Number of elements in the quotient ring $\mathbb{Z}[X]/(X^2-3, 2X+4)$ 
I had to calculate the number of elements of this quotient ring:
  $$R = \mathbb{Z}[X]/(X^2-3, 2X+4).$$

This is what I've got by myself and by using an internet source:
Writing the ring $R = \mathbb{Z}[X]/(X^2−3, 2X+4)$ as $\mathbb{Z}[X]/I$, we note that $I $ contains $2(X^2−3)−(X−2)(2X+4) = 2$. We then note that the generator $2X + 4$ is actually superfluous since $2X + 4 = 2(X + 2)$. 
Now we can write $R = \mathbb{Z}[X]/(X^2 − 3, 2) = \mathbb{Z}[x]/((X+1)^2, 2)$, because $(X+1)^2=X^2+2X+1=X^2-3+2X+4$. The internet source states now the following:
$$ R \cong (\mathbb{Z}/2\mathbb{Z})[\alpha] \quad \text{with} \quad \alpha = X+1$$
I guess that $\mathbb{Z}/2\mathbb{Z}[\alpha]$ represents the set of dual numbers of the field $\mathbb{Z}/2\mathbb{Z}$. I see that $\alpha^2=0$, but what exactly implies the isomorphism? And does this mean that the quotient ring $R$ contains four elements?
If necessary, you can take at the site I used. (It's about page 5, exercise 4.3a.) http://www.math.umn.edu/~musiker/5286H/Sol1.pdf 
I thank you in advance for your answers.
 A: Here is an alternative approach. Remember, for ideals $I,J \trianglelefteq R$ we have $R/(I+J) \cong (R/I)/\overline{J}$ where $\overline{J}$ denotes the embedding of $J$ in $R/I$.
In this case we have $\mathbb{Z}[X]/(X^2-3,2X+4) \cong (\mathbb{Z}[X]/(X^2-3))/(2 \overline{X}+4) \cong \mathbb{Z}[\sqrt{3}]/(2 (\sqrt{3}+2))$. In $\mathbb{Z}[\sqrt{3}]$ the element $2+\sqrt{3}$ is a unit (with inverse $2-\sqrt{3}$), so $\mathbb{Z}[\sqrt{3}]/(2 (\sqrt{3}+2)) = \mathbb{Z}[\sqrt{3}]/(2) \cong \mathbb{Z}[X]/(X^2-3,2) \cong \mathbb{F}_2[X]/(X^2+1)$.
The latter is obviously a two-dimensional $\mathbb{F}_2$-vector-space so the number of elements is clear.
Of course, instead of considering the first line of isomorphisms you could also show the equality $(X^2-3,2X+4) = (X^2-3,2)$ directly. Nevertheless the correspondence to $\mathbb{Z}[\sqrt{3}]$ is useful, because it gives us an idea how to show this.
A: You almost have it. To get representatives of the classes in the quotient ring divide by $X^2-3$ and take the remainders as those representatives. Notice you can do long division by $X^2-3$ over the integers because the leading coefficient is $1$. What remainders do you get? Then the mod $2$ reduces the list further, from an infinite list to a finite list.
