Is there a number field of degree n whose ring of integers is a unique factorization domain?

Question

For every $n$, can we find a number field of degree $n$ whose ring of integers is a unique factorization domain?

As a Dedekind domain is a UFD iff it is a PID, this is equivalent to asking the following: For every $n$, can we find a number field of degree $n$ with class number 1.

Answer 1

It seems that the answer is not known, see here. It has not even been proven that there are infinitely many number fields with class number $1$. It is tempting to look for a family of number fields, like for cyclotomic fields $\Bbb Q(\zeta_n)$ of degree $\phi(n)$. However, there the class number is equal to $1$ only for some "small" $n$, i.e., we have $n\le 90$. And, of course, $\phi(n)$ does not cover all positive integers.

Answer 2

Several Comments:

Comment 1:

Suppose you consider all fields with the following parameters fixed:

1. The signature $(r_1,r_2)$ of $F$ (and thus the degree $r_1 + 2 r_2$).

2. The Galois group $G$ of the Galois closure of $F$.

Then there are very well studied conjectures (due originally to Cohen-Lenstra, Cohen-Lenstra-Martinet, and now many others) predicting the distribution of the $p$-part of the class group for any "good" prime $p$. (If $p$ divides neither $|G|$ nor the order of the group of roots of unity on $F$ then $p$ is certainly good). Without going into this very nice story here, the upshot is that, providing there is at least one real place, there is a (conjecturally!) positive probability that the class group is not divisible by any good prime. For example, if one considers all real quadratic fields, the probability that the class group is a power of $2$ (see https://openaccess.leidenuniv.nl/bitstream/handle/1887/2137/346_069.pdf?sequence=1) has a formula (explicitly in terms of some infinite products) which turns out to be about 75.446%. If $C$ is the class group of a real quadratic field, then $C/2C$ is controlled by genus theory. In particular, if the discriminant is prime, then the class group is always odd. But one can guess that these fields behave like general fields (there is no reason to think otherwise and this is all conjectural) and so the same proportion of real quadratic fields with prime conductor has class number one. This accords well with experiment. Similarly, one would certainly predict the existence of infinitely many fields of signature $(r_1,r_2)$ with $r_1 > 0$ of class number one. In particular, it is certainly expected (but completely open) that the answer to your problem is "yes"

Comment 2:

The other answer mentions the case of cyclotomic fields $\mathbf{Q}(\zeta_n)$. These fall into the more general class of CM fields (totally imaginary quadratic extensions of totally real fields) which includes imaginary quadratic fields. In this case, one (Stark) conjectures that there are only finitely many such fields of class number one. This is known for Galois CM fields, and in particular known for cyclotomic fields. One can even write down all abelian CM fields with class number one. It turns out the hardest case is the case of imaginary quadratic fields!

Comment 3: Some random comments. Weber conjectured that the class numbers of the totally real fields $\mathbf{Q}(\zeta_{2^n} + \zeta^{-1}_{2^n})$ have class number one where $\zeta_{2^n}$ is a $2^n$th root of unity for all $n$. Unknown, but it is true that the class number of these fields (for all $n$) is prime to $p$ for any prime $p < 10^9$ (see https://arxiv.org/pdf/1410.2921.pdf)

I believe the largest degree field which has been proven (unconditionally) to have class number one has degree $120$ (https://arxiv.org/pdf/1606.09320.pdf). This paper (https://arxiv.org/pdf/1407.2373.pdf) also has some nice examples, including $\mathbf{Q}(\zeta_{151} + \zeta^{-1}_{151})$ of degree $75$, and $\mathbf{Q}(\zeta_{241} + \zeta^{-1}_{241})$ of degree $120$ conditional on GRH.