Prove that the ring of algebraic integers in is .
Recall that an element in an algebraic extension of is called an algebraic integer if its minimal polynomial over has only integer coefficients.
First we will show that the minimal polynomial of every element of has only integer coefficients. To that end, let . If , then certainly is an algebraic integer, since its minimal polynomial is simply . Suppose ; in this case , and so its minimal polynomial has degree at least 2. Note that is a root of , and ; so has degree exactly 2, and in fact is its minimal polynomial. So is an algebraic integer.
Conversely, suppose is an algebraic integer. If , then is an algebraic integer, and so . Certainly then . Suppose now that . If , then is a root of , which is irreducible over and hence the minimal polynomial of . So . If is not an integer, then neither is ; hence , and so . So we assume .
Now , so the minimal polynomial of over has degree at least 2. Since is (evidently) a root of , this must be the minimal polynomial of . Since is an algebraic integer, we have .
Let us say and , where and . Then (rewriting) we have , so that . Let be a prime factor of ; then . Since and are relatively prime, , and in fact . So for some . However, if , then we have , a contradiction since and are relatively prime. So or . Suppose .
We also have that , so that . Say ; rearranging, we have . Let be a prime factor of . Again because and are relatively prime, we have , so that . So for some . If , then we have 2|c$, a contradiction. So or . Suppose .
Now , so that for some . In particular, mod 4. Note that the squares mod 4 are precisely 0 and 1, and the only solutions of this equivalence are . In any case, and are either 0 (a contradiction) or divisible by 2 (a contradiction since they are relatively prime to and , respectively.) Now . Again, we have for some integer , so that mod 4, a contradiction. So .
Thus as desired.
Hence, the ring of algebraic integers in is precisely .