NeukirchPosted on June 1, 2019
Here are my exercise solutions and notes for Neukirch’s Algebraic Number Theory.
1.1. Recall that for . Therefore for . Moreover, for all . Thus if we must have and therefore .
1.2. Suppose that for with a unit and relatively prime. Since is a UFD we may write out the prime factorizations and with units and the ’s and ’s prime. Since are relatively prime, the ’s and ’s are all distinct primes. Therefore, since , we must have and for all . I.e., we have and where and .
1.3. Observe that if for , then for . Therefore by Exercise 1.2 we have with a unit and . Therefore . So and . From Exercise 1 we see that the only units in are . We cannot have as . Thus we may ignore the as it just swaps the sign.
Observe that if then we we would have , a contradiction. If are both odd, then and . Thus , a contradiction.
1.4. Recall that an ordering on a ring satisfies implies for all , as well as and implies for all . The first implication shows that for all ; in particular . If , then , a contradiction. However, if then and .
1.5. Observe that we may define a norm on by where if we have . This gives . Again we see that and thus an element is a unit if and only if . The only elements of with norm are .
1.6. After struggling for a while and digging a bit on math stack exchange, it appears that this question is fairly difficult for its place in the book. Perhaps I will revisit this question after progressing further.