TITLE: How to prove an algebraic integer.
QUESTION [5 upvotes]: How can you prove that something is an algebraic integer? I have two examples:
1) $\frac{\sqrt{p} + \sqrt{q}}{2}$ , where $p$ and $q$ are integers congruent to $3 \pmod 4$
2) $\sqrt{\frac{\sqrt{15} + \sqrt{3}}{\sqrt{6}}}$
Any help would be much appreciated.

REPLY [3 votes]: As a last resort, by exhibiting an appropriate polynomial:
With $\alpha=\frac{\sqrt p+\sqrt q}{2}$, we have $\alpha^2=\frac{p+q}4+\frac{\sqrt{pq}}{2}$.
Note that therefore $(4\alpha-(p+q))^2-4pq $ is some integer polynomial having $\alpha$ as root.
However, it is not monic. Can you see that (after expanding) dividing out the leading coefficient is possible - because of the special assumptions about $p$ and $q$?
In the second problem I'd suggest to cancel $\sqrt 3$ in the fraction before proceeding.