# Square root of an algebraic integer is an algebraic integer

For some $\alpha\in\mathbb C$ let $E=\mathbb Q(\alpha)$ be a number field and $\mathcal I$ its ring of integers. Suppose $(a_1+b_1\alpha)\cdots(a_k+b_k\alpha)=\beta^2$ for some $\beta\in E$ and $a_i,b_i\in\mathbb Z$. Then it follows that $\beta\in\mathcal I$.

My initial idea: $\beta^2$ is the product of algebraic integers, so it follows that $\beta^2\in\mathcal I$ and therefore $\beta^2$ is the root of some monic polynomial $p(x)$. But now I am struggling can construct another polynomial $q(x)$ such that $\beta$ is a root.

• Like $q(x)=p(x^2)$? :) Aug 20, 2013 at 12:35
• $\mathcal{I}$ is integrally closed if I remember correctly. $\beta$ is certainly integral over $\mathcal{I}$. Aug 20, 2013 at 12:36
• Aha, sometimes you just don't see the forest for the trees. Aug 20, 2013 at 12:39

$$p(x)=\sum_{k=0}^na_kx^k\implies 0=p(\beta^2)=\sum_{k=0}^na_k\beta^{2k}\implies \beta\;\text{is the root of}$$
$$q(x):=a_0+a_1x^2+a_2x^4+...+a_nx^{2n}=\sum_{k=0}^na_k(x^2)^k$$