Exercise 4.4.3

Proof.

(a)

$\text{∙}$ Following this definition, suppose that $p(\alpha )=0$, where $p\in \mathbb{Z}[x]$ is monic.

Write $P\in \mathbb{Q}[x]$ the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Then $P$ divides $p$ in $\mathbb{Q}[x]$: there exists $q\in \mathbb{Q}[x]$ such that $p=\mathit{Pq}$.

By Gauss Lemma, Proposition A.3.2 of appendix A, there exists $\delta \in {\mathbb{Q}}^{\text{∗}}$ such that $\tilde{P}=\mathit{\delta P}$ and $\tilde{q}={\delta }^{-1}q$ have integer coefficients. So $p=\tilde{P}\tilde{q},\tilde{P},\tilde{q}\in \mathbb{Z}[x]$.

As $p$ is monic, $\pm \tilde{P},\pm \tilde{q}$ are also monic. Possibly by multiplying $\delta$ by $-1$, we can so suppose that $\tilde{P},\tilde{q}$ are monic. Thus $P=\tilde{P}$, and so $P\in \mathbb{Z}[x]$.

$\text{∙}$ The converse is straightforward: If the minimal polynomial $P$ of $\alpha$ over $\mathbb{Q}$ has integer coefficients, $P$ is an example of monic polynomial such that $P(\alpha )=0$ , so $\alpha$ is an algebraic integer.

Conclusion: $\alpha$ is an algebraic integer iff the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has integer coefficients.

(b)

$\omega ∕2$ is a root of $x^2+\frac{1}{2}x+\frac{1}{4}$, and $f=\omega ∕2\notin \mathbb{Q}$, thus $x^2+\frac{1}{2}x+\frac{1}{4}$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Since $f\notin \mathbb{Z}[x]$, by part (a), $\omega ∕2$ is not an algebraic integer.