangle}),\text{Nil}(\frac{\Bbb Z_2}{\langle x^2\rangle})$ are $\it{non}$-$\text{trivial}$. $$\therefore\;\frac{\Bbb Q[x]}{\langle x^2\rangle}\not\approx\frac{\Bbb Q[x]}{\langle x^2-1\rangle}\;\land\;\frac{\Bbb Z_2[x]}{\langle x^2\rangle}\not\approx\frac{\Bbb Z_2[x]}{\langle x^2-1\rangle}$$
