# How to prove an algebraic integer.

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.

-

Just to be clear, an algebraic integer is a root of a monic polynomial with integer coefficients.

First show that the given numbers are algebraic, by finding polynomials of which they are roots (this will involve squaring things, moving all terms without square roots to one side of the equation, and squaring again). Then you divide by the coefficient of the leading term, and check whether all the coefficients are still integers.

I can be more explicit if necessary, after you have attempted the above.

-
Great, that helps a lot. Thank you! – J.B May 10 '13 at 10:44

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.

-