Algebraic integers divided by a prime Denote the set of all algebraic integers by $A$ (thus $A$ is a subset of $\mathbb C$). Denote by $A'$ the set of all $\alpha\in A$ such that $\frac{\alpha}{p}\not\in A$ for all primes $p$. Question : if $\alpha\in A'$, does it automatically follow that the powers of $\alpha$ are in
$A'$ also ? 
My algebraic number theory is a little bit rusty, so I may have missed something easy on this one.
 A: No. For example, $\sqrt 2$ is an algebraic integer, whilst for any $p$, $\frac {\sqrt2}p$ is a root of $$ p^2x^2-2$$ so is not an algebraic integer. However $$\frac {(\sqrt 2)^2}2=1.$$

Whilst this answer may seem trivial, it actually reveals something: the idea of ramification. Let $\alpha$ be an algebraic integer and let $K=\mathbb Q(\alpha)$. Suppose that $\alpha\in A'$. For any rational prime $p$, we can factorise
$$p\mathcal O_K=\mathfrak p_1^{e_1}\cdots \mathfrak p_g^{e_g}$$ where the $\mathfrak p_i$ are prime ideals of the ring of integers $\mathcal O_K$. Suppose that $e_i=1$ for all $i$ - i.e. that the prime $p$ is unramified in the extension. Suppose , moreover, that $\frac{\alpha^m}p$ is an algebraic integer for some $m$ with $m$ minimal. Then $$\alpha^m\in p\mathcal O_K=\mathfrak p_1\cdots \mathfrak p_g.$$ So $$\mathfrak p_1\mid\alpha^m\mathcal O_K=(\alpha\mathcal O_K)^m$$but $$\mathfrak p_1\nmid \alpha \mathcal O_K,$$contradicting the primarily of $\mathfrak p_1$. So the prime $p$ must be ramified in the extension. 
