Quote by Jarle
[tex]y= p+q\sqrt{2}[/tex] is a general element of [tex]\mathbb{Q}(\sqrt{2})[/tex] where p and q are rational. First you will need to show that if y is a solution to a polynomial over Z, then it is a solution to a polynomial over Z of degree 2. Then find necessary and sufficient conditions for y such that y is the solution of such a polynomial (use the fact that p and q are rational).

This works for quadratic and some cubic extensions. For bigger extensions, this gets really nasty quite quickly, so you're better off using some more theory (differents, discriminants ,etc.)
In fact it's quite easy to see by simple calculations what is ring of integers in Q(sqrt(d)), but not so simple to find integral closure in Q(17^(1/3)) with simple calculations like this. (the example I gave above).