= Solution
An <algebraic integer> is a complex number satisfying a monic polynomial with integer coefficients. Let $f\in\mathbb Q[x]$ be the monic <minimal polynomial> of $\alpha$. It is irreducible over $\mathbb Q$. If $H\in\mathbb Z[x]$ is a monic polynomial vanishing at $\alpha$, then $f$ divides $H$ over $\mathbb Q$. <Gauss lemma for polynomials> implies that monic rational factors of a monic integer polynomial have integer coefficients. Thus $f\in\mathbb Z[x]$, and its irreducibility over $\mathbb Q$ implies irreducibility in $\mathbb Z[x]$.
For $h\in\mathbb Z[x]$ with $h(\alpha)=0$, division by the monic $f$ works in $\mathbb Z[x]$, giving $h=qf+r$ with $\deg r<\deg f$. Since $r(\alpha)=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. Therefore
$$
\boxed{I=(f),\qquad\mathbb Z[\alpha]\cong\mathbb Z[x]/(f).}
$$
Writing $n=\deg f$, division by $f$ expresses every element uniquely as $a_0+a_1\alpha+\cdots+a_{n-1}\alpha^{n-1}$ with integer coefficients. A relation among these powers would be a nonzero vanishing polynomial of degree less than $n$. Hence \b[$1,\alpha,\ldots,\alpha^{n-1}$ form a <basis> of the <free module> over $\mathbb Z$ $\mathbb Z[\alpha]$].
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