Thread: Integral closure View Single Post
P: 39
 Quote by hochs 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).