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Although I've marked this "under construction," I welcome help from others regarding the confusing issue of distinct multiple factorizations in those fields that are not UFDs. Alonso del Arte 17:47, 16 September 2012 (UTC)

In ${\displaystyle \scriptstyle \mathbb {Z} [{\sqrt {-5}}]\,}$, 6 has two irreducible factorizations (i.e. factorizations into irreducible elements), i.e. ${\displaystyle \scriptstyle 6\,=\,2\cdot 3\,=\,(1-{\sqrt {-5}})(1+{\sqrt {-5}}\,}$), so the [commutative] ring [with unity] ${\displaystyle \scriptstyle \mathbb {Z} [{\sqrt {-5}}]\,}$ is not a UFD. Prime elements must obey the rule ${\displaystyle \scriptstyle p\,|\,ab\implies p\,|\,a{\rm {~or~}}p\,|\,b\,}$ (true only if factorizations into irreducible elements are unique, in which case we have a prime factorization, i.e factorization into prime elements).— Daniel Forgues 05:02, 22 September 2012 (UTC)

I reaffirm that I welcome help on this one. I find the issue of associates confusing whether UFD or not. Alonso del Arte 19:36, 19 September 2012 (UTC)

Associates are subsets of [rational, Gaussian, Eisenstein, ...] integers where each element can be obtained by any other element by multiplication by a unit.
In ${\displaystyle \scriptstyle \mathbb {Z} \,}$, the rational integers ${\displaystyle \scriptstyle -n,\,+n\,}$ are associates (they come in sets of 2 elements since there are 2 units, e.g. ${\displaystyle \scriptstyle +1,\,-1\,}$) of each other.
In ${\displaystyle \scriptstyle \mathbb {Z} [i]\,}$, the Gaussian integers ${\displaystyle \scriptstyle +z,\,iz,\,-z,\,-iz\,}$ are associates (they come in sets of 4 elements since there are 4 units, e.g. ${\displaystyle \scriptstyle +1,\,i,\,-1,\,-i\,}$) of each other.

Daniel Forgues 04:39, 22 September 2012 (UTC)

I know the definition of associates and I know the definition of units. That doesn't mean I am not constantly confused by the multitude of units and associates in real fields. Alonso del Arte 16:00, 23 September 2012 (UTC)

## Contents

This page should be moved to quadratic integer rings. A new page should be written for quadratic number fields, in which all the numbers of ${\displaystyle \scriptstyle \mathbb {Q} ({\sqrt {D}})\,}$ are of the form ${\displaystyle \scriptstyle a+b{\sqrt {D}},\ a,\,b\,\in \,\mathbb {Q} \,}$ and ${\displaystyle \scriptstyle D\,}$ is squarefree.— Daniel Forgues 04:21, 22 September 2012 (UTC)

The page quadratic number fields should be moved to quadratic integer rings (i.e. ring of integers of quadratic number fields), and the page quadratic number fields should be rewritten to emphasize about the field properties. I did lots of edits, please help in replacing number fields by integer rings, then I still have to review the whole contents... Thanks.— Daniel Forgues 06:24, 22 September 2012 (UTC)

## The "wrong" rings

This article describes the wrong rings when d is 1 mod 4. In that case we want ${\displaystyle \mathbb {Z} [(1+{\sqrt {d}})/2]}$. In particular the article currently claims that ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ is a unique factorization domain. This is incorrect. For example ${\displaystyle (1+{\sqrt {13}})^{2}=2(7+{\sqrt {13}})}$. Since 2 is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ but does not divide ${\displaystyle 1+{\sqrt {13}}}$, we see that ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$ is not a UFD. This also shows that 2 is not prime in ${\displaystyle \mathbb {Z} [{\sqrt {13}}]}$. — Eric M. Schmidt 08:45, 17 February 2013 (UTC)

