If $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$ has class number $2$ or higher, does that mean $\sqrt{d}$ is irreducible but not prime?

Question

Actually, I do know of exceptions to this, that's when $d$ is prime in $\Bbb{Z}$, e.g., $\Bbb{Z}[\sqrt{-79}]$ has class number $5$ and $\mathcal{O}_{\Bbb{Q}(\sqrt{79})}$ has class number $3$, but obviously $\sqrt{-79}$ and $\sqrt{79}$ have prime norms and are thus primes in their respective domains.

But if $d$ is a negative composite number, say $d = -pq$ (where $p$ and $q$ are positive primes) then obviously $p$ and $q$ are both irreducible but neither is prime. And what if $d = pq$ or some other positive composite number?

I've worked through a few specific examples. Like in $\Bbb{Z}[\sqrt{-10}]$, norms $2$ and $5$ are impossible, so $10$ has two distinct factorizations. And in $\Bbb{Z}[\sqrt{10}]$, it is easy to see that $2$ is irreducible, though it takes a little more work to show that $5$ is irreducible as well.

Obviously this is too laborious, especially if the smallest counterexample is some number with a thousand prime factors.

Also, I'm not sure why last night I was looking up "real quadratic fields of generalized ERD-type" (a paper by Xianke and Fuhua only left me more confused). Does anyone even know what ERD-type is?

But my main question is this: can $d$ be composite in $\Bbb{Z}$, $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$ has class number greater than $1$, yet $\sqrt{d}$ is composite in $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$?

Answer 1

Also, I'm not sure why last night I was looking up "real quadratic fields of generalized ERD-type" (a paper by Xianke and Fuhua only left me more confused). Does anyone even know what ERD-type is?

Based on Sloane's A051990, I would guess it's $\mathcal{O}_{\mathbb Q(\sqrt d)}$ such that $d = r^2 k^2 + k$ with $k$ odd. Hopefully someone else can give a definitive answer to that side question.

But my main question is this: can $d$ be composite in $\Bbb{Z}$, $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$ has class number greater than $1$, yet $\sqrt{d}$ is composite in $\mathcal{O}_{\Bbb{Q}(\sqrt{d})}$?

Absolutely. If the class number is 1 and $d = pq$, then we know that $\langle p \rangle$ can be factorized as principal ideals, and the same goes for $\langle q \rangle$. Just because the class number is 2 or greater doesn't mean that still can't happen, though the evidence of small numbers suggests that more frequently than not it doesn't.

So for example, as you noted in a comment, $(4 - \sqrt{14})(7 + 2\sqrt{14}) = \sqrt{14}$. Looking at ideals, we see that $\langle 2 \rangle = \langle 4 + \sqrt{14} \rangle^2$ and $\langle 7 \rangle = \langle 7 + 2\sqrt{14} \rangle^2$, so then $\langle \sqrt{14} \rangle = \langle 4 + \sqrt{14} \rangle \langle 7 + 2\sqrt{14} \rangle$. You know about ideals, right?

Okay, so with 10, we see that $\langle 2 \rangle = \langle 2, \sqrt{10} \rangle^2$ and $\langle 5 \rangle = \langle 5, \sqrt{10} \rangle^2$. Not principal ideals, so we can't pick and choose factors of the generating numbers to factorize $\sqrt{10}$ as a number.

Compare 34. Check that $(1 - \sqrt{34})(1 + \sqrt{34}) = -33$, yet 3 and 11 are both irreducible, so $\mathbb Z[\sqrt{34}]$ must have a class number of at least 2.

But $\langle 2 \rangle = \langle 6 + \sqrt{34} \rangle^2$ and $\langle 17 \rangle = \langle 17 + 3\sqrt{34} \rangle^2$, so $\langle \sqrt{34} \rangle = \langle 6 + \sqrt{34} \rangle \langle 17 + 3\sqrt{34} \rangle$, and then to factorize $\sqrt{34}$ it's a simple matter of shopping the signs.

I don't know of any clever way to go from, say, $\langle 2, \sqrt{34} \rangle$ to $\langle 6 + \sqrt{34} \rangle$, and the textbooks are no help, since their authors generally prefer not to see the trees from the forest. All I can say is try to solve equations like $x^2 - pqy^2 = \pm p$.

Cross-checking A051990 with A029702 (when you get to 79 you'll need to also look at A029703), we find that in $\mathbb Z[\sqrt{51}]$ the ideal $\langle 3 \rangle$ does not split into principal ideals, so your insomniac theory about ERD-type rings might just be a red herring.

Answer 2

Take $\mathbb{Q}(\sqrt{55})$ then

$$\sqrt{55}=(15-2\sqrt{55})(22+3\sqrt{55})$$

And the class number is $>1$.

Answer 3

You seem to not know about ideals. I could use that to berate you for your ignorance, but that would for the most part just expose my own ignorance, pedantry and impoliteness.

Besides, I am also guilty of not looking far enough to find counterexamples, when it turns out that I didn't have to go too far.

