= Solution
A complex number is an <algebraic number> over $\mathbb Q$ when it is a root of a nonzero rational polynomial. Its <minimal polynomial of an algebraic element> is the unique monic polynomial in $\mathbb Q[X]$ of smallest degree annihilating it; polynomial division shows that it divides every rational polynomial annihilating the element, and it is irreducible.
Consider the surjective evaluation <ring homomorphism> $\operatorname{ev}_\alpha:\mathbb Z[X]\to\mathbb Z[\alpha]$. Certainly $(f)\subseteq\ker\operatorname{ev}_\alpha$. Because the nonconstant $f$ is irreducible over $\mathbb Z$, its content is a unit: otherwise its content and its nonconstant primitive part give a factorization into nonunits. By the <Gauss lemma for polynomials>, $f$ is irreducible in $\mathbb Q[X]$ and is a scalar multiple of the minimal polynomial.
If $g\in\mathbb Z[X]$ vanishes at $\alpha$, then $f$ divides $g$ in $\mathbb Q[X]$. Primitive divisibility in <Gauss lemma for polynomials> implies division in $\mathbb Z[X]$ as well. Explicitly, write the rational quotient as $(a/b)h$ with $h$ primitive integral and $\gcd(a,b)=1$; the primitive product $fh$ shows that integrality of $(a/b)fh$ forces $b=1$. Thus the kernel is precisely $(f)$, and the <first isomorphism theorem for rings> yields
$$
\boxed{\mathbb Z[X]/(f)\cong\mathbb Z[\alpha],\qquad[g]\mapsto g(\alpha).}
$$
No monicity of $f$ is needed; this is an isomorphism of rings, not an assertion that either ring is a field.
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