 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 $\mathrm{deg}r<\mathrm{deg}f$. Since $r(\alpha )=0$, minimality forces $r=0$. Conversely each multiple of $f$ vanishes at $\alpha$. ThereforeWriting $n=\mathrm{deg}f$, 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 $1,\alpha ,\dots ,{\alpha }^{n-1}$ form a basis of the free module over $\mathbb{Z}$ $\mathbb{Z}[\alpha ]$.