Okay, so, as you have already determined, if $p$ is a prime in $\mathbb{Z}$, then $\sqrt{p}$ is irreducible and prime in $\mathcal{O}_{\mathbb{Q}(\sqrt{p})}$. Also, the ideal $\langle p \rangle$, or $\mathfrak{P}$ if you prefer, is principal and ideal.

An ideal $\langle n \rangle$ is principal if $n$ represents a single number in the particular domain we're looking at.

The ideal $\langle m, n \rangle$, which consists of all numbers of the form $mx + ny$ (where $m, x, y$ are also numbers in the relevant domain) is not a principal ideal.

However, factorization of an ideal into ideals is unique, even if factorization of a number is not. Also keep in mind that ideals may be written differently but are the same, at least in commutative algebra. So $\langle m, n \rangle = \langle n, m \rangle$.

Also, a principal ideal can be written so that it doesn't look like a principal ideal. For example, $\langle 3, 9 \rangle$ is in fact just $\langle 3 \rangle$.

Much more interestingly, though, this enables us to write ideals in a very general way. For example, if $p$ is an odd prime in $\mathbb{Z}$, then in $\mathbb{Z}[\sqrt{2p}]$ we can say that $\langle 2 \rangle = \langle 2, \sqrt{2p} \rangle^2$, whether that boils down to a principal ideal or not.

Maybe you looked at $\mathbb{Z}[\sqrt{6}]$ and observed that $\langle 2 \rangle = \langle 2, \sqrt{6} \rangle^2 = \langle 2 + \sqrt{6} \rangle^2$. The case of $\langle 3 \rangle$ is similar and so the factorization of $\langle \sqrt{6} \rangle$ looks a lot like the factorization of $\sqrt{6}$.

And as you've certainly noticed, in $\mathbb{Z}[\sqrt{10}]$ we see that $\langle 2 \rangle = \langle 2, \sqrt{10} \rangle^2$ but we can't take that step to a principal ideal like we did in $\mathbb{Z}[\sqrt{6}]$ because no such ideal exists.

But that has to do more with the properties of $d$ itself than the class number of $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$.

For example, if $d$ is a multiple of $10$, then we already know that $2$ is irreducible but not prime, even though $\langle 2 \rangle$ is the square of a prime, non-principal ideal. And so $\sqrt{10}$ as a number has two distinct factorizations, but the ideal $\langle \sqrt{10} \rangle$ has a unique factorization as a product of squared prime ideals.

Mr. Soupe has already gone over how $34$ is in this regard more like $6$ than like $10$. Not every ideal in $\mathbb{Z}[\sqrt{34}]$ is principal, but the ideals in the factorization of $\langle \sqrt{34} \rangle$ all are.

And that's what your question really boils down to: can $\langle \sqrt{d} \rangle$ be factorized into principal ideals? If $\mathcal{O}_{\mathbb{Q}(\sqrt{p})}$ is a principal ideal domain (that is, class number $1$) then the answer is of course yes. But if not, the answer is... maybe.

Answer 4

I think I have found a necessary condition, I'm fairly confident that it is necessary. That it is sufficient, I am quite less confident, because it could be the case that I have stopped checking one or two short of the counterexample.

Given $d$ being the product of two or more distinct positive primes $p_i$, if for any $p_i$ we find that neither $p_i$ nor $-p_i$ is a quadratic residue modulo $p_j$ for any $j \neq i$, then $d$ as number must have two distinct factorizations in $\mathcal{O}_{\textbf{Q}(\sqrt{d})}$, and the ideal $\langle d \rangle$ has a unique factorization into ideals but only some, if not none, of those ideals are principal.

But if either $p_i$ or $-p_i$ is a quadratic residue modulo $p_j$ for each $j \neq i$, then $d$ probably only has one distinct factorizations in $\mathcal{O}_{\textbf{Q}(\sqrt{d})}$, and perhaps, but not always, that is a unique factorization domain.

For example, in $\textbf{Z}[\sqrt{6}]$, we find that $2$ is not a quadratic residue modulo $3$, but $4 \equiv -2$ is, and so $$(-1)(2 - \sqrt{6})(2 + \sqrt{6})(3 - \sqrt{6})(3 + \sqrt{6}) = 6,$$ but in $\textbf{Z}[\sqrt{10}]$, neither $2$ nor $3$ are quadratic residues modulo $5$. Psychologically, this is not surprising because we know the former is a norm-Euclidean while the latter is not even a unique factorization domain.

More interesting is $\textbf{Z}[\sqrt{34}]$, where we find that $6^2 \equiv 2 \pmod {17}$, and therefore $$(-1)(6 - \sqrt{34})(6 + \sqrt{34})(17 - 3 \sqrt{34})(17 + 3 \sqrt{34}) = 34,$$ and so it becomes obvious that both $\langle 6 + \sqrt{34} \rangle$ and $\langle 17 + 3 \sqrt{34} \rangle$ are principal ideals.

Likewise with $\textbf{Z}[\sqrt{55}]$ we find that $4^2 \equiv 5 \pmod {11}$. I have checked up to $\textbf{Z}[\sqrt{195}]$ but have doublechecked only some of those calculations.