Yesterday I asked a question and Franz Lemmermeyer answered my question in a comment. After that I came to another guess, but I can not prove or disprove it.
Let $L/\mathbb{Q}$ be a cyclic Galois extension of prime degree $r$. Suppose $p$ is a prime that is neither ramified nor inert. So we can conclude that it is split, and if $\mathfrak{p}$ is a prime above $p$, then $p\mathcal{O}_L=\prod\sigma(\mathfrak{p})=\mathfrak{p}_1 \cdots \mathfrak{p}_r$, where $\sigma$ runs over elements of Galois group, and all of $\sigma(\mathfrak{p})$'s are distinct. I am curious about some facts about the classes of $[\sigma(\mathfrak{p})]$'s, in the ideal class group. I do not know how they behave, but my observations evoke these two childish conjectures:
Assume that $\mathcal{Cl(O}_L) \cong \dfrac{\mathbb{Z}}{r\mathbb{Z}}$. Then $[\sigma(\mathfrak{p})]=[\mathfrak{p}]$, in the ideal class group.
I hope at least the first statement is correct.
Assume that $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{r\mathbb{Z}})^k$. Then $[\sigma(\mathfrak{p})]=[\mathfrak{p}]$, in the ideal class group.
In the case of this problem (Galois extension of prime degree), this relation is clearly satisfied for any ramified or inert prime, so we can eliminate the assumption of "splits completely" from these two statements.
What I can do:
- If $r=2$, and $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{2\mathbb{Z}})^k$, then for any prime $p$ that splits completely ($p\mathcal{O}_L=\mathfrak{p}_1 \mathfrak{p}_2$), it is very easy to show that $[\mathfrak{p}_1]=[\mathfrak{p}_2]$.
- [This is very very special and ridiculous]: Let $\alpha$ be a root of the polynomial $x^3-21x-35=0$, and let $L:=\mathbb{Q}(\alpha)$, then $r=3$, and $\mathcal{Cl(O}_L) \cong \dfrac{\mathbb{Z}}{3\mathbb{Z}}$, then for $p=5, 11$ (!!!), I can verify that $p$ splits completely ($p\mathcal{O}_L=\mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$), and $[\mathfrak{p}_1]=[\mathfrak{p}_2]=[\mathfrak{p}_3]$.
I strongly feel that even if these two statements are wrong, then the assumption of "$[\sigma(\mathfrak{p})]=[\mathfrak{p}]$" for "every prime $p$ (that splits completely)", implies that $\mathcal{Cl(O}_L) \cong (\dfrac{\mathbb{Z}}{r\mathbb{Z}})^k$.
Best Answer
If $M$ is the Hilbert class field of $L$, $G = \mathrm{Gal}(M/\mathbf{Q})$ and $H = \mathrm{Gal}(M/L)$, then the conditions listed imply (by class field theory) that
$$G = H \rtimes G/H = (\mathbf{Z}/r)^k \rtimes (\mathbf{Z}/r),$$
and moreover, the corresponding order $r$ automorphism $\varphi$ of $H = (\mathbf{Z}/r)^k$ may be identified with the action of a generator $\mathrm{Gal}(L/\mathbf{Q})$ on the class group of $L$. Here the fact that $G \rightarrow G/H$ splits follows from Minkowski: the extension $L/\mathbf{Q}$ must be ramified at some prime $q$, and then the inertia group $I_q \subset G$ has order $r$ and projects isomorphically onto $G/H$, giving the splitting.
Suppose the automorphism is non-trivial. Then there are certainly elements in $H$ which are not fixed by this automorphism. There is an isomorphism $\mathrm{Cl}(L) \rightarrow H$ given by the Artin map. By the Cebotarev density theorem, we deduce that there are infinitely many primes $p$ totally split in $K$ such that the Frobenius element associated to $\mathfrak{p} | p$ is not fixed by $\varphi$, and hence that $\mathfrak{p}$ inside the class group is not fixed by $\varphi$. Hence:
Note that without any assumption on the class group, the group $G$ is a semi-direct product (the splitting argument above applies). The condition that $\varphi$ is trivial then implies that
$$G = H \times G/H.$$
The ramification at any prime $q$ if non-trivial must be cyclic of order $r$. But now again by Minkowski this implies that $G = (\mathbf{Z}/r)^{k+1}$ and so:
The action of $\varphi$ is trivial if and only if $G = (\mathbf{Z}/r)^{k+1}$ assuming only that $L/\mathbf{Q}$ is cyclic of order $r$.
