Check whether two given given rings are isomorphic Examine with justification whether $\Bbb Q[X]$/$[X^2-X]$ and $\Bbb Q$$\times$$\Bbb Q$ are ring isomorphic.
My approach:
say if we take any polynomial $aX^3+bX^2+cX+d$ in $\Bbb Q[X]$, so in $\Bbb Q[X]$/$[X^2-X]$ it will be          
$aX^3+bX^2+cX+d$
=$aX^3-aX^2+bX^2-bX+cX+d+aX^2+bX$
=$aX(X^2-X)+b(X^2-X)+cX+d+aX^2-aX+bX+aX$
= $aX+bX+cX+d$,   as $X^2-X$ is the kernel.
Hence by simple induction method, it can be shown any polynomial $a_0+a_1X+a_2X^2+...+a_nX^n$  in $\Bbb Q[X]/[X^2-X]$ can be reduced to $(a_1+a_2+...+a_n)X+a_0$.
Now let us assume there exists an isomorphism $\mathcal L$ between this to rings.
Say,    $\mathcal L$($cX$) =$a$,  where $cX$ $\in$ $\Bbb Q[X]$/$[X^2-X]$, and $a$ $\in$ $\Bbb Q$$\times$$\Bbb Q$ .
Now, $\mathcal L$($cX$$\times cX$)=  $\mathcal L$($c^2X^2$)= $\mathcal L$($c^2X$)= $a^2$.
But $\mathcal L$($c^2X$)=$\mathcal L$($cX$) $\mathcal L$($c$)=$a^2$, which implies $\mathcal L$($c$)=$a$.
Now ofcourse $c$ $\in$ $\Bbb Q[X]$/$[X^2-X]$ and we get  $\mathcal L$($cX$)= $\mathcal L$($c$)=$a$.
So these two rings cannot be isomorphic.
Is my approach correct?
 A: Your only error is here:

But $\mathcal L$($c^2X$)=$\mathcal L$($cX$) $\mathcal L$($c$)=$a^2$, which implies $\mathcal L$($c$)=$a$.

$a \mathcal L(c) = a^2$ does not imply $\mathcal L (c) = a$. It would if $a$ were invertible, but $\mathbb Q \times \mathbb Q$ is not a field; there are noninvertible elements (e.g., $(0,1)$).
Thus your proof shows that if an isomorphism $\mathcal L$ exists, $\mathcal L(cX)$ cannot be invertible. We can get some more conditions on the isomorphism $\mathcal L$.
In the particular case $c=1$, we have $\mathcal L(X) = \mathcal L(X^2) = (\mathcal L(X))^2$. There are only four members of $\mathbb Q \times \mathbb Q$ which are their own square: $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. Since we also know $\mathcal L(X)$ is not invertible (by your argument), it can't be $(1,1)$. Also , as a ring homomorphism, $\mathcal L$ must map zero element to zero element; thus $\mathcal L(0) = (0,0)$, so $\mathcal L(X) \neq (0,0)$. We have left two symmetric options: $L(X) = (0,1)$ or $L(X) = (1,0)$. Combining this with the fact that $L(1) = (1,1)$ (since a ring homomorphism must map multiplicative identity to multiplicative identity) determines the isomorphism; you can now check that these two both give isomorphisms (it suffices to check one option, and the other must work by symmetry), and, if you want, you can check that they match the Chinese Remainder Theorem isomorphisms rschwieb pointed to.
A: No, it's not correct.
Using the ideals $I=(x)$ and $J=(1-x)$ in $\mathbb Q[x]$, we have, by the Chinese Remainder Theorem, that $\mathbb Q[x]/(I\cap J)\cong \mathbb Q[x]/I\times \mathbb Q[x]/J$.

But $\mathcal L$($c^2X$)=$\mathcal L$($cX$) $\mathcal L$($c$)=$a^2$, which implies $\mathcal L$($c$)=$a$.

And why should that be? You could only cancel like that if $a$ is a regular element (in the sense that $ax=0$ implies $x=0$), and we already know it is not (since $a\mathcal L(c^{-1}(x-1))=0$).
