Writing a ring $\mathbb Z[\alpha]$ as a quotient ring Context.
If have an algebraic element $\alpha$ over $\mathbb Q$, and I want to write $\mathbb Z[\alpha]:=\{a+\alpha b,\ a,b\in \mathbb Z\}$ as a quotient ring of the form $\mathbb Z[X]/I$.

Is the following approach correct?

Let $\pi$ be an irreducible of $\mathbb Z[X]$ such that $\pi(\alpha)=0$.
Then let's consider the function
$$ \begin{matrix}\varphi\colon& \mathbb Z[X] & \to & \mathbb Z[\alpha] \\ &P& \mapsto& P(\alpha).\end{matrix}$$
The function $\varphi$ is a surjective ring morphism.
Plus, if $\varphi(P)=0$, then let's do the euclidean division (in $\mathbb Q[X]$) of $P$ by $\pi$:
$$P=Q\pi + R.$$
So we have $Q(\alpha)\pi(\alpha)+R(\alpha)=0$, so $R(\alpha)=0$ since $\pi(\alpha)=0$.
So $P\in (\pi)$ where $(\pi)=\pi\mathbb Z[X]$ the ideal generated by $\pi$.
Reciprocally, if $P\in (\pi)$ we obviously have $P\in \mathrm{ker}(\varphi)$.
Then,
$$\mathbb Z[X]/(\pi)\simeq \mathbb Z[\alpha].$$

Edit.
Thanks to a comment, I should assume certain conditions on $\alpha$ which would assure that $\mathbb Z[\alpha]$ is a ring. It seems that $\alpha$ is algebraic of degree $2$ is sufficient, so I will be assuming this.

The way you have defined it, it is not even a ring
