Help with Problem in Abstract Algebra

Question

Let $K$ be an algebraic number field and $O_K$ its ring of integers. Let $P$ be a prime ideal of $O_K$. Prove that $G = \{a+P \mid a \in O_K, a \not\in P\}$ is a cyclic group with respect to multiplication. What is the order of $G$?

I have been stuck on this problem for some time now. Proving that $G$ is a group was fairly straightforward but I am unsure how to prove that it is cyclic. This problem was in the Norms of Ideals chapter in my book but I am not sure how to apply concepts learned from that chapter for this problem. Any help is appreciated. Thanks.

Isn't $G$ equal to the group of units in $O_K/P = k(P)$? The fact that such a group is known to be cyclic when $k(P)$ is finite might help. — Daniel, Feb 16 '22 at 03:01
Since $P$ is prime ideal, $O_K/P$ is a finite field and hence $(O_K/P)^{*}$ forms a cyclic group with respect to multiplication — MAS, Feb 16 '22 at 07:32

score 1 · Answer 1 · answered Feb 16 '22 at 03:05

This rest on the following two facts about the ring $\mathcal O $. (i) All nonzero prime ideals in this ring are also maximal ideals. (2) For any ideal $I$ the quotient ring $\mathcal O/I$ is a finite ring (the number of elements being called the norm of the ideal).

Now use this along with the following: (3) For any finite field the nonzero elements form a cyclic group under multiplication. (A proof of this uses the property of Euler phi function)

Help with Problem in Abstract Algebra

1 Answers1