2

Somebody can to help me in the following problem:

Let R be the following subring of the complex numbers: $$R = \left\{\frac{z_1}{2}+\frac{z_2\sqrt{-19}}{2} : z_1,z_2\in\mathbb{Z}, \;\text{with the same parity}\right\}.$$ Then R is a principal ideal domain that is not a Euclidean domain.

Note: A ring for me is commutative and with identity.

Walner
  • 1,082
  • 1
    Duplicate of this question. – Key Ideas Jun 10 '13 at 01:29
  • 1
    What is the question here? This statement is true. – Stahl Jun 10 '13 at 01:29
  • 1
    There is a general problem with asserting that an algebraically-closed ring of algebraic integers is a PID but not Euclidean, unless the notion of "Euclidean" is constrained to mean something like "with the usual norm". Namely, I seem to recall that under GRH or similar every such ring that is a PID has some "norm" that makes it Euclidean. So please do append "not Euclidean with respect to its usual norm", or some similar. Otherwise the question is too hard. – paul garrett Jun 10 '13 at 01:36
  • I don't get to do this exercise. Do you can to help me? – Walner Jun 10 '13 at 01:36
  • This is a problem of introductory book in abstract algebra. I think that this is only a little trick that I don't know. – Walner Jun 10 '13 at 01:41
  • @paulgarrett Thee is no such problem here - see the proof I linked to, by H. Lenstra, which uses the universal side divisor criterion to deduce that no Euclidean function exists. This does not contradict Weinberger's 1973 result (assuming GRH) since that applies only to number rings with infinitely many units. – Key Ideas Jun 10 '13 at 02:02
  • @KeyIdeas, Ah, indeed, good. – paul garrett Jun 10 '13 at 12:21
  • Guys, I see that somebody already ask this question. But this don't mean that I don't have doubts. If somebody can to help me, is it wellcome. – Walner Jun 11 '13 at 00:05

0 Answers0