Prime ideals in real quadratic fields being principal depends only on the residue class mod D of its norm? Let $K$ be a real quadratic number field of discriminant $D$. Then it is well known that each prime $p \in \mathbb Z$ splits completely in $\mathcal O_K$ if and only if $\chi_D(p)=1$. Splitting completely means that $(p)=PP'$ with $P \ne P'$ being prime ideals of $\mathcal O_K$.
Now I found out by doing a computer check that for completely splitting primes with $p \equiv q \pmod D$ we have that the corresponding ideal $P$ is principal if and only if $Q$ is. Is that actually true? How would you prove that?
 A: Let $H$ be the class field of $K$ (in particular, it is totally real) and $p$ be a prime number that splits in $K$, write $p=PP’$. We have a canonical isomorphism $Gal(H/K) \cong Cl(K)$; fot this isomorphism, $P$ is trivial in $Cl(K)$ iff $P$ is totally split in $H$ iff $Frob_P \in Gal(H/K)$ is trivial.
Now $Frob_P\in Gal(H/K)$ is the element $g$ of this group such that $g$ preserves all primes of $H$ above $P$, and $g$ acts on the residue field at $P$ by $x \longmapsto x^{N(P)}=x^p$.
But $Frob_P \in Gal(H/\mathbb{Q})$ as well, and because $N(P)=p$, it is a Frobenius at one (any) of the primes of $H$ above $P$, so its conjugacy class is that of $Frob_p \in Gal(H/\mathbb{Q})$.
In other words, for primes $p$ totally split in $K$, the “splitting prime” is principal in $K$ iff $Frob_p \in Gal(H/\mathbb{Q})$ is trivial.
Let $C=\mathbb{Q}(e^{2i\pi/D})$, then $C \supset K$ and $C/K$ is abelian. Consider a prime $p$ totally split in $K$, and $P$ one of its factors. Then $Frob_p \in Gal(HC/\mathbb{Q})$ lies in fact in $Gal(HC/K)$ (which is abelian), and is a Frobenius at $P$.
Now, we have an injection $Gal(HC/\mathbb{Q}) \rightarrow Gal(H/\mathbb{Q}) \times (\mathbb{Z}/(D))^{\times}$, with image $U$, mapping $Frob_p$ to $Frob_p$ on the first coordinate and $p$ mod $D$ on the second one.
By Cebotarev, the conjugacy classes of images of Frobenius cover $U$. Note that $p$ being split in $K$ can be seen as a constraint on $p$ mod $D$ – let $S$ be the corresponding subgroup.
Your question thus amounts to: given $g,h \in U$, such that $p_2(g)=p_2(h)$ corresponds to a split prime, must we have $p_1(g)=e$ iff $p_1(h)=e$? The answer is yes iff for each $s \in S$, $p_1(p_2^{-1}(s) \cap U)$ is the identity iff it contains the identity.
In other words, if $U’ \subset U$ is the (abelian) image of $Gal(HC/K)$, when your answer is yes, $p_1(\ker{p_2})$ is trivial, and, by translation, $p_1$ is trivial on the fibers of $p_2$. This condition is also sufficient to yield a positive answer to your question.
In other words: your statement is equivalent to the following claim: in $Gal(HC/K)$, the class of an element in $Gal(C/K)$ determines the class of this element in $Gal(H/K)$, ie $Gal(HC/K) \rightarrow Gal(C/K)$ injective, is $H \subset C$.
In other words, when your statement holds, $H/\mathbb{Q}$ is abelian. But we have an exact sequence of groups $0 \rightarrow Cl(K) \rightarrow Gal(H/\mathbb{Q}) \rightarrow Gal(K/\mathbb{Q}) \rightarrow 1$ where the conjugation action is by $-1$ – in particular, if $H/\mathbb{Q}$ is abelian, then $Cl(K)$ is a $\mathbb{F}_2$-vector space.
So this shouldn’t work for the number field 2.2.229.1 (see the LMFDB – it is $\mathbb{Q}(\sqrt{229})$), and whose class field should be 6.6.12008989.1 (see the LMFDB again – it’s not spelled out but it has the right degree, signature and ramification). Maybe you can check this case?
A particular case where the result holds is when $O_K$ is a PID, so that $H=K$. This happens for many “small” real quadratic fields, which could be the cases that you’ve studied?
