Let $c$ be an integer, not necessarily positive and $|c|$ not a square. Let $\mathbb{Z}[\sqrt{c}]$ be the set of complex numbers $$a+b\sqrt{c}, \quad a, b\in \mathbb{Z},$$ which form a subring of the ring $\mathbb{C}$ under the usual addition and multiplication.

Are the following questions completely solved?

  1. For what $c$ is $\mathbb{Z}[\sqrt{c}]$ a Euclidean domain?

  2. For what $c$ is $\mathbb{Z}[\sqrt{c}]$ a UFD (unique factorization domain) but not Euclidean ?

  3. For what $c$ is $\mathbb{Z}[\sqrt{c}]$ not a UFD ?

I know that for $c=-1$, Question 1 has a positive answer; for $c=-5$, Question 3 has a positive answer.

  • 6
    $\begingroup$ Do you really mean to be writing about the ring Z[sqrt(c)] or do you want to write about the ring of integers of Q(sqrt(c))? $\endgroup$ – KConrad Jan 10 '11 at 19:32
  • $\begingroup$ I guess your question reduces to whether I want $|c|$ to be squarefree or not. No, that wasn't in my mind when I wrote this. But is it true that if $|c|$ is not squrefree, then $\mathbb{Z}[\sqrt{c}]$ will not be UFD? $\endgroup$ – TCL Jan 10 '11 at 20:09
  • 1
    $\begingroup$ TCL - yes. UFDs are integrally closed, so you need c squarefree and, if $c=1$ mod 4, look at $\mathbb{Z}\left[\frac{1+\sqrt{c}}{2}\right]$. $\endgroup$ – George Lowther Jan 10 '11 at 21:28
  • 1
    $\begingroup$ @TCL: For your question, exactly as you wrote it, Franz Lemmermeyer gave a detailed answer. @George Lowther: It seems TCL is actually interested in Z[sqrt(c)], regardless whether it is the maximal order or not (which intially several people, including me, doubted). So, there seems no need to consider a different ring for 1 mod 4. $\endgroup$ – user9072 Jan 10 '11 at 22:26

A quadratic order has unique factorization (or is Euclidean) only if it is the maximal order of a number field (it must be integrally closed); for your examples, this holds if and only if $c$ is a squarefree integer congruent to $2$ or $3$ modulo $4$.

  1. It is known, as others have pointed out, which quadratic fields are Euclidean with respect to the absolute value of the norm. In the complex quadratic case, every Euclidean ring is norm-Euclidean (Motzkin); in the real quadratic case, there are examples (by students of Murty, namely Clark and Harper) of Euclidean rings that are not norm-Euclidean.

  2. By a result of Weinberger, every real quadratic field with unique factorization is Euclidean assuming the generalized Riemann hypothesis.

  3. The ring of integers in a quadratic number field is not a UFD if its class number is nontrivial; it is easy to construct examples by making $c$ a product of at least three primes. It is believed but not known that infinitely many quadratic number rings have class number $1$.

See Simachew or my own survey for references.

| cite | improve this answer | |
  • $\begingroup$ Franz's survey is fantastic. By the way Franz, the first student of Murty you mention is David Clark: no "e". (Also no relation to me.) $\endgroup$ – Pete L. Clark Jan 10 '11 at 18:43

For question #1, see this sequence, which contains all values $c$ such that $\mathbb{Z}[\sqrt{c}]$ is Euclidean: http://oeis.org/A048981

Note that these are the only such quadratic fields which are Euclidean.

For question #3, it is not solved completely since it is not known which values of $c>0$ produce a unique factorization domain. However, the problem is solved for $c<0$ completely via the Stark-Heegner theorem. The problem in general dates back to Gauss, and is known as the class number problem. Wikipedia has good info.

Using the answers to these, we can answer #2 since there are only a finite number of Euclidean quadratic fields.

| cite | improve this answer | |
  • 4
    $\begingroup$ Please be more careful and write that those fields are norm-Euclidean, not just Euclidean. It's known that GRH implies the ring of integers of any number field with an infinite unit group (e.g., real quadratic field) which has class number 1 is a Euclidean domain in the sense of having some Euclidean function, but that might not be the norm function. The sequence you refer to is the norm-Euclidean quadratic fields, so it does not directly address the original questions about Euclidean quadratic fields (in the real quadratic case). $\endgroup$ – KConrad Jan 10 '11 at 19:31

This is more a comment to the question (which I cannot do).

As written (intentionally?) [ADDED: Apparently, it was intentional.] the specified rings are not always the (full) ring of algebraic integers of the field $\mathbb{Q}(\sqrt{c})$ (see, e.g., https://en.wikipedia.org/wiki/Quadratic_integer for details).

In these cases the rings in question are not integrally closed and thus not UFDs, even if the class number of the field is one and thus the (full) ring of algebraic integers would be a UFD (see, e.g., https://en.wikipedia.org/wiki/Class_number_problem ).

Possibly, one needs to take this into account too, when using the list, mentioned in another answer, where the rings are Euclidean.

ADDED: Franz Lemmermeyer's answer is considerably more complete than this remark.

| cite | improve this answer | |

Question 1 is slightly ambiguous. Does "euclidean" mean euclidean with respect to the obvious norm, or does it mean euclidean with respect to some norm?

For example, Z[$\sqrt{14}$] is not euclidean w.r.t. the usual norm, but I'm not sure it's known whether there's some other norm that makes it euclidean.

| cite | improve this answer | |
  • 3
    $\begingroup$ Actually this is an important point. The Euclidean quadratic fields which I referred to in my response are ones whose standard norm${}^\ast$ is a Euclidean function. There are indeed quadratic fields whose integer ring is Euclidean (there exists a Euclidean norm), yet the standard norm is not a Euclidean function. $\mathbb{Z}[\sqrt{69}]$ is one such example, and it is an open problem if there are an infinite number of them. ${}^\ast$ Given $\mathbb{Z}[\sqrt{d}]$, [\vert a+b\sqrt{d}\vert=\prod_{\rho\in\operatorname{RootsOf}_x(x^d - 1)}(a+b\rho)] $\endgroup$ – Quadrescence Jan 10 '11 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.