The article is still marked as being under construction. For the specific case of Z[sqrt(13)], what would you say is the optimal unit? Off hand I can find 119/2 + 33sqrt(13)/2, but it seems like there should be a better one. Alonso del Arte 20:19, 17 February 2013 (UTC)
Is this sequence right? It contains 5, 13, 17, 21, 29, 33, 37, 41, 53, 57, ...
A003172 Q(sqrt n) is a unique factorization domain (or simple quadratic field).
{2, 3, 5, 6, 7, 11, 13, 14, 17, 19, 21, 22, 23, 29, 31, 33, 37, 38, 41, 43, 46, 47, 53, 57, 59, 61, 62, 67, 69, 71, 73, 77, 83, 86, 89, 93, 94, 97, 101, 103, 107, 109, 113, ...}
Daniel Forgues 01:38, 18 February 2013 (UTC)
I guess it is right with the proviso that Q(sqrt n) stands for Q((1 + sqrt n)/2) when n = 1 (mod 4). — Daniel Forgues 02:35, 18 February 2013 (UTC)
That's precisely what I mean by "one will occasionally encounter the latter when the former is meant." And not just in the OEIS, but in other websites, and books, too. Alonso del Arte 04:22, 18 February 2013 (UTC)
According to Mathworld a fundamental unit for (the integers in) Q(sqrt(13)) is 1/2 (3 + sqrt(13)). Re A003172: A more accurate name would be "The ring of integers of Q(sqrt n) is a unique factorization domain", though I think it is fairly common to attribute things to the field Q(sqrt n) rather than to the subring of integers. A standard notation for the ring of integers in Q(sqrt n) is ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {n}})}}$, which I find rather clunky. Perhaps in the article use something like ${\displaystyle {\mathcal {R}}({\sqrt {n}})}$? Much less clunky and it can't be confused with anything else. Or perhaps there's another more standard alternative.
By the way, Q(sqrt n) is the same field as Q((1 + sqrt n)/2), so replacing one with the other would not change the meaning. Eric M. Schmidt 05:59, 18 February 2013 (UTC)
The article currently claims that, e.g., ${\displaystyle {\frac {1+{\sqrt {5}}}{2}}\in \mathbb {Z} [{\sqrt {5}}].}$ Sigh... it's in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {5}})}}$ but obviously not in ${\displaystyle \mathbb {Z} [{\sqrt {5}}].}$
Charles R Greathouse IV 08:00, 18 February 2013 (UTC)
Yes, the article does explain near the beginning that its notation isn't strictly correct. I think the article's current convention is inadvisable. It conflicts with standard notation. Also, since the right rings in the d = 1 mod 4 case are not what most people would guess, it is pedagogical mistake to deemphasize this by ignoring it in the notation.
I will now start editing the article to be correct in this regard. Eric M. Schmidt 23:03, 18 February 2013 (UTC)
By the way I just added sequence A219361 which is related to this article. Charles R Greathouse IV 20:42, 19 February 2013 (UTC)

## Splitting the page in two?

I suggest

Main article page: Imaginary quadratic integer rings

Main article page: Real quadratic integer rings

Daniel Forgues 21:49, 21 February 2013 (UTC)

I don't really think there's enough content to justify the split. The page is mostly long just because of the tables. What does everyone else think? Charles R Greathouse IV 22:35, 21 February 2013 (UTC)
I say yeeeeeaaaaaaahhhh... but not yet. I'm going to try to figure out all the proofs and then compare what I get to Niven-Zuckerman. Once we have valid proofs on all the theorems here, we can start figuring out how to do the spit. Alonso del Arte 03:05, 22 February 2013 (UTC)
If we split at some point this page should probably remain to cover general aspects, with those two pages covering only their specific features (and pointing back here for the rest). So I guess that would be splitting the page in three? But I still think there needs to be more content before splitting. Charles R Greathouse IV 04:56, 22 February 2013 (UTC)

Should I split the page in two?

Daniel Forgues 00:26, 7 May 2014 (UTC)

Please hold off on that until all the proofs are filled in. Niven-Zuckerman gives a proof of Theorem R1 which I haven't fully comprehended yet. Alonso del Arte 03:38, 7 May 2014 (UTC)
OK, I'll wait then. Please tell me if/when you agree to split the article later on, and whether you want me to do it... — Daniel Forgues 04:51, 9 May 2014 (UTC)
Once we have all the theorems and proofs filled in, we'll able to decide how to do the split. But basically, what applies to both stays on this page, what only applies to real rings or only imaginary rings gets branched off. Let's say the proof for Theorem R1 doesn't depend at all on the proof for Theorem I1: in that case we can put R1 on the page for the page for real rings, otherwise we keep it on this page. Alonso del Arte 01:12, 10 May 2014 (UTC)

## Notation

${\displaystyle {\mathcal {O}}_{\mathbb {Q} {\sqrt {D}}}}$ or ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {D}})}}$? I see the notation with the parentheses being used. Here is an example (e.g. on page 5) using ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {D}})}}$

http://arxiv.org/abs/1202.1756

Daniel Forgues 01:11, 25 February 2013 (UTC)

Always the latter, as I commented elsewhere. Parentheses are needed to denote a field. - Charles R Greathouse IV 02:22, 27 February 2013 (UTC)