I have troubles in solving the first question of the exercise: Let $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ a polynomial in $\mathbb{Z}[x]$, and let $p$ be a prime divisor of $a_0$. Let $p^r$ be the exact power of $p$ dividing $a_0$ and suppose $p^r$ divides all $a_k$. Assume moreover that $f$ is irreducible over $\mathbb{Q}$ and let $\alpha$ be a root of $f.$ Let $K=\mathbb{Q}(\alpha).$ Prove that (p^r) is the $n-$th power of an ideal in $\mathcal{O}_K.$
Following the hint, I have proved that $\alpha^n=p^r\beta$ with $(\beta)$ and $(p)$ comaximal. Then $(\alpha)^n\subset(p^r).$ how can I prove the other containment?
It need not be the case that $(p^r)=(\alpha)^n$.
But the result to be proved is that the ideal $(p^r)$ is the $n$-th power of some ideal of $\mathcal{O}_K$.
It can be shown as follows . . .
Let \begin{align*} (p^r)&=P_1^{v_1}\cdots P_j^{v_j}\\[4pt] (\beta)&=Q_1^{w_1}\cdots Q_k^{w_k}\\[4pt] \end{align*} be the unique factorizations in $\mathcal{O}_K$ of the ideals $(p^r)$ and $(\beta)$ into prime ideals.
From $(p,\beta)=(1)$, it follows that the sets $\{P_1,...,P_j\}$ and $\{Q_1,...,Q_k\}$ are disjoint.
In the unique factorization of $(\alpha^n)$ into prime ideals, the exponent for each prime ideal is a multiple of $n$, hence, from $\alpha^n=p^r\beta$, it follows that each of the exponents $v_1,...,v_j$ is a multiple of $n$.
Thus, we get \begin{align*} (p^r)&=P_1^{v_1}\cdots P_j^{v_j}\\[4pt] &=P_1^{nu_1}\cdots P_j^{nu_j},\;\text{for some}\;u_1,...,u_j\in\mathbb{Z}^+\\[4pt] &=(P_1^{u_1}\cdots P_j^{u_j})^n\\[4pt] \end{align*} so $(p^r)$ is the $n$-th power of an ideal of $\mathcal{O}_K$, as was to be shown.