How to find the invertible elements of $\Bbb{Q}[X]$ mod $X^2$ 
I am looking for some hints for finding all the invertible elements of $\Bbb{Q}[X]$ mod $X^2$.

Thank you very much in advance.
 A: The ring $\mathbb{Q}[X]/(X^2)$ is local, that is, it has a unique maximal ideal.
An element of the maximal ideal is not invertible (prove it). What about the elements not in the maximal ideal?
A: In this ring, $\overline{X}$ is a nilpotent, because $\overline{X}^2 = 0$. Nilpontents behave in many ways like infintiesimals; many tools from calculus become available, such as power series:
$$ (1 - x)^{-1} = 1 + x + x^2 + \ldots $$
which truncate to finite sums. And you can check
$$ \overline{(1-X}) \overline{(1+X)} = \overline{1 - X^2} = \overline{1}$$
Any element of this ring is of the form $\overline{a + bX}$, so you can adapt this approach for computing inverse (and seeing when it fails).

However, the specific problem of finding inverses has a neat trick:
$$ (a + \epsilon) \cdot (a - \epsilon) = a^2 - \epsilon^2 $$
which gives you an algorithmic way to keep multiplying in terms to keep increasing the order of the infinitesimal adjustment until it goes away entirely. Of course, here, it goes away after one step, since $\overline{X}^2 = \overline{1}$.

There is a homomorphism $\mathbb{Q}[X] / X^2 \to \mathbb{Q}$ defined by sending $X \mapsto 0$. Thus, if $\overline{a + b X}$ is a unit, so is $a+b0 = a$.
You now have enough tools to prove the converse.