So now your first question is equivalent to asking if $H = (\mathbf{Z}/r)^k$ implies that $G = (\mathbf{Z}/r)^{k+1}$. There is no a priori reason to suppose this is true. However, as you noted, the result is true in your formulation by a quite easy argument for $r=2$. That means that $G$ must always be abelian in this case. Thus we should be a little careful. Note that one can write down all automorphisms of order $r$ on vector spaces over $\mathbf{F}_r$ - they have a Jordan decomposition where the block sizes are at most $r$. Hence we can even write $G$ explicitly as
$$G = \langle s,x_{i,j} | s^p, x^r_{ij}, [x_{i,j},x_{k,l}], [s, x_{i,j}] = x_{i,j-1} \rangle $$ where
Exercise: If $x \in H$, the element $sx$ has order $p$ if and only if $x$ lies in the subgroup generated by $x_{i,j}$ for $j \le \max\{m(i),r-1\}$.
This follows easily enough from the identity (true by induction, multiply by $sm$ on the right)
$$(sx)^m = \varphi(x) \varphi^2(x) \ldots \varphi^m(x) s^m x,$$
and hence
$$(sx)^r = x \varphi(x) \ldots \varphi^{r-1}(x) = (I + \varphi + \ldots + \varphi^{r-1})(x) = (\varphi - 1)^{r-1}(x),$$
where the polynomial identity $(1+X+\ldots X^{r-1}) = (X-1)^{r-1}$ follows by multiplying both sides by $(X-1)$ and then using Frobenius to see $(X-1)^r = X^r - 1$.
The ramification group $I_q$ at any prime $q$ is either trivial or provides a splitting $G/H \rightarrow G$. Each such group has an element which maps in $G/H$ to $s$. Hence the inertia groups $I_q$ are generated by elements of order $r$ of the form $ s x$. By the computation above, they are thus given by $s x $ where $x$ is a product of elements of the form $x_{i,j}$ for $j < r$. It follows that the images of the inertia groups $I_q$ for all $q$ lies inside the group generated by $s x_{i,j}$ for all $i$ and all $j < r$. Suppose that $m(i) = r$ for some $i$. Then the quotient obtained by setting $s = 0$ and $x_{i,j} = 0$ for all $i$ and $j$ except for $x_{i,m(i)}$ gives a quotient group $(\mathbf{Z}/r)$ where by this analysis the inertia groups are all trivial. But that means the corresponding extension is everywhere unramified which contradicts Minkowski. To summarize:
If $r = 2$ this means the Jordan blocks all have size $1$ and $\varphi$ is trivial and $G = (\mathbf{Z}/2)^{k+1}$ is abelian. However, if $r = 3$, there is no a priori obstruction, however, to
$$\mathrm{Gal}(M/\mathbf{Q}) = \langle x, y , s | x^3, y^3, [x,y], s^3, [s,x], [s,y]=x \rangle$$
occurring as a $G$. Here $G$ is the unique non-abelian group of order $27$ all of whose elements have order $3$ (the $3$-Sylow of $\mathrm{GL}_3(\mathbf{F}_3)$. (With the notation above, $x = x_{1,1}$ and $y = x_{1,2}$ and $m(1) = 2$.) The abelianization of $G$ is $(\mathbf{Z}/3)^2$, so the corresponding cyclic extension (by genus theory, omitted) should be ramified at exactly two primes which are either $3$ or $1 \bmod 3$.
Hence this suggests looking exactly for the following:
Now you can go to this website:
https://hobbes.la.asu.edu/NFDB/
And impose the following restrictions
This returns (for example) the extension $x^3 - 219 x - 1241$ with class group $(\mathbf{Z}/3)^2$ (as given by the website), and thus this extension will lead to a counter-example to your guess.
To be completely explicit, the ring of integers is $\mathbf{Z}[x]$, and
$$11 = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3 = (x-1,11)(x-4,11)(x-6,11).$$
Now one can check by one's favourite method that these elements are distinct in the class group.