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Is there any general idea about for which $d$, $\mathbb Z[\sqrt d]$ a principal ideal domain (PID)?

As for example $\mathbb Z[\sqrt{-1}]$ and $\mathbb Z[\sqrt 2] $ are PIDs, but $\mathbb Z[\sqrt{-5}] $ is not a PID.

user26857
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Sankha
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    [Real case](http://en.wikipedia.org/wiki/List_of_number_fields_with_class_number_one) and [Gauß conjecture](http://en.wikipedia.org/wiki/Class_number_problem#Gauss.27s_original_conjectures). It is generally not known. – awllower Mar 05 '13 at 09:52
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    Note that ${\bf Z}[\sqrt d]$ is not the full ring of integers in ${\bf Q}(\sqrt d)$ if $d\equiv1\pmod4$ (also if $d$ is not squarefree) and doesn't necessarily have the same ideal structure as the full ring of integers. – Gerry Myerson Mar 05 '13 at 12:28
  • I could not get what is meant by ''$\mathbb Z [\sqrt d]$ is not the full ring of integers in $\mathbb Q [\sqrt d]$ if $d\equiv 1 (mod 4)$ ''.Why it is not so? I just know only that $\mathbb Q [\sqrt d]$ is a quotient field of $\mathbb Z [\sqrt d]$ for all $d$.So if you please explain your point more clearly then it will be good for me.@ Gerry Myerson – Sankha Mar 06 '13 at 06:46
  • Take $d=5$. $(1+\sqrt5)/2$ is in ${\bf Q}(\sqrt5)$ and is an algebraic integer, but is not in ${\bf Z}[\sqrt5]$. Take $d=8$. $\sqrt2$ is in ${\bf Q}(\sqrt8)$ and is an algebraic integer, but is not in ${\bf Z}[\sqrt8]$. When you look at the links from @awllower, check to see whether they are about ${\bf Z}[\sqrt d]$, or whether they are about the full ring of algebraic integers in ${\bf Q}(\sqrt d)$, which is a bigger ring in the cases I mentioned. – Gerry Myerson Mar 06 '13 at 12:33

1 Answers1

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As stated your question is too general for us to have an answer - it's an open problem, but from what you said after that I think you meant to ask this more specific question instead: for negative integers $d$, when is the ring of integers of $\mathbb{Q}(\sqrt{d})$ a PID? To that we have an answer but the proof is highly non-trivial.

First note that this ring of integers is $\mathbb{Z}[\omega]$ where $\omega=\frac{1+\sqrt{d}}{2}$ if $d\equiv 1 \bmod 4$, and $\omega=\sqrt{d}$ if $d\equiv 2,3 \bmod 4$.

Then, due to Stark, 1966, we know that $\mathcal{O}_d$ is a PID $\iff d=$-1, -2 , -3 , -7, -11, -19, -43, -67, or -163.

Stark's proof is hard. But it is easier in the cases where you can show it's a PID by showing it's a Euclidian domain, which (for $d< 0$) is true $\iff d=-1,-2,-3,-7,$ or $-11$. Getting that part has to do with using the field norm and doing a lot of arithmetic. You should be able to find it in textbooks.

j0equ1nn
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    Stark's result covers the **negative** square-free values of $d$, no? Also, I'm fairly sure that last number in the list is $-163$ not $-143$. – Jyrki Lahtonen Jul 23 '13 at 05:37
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    @JyrkiLahtonen: I made those changes. – KCd Feb 25 '14 at 00:28
  • You are correct, but in some books (especially hyperbolic geometry, where imaginary quad extensions are much more relevant than real ones) they define $\mathcal{O}_d$ as $\mathbb{Q}(\sqrt{-d})$ despite the number theory conventions. Having been reading stuff like that, that's where my positive $d$s came from. Thanks for fixing it though. – j0equ1nn Mar 03 '14 at 18:59
  • But 4=2.2=(1+√-3)(1-√-3) in Z[√-3]. Doesn't it shows that Z[√-3] is not a UFD, hence not a PID either? – OppositeObserver Apr 19 '17 at 08:12
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    @OppositeObserver That's the non-maximal order inside the ring of integers since that's $1\pmod 4$, rather than being $\mathcal{O}_{-d}$. – Earth Cracks May 21 '19 at 00:19