Is $\Bbb Q[x]/(x^2+x)$ isomorphic to $\Bbb Q[x]/(x^2-x)$? It seems the statement is true, but I have no idea how to prove it 
I try to let $f=(x^2+x)Q(x)+\bar f=(x^2-2)P(x)+\bar f'$
Then I construct a function $\phi:\Bbb Q[x]/(x^2+x) \to  \Bbb Q[x]/(x^2-x)$ by 
$\phi(\bar f+(x^2+x))=\phi(\bar f'+(x^2-x))$
Is this function well-defined? Am I going in the right way?
If not, any hints for constructing such function?
Thank you!
 A: Map $x+(x^2+x)\mapsto -x+(x^2-x)$
A: Hagen's answer is great, if you prefer an alternative note that the Chinese Remainder theorem gives

$$\Bbb Q[x]/(x(x+1))\cong \Bbb Q\oplus\Bbb Q\cong \Bbb Q[x]/(x(x-1))$$

as rings since $(x), (x+1), (x-1)$ are prime ideals.
Edit The generalized CRT says that if you have an ideal $I$ which is the product of coprime prime-power ideals i.e. $I=\mathfrak{p}^n\mathfrak{q}^m$ and $\mathfrak{p}^n+\mathfrak{q}^m=R$ then $R/I\cong R/\mathfrak{p}^n\oplus R/\mathfrak{q}^m$.
In your case the ideals $(x), (x+1)$ are clearly coprime since $1\in (x)+(x+1)$ since $1= 1\cdot(x+1)+(-1)\cdot (x)$ and similarly for $(x), (x-1)$. But then
$$\begin{cases}\Bbb Q[x]/x\cong \Bbb Q\\
 \Bbb Q[x]/(x-1)\cong \Bbb Q \\ 
\Bbb Q[x]/(x+1)\cong\Bbb Q\end{cases}$$
are all clear by maps
$$\begin{cases}x\mapsto 0\\ x\mapsto 1\\ x\mapsto -1\end{cases}$$
and the first isomorphism theorem.
A: Hint $ $ $r$ is a root of $f(x) = x^2+x $ iff $\,-r$ is a root of $f(-x) = x^2-x$
A: $x^2-x = (x-1)^2 + (x-1)$, so if you call $y = x-1$, $\Bbb Q[x]/(x^2-x) = \Bbb Q[x]/(y^2+y) \cong \Bbb Q[y]/(y^2+y)$
