I think the following is an example of a cyclic extension of degree $4$ over the field
$K=\Bbb{F}_3((x))$ with $e=f=2$.
Let $i$ be a solution of $i^2+1=0$ in the extension field $\Bbb{F}_9$. Let
$$
u=\sqrt{(1+i)x},\quad v=\sqrt{(1-i)x}.
$$
It follows that $uv=\pm ix\notin K$. Similarly $u^2,v^2\notin K$, but $u^2+v^2=-x\in K$. This implies that
$$
p(T)=(T^2-u^2)(T^2-v^2)=T^4+xT^2-x^2\in K[T]
$$
is irreducible. None of the zeros $\pm u,\pm v$ are in $K$, ruling out linear factors, and no product of two zeros is in $K$ ruling out quadratic factors.
The coefficients $a=x$ and $b=-x^2$ satisfy the condition described here, part (b). Neither $b=-x^2$ nor $a^2-4b=-x^2$
is a square in $K$ but their product is. Therefore the Galois group of $p(T)$ over $K$
is cyclic of order four.
We easily see that $L=K(u)$ is the splitting field. The element $i=(u^2/x)-1\in L$
as is $v=ix/u$. So $L$ has a subfield isomorphic to $\Bbb{F}_9$. This forces the inertia degree to be divisible by two. On the other hand $u^2=(1+i)x$ forces the ramification degree $e$ to be at least two also. As $[L:K]=ef=4$, the conclusion is that we must have $e=f=2$.
It is a bit unnerving that the leading coefficient of $u$ appears to be $\sqrt{1+i}\notin\Bbb{F}_9$. The explanation is surely that $\sqrt x\notin L$! Mentioning this because something similar may allow us to generalize this construction to other $(p,\ell)$ pairs? A key ingredient seems to be that $L=\Bbb{F}_9((u))$, but $u$ is not really a fractional power of the original local parameter $x$. We have this extra twist coming from the coefficient $1+i$ under the square root. Similar twisting may give us extensions with $e=f=\ell$ whenever $\ell\mid p-1$, but that needs a bit more work.
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:
Your conclusion is true if and only if (under the assumptions) the Hilbert class field $M$ is abelian, and in particular $G = (\mathbf{Z}/r)^{k+1}$.
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
- $i$ ranges over $i = 1,\ldots, n$ for some $n$,
- $j = 1, \ldots, m(i)$, where $m(i) \le r$,
- by convention $x_{i,0} = e$.
- $\sum_{i=1}^{n} m(i) = k$.
- $\varphi(x_{i,j}) = s x_{i,j} s^{-1} = x_{i,j} x_{i,j-1}$.
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:
The only $\varphi$ which can arise must have Jordan blocks of size at most $r-1$, not simply $r$ as follows from group theory.
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:
look for cyclic degree $3$ fields ramified at two primes $p$ and $q$ whose class group is $9$.
Now you can go to this website:
https://hobbes.la.asu.edu/NFDB/
And impose the following restrictions
- degree $3$ extensions
- discriminant in the range 1..1000000 (can be increased)
- Galois group is a p-group
- $h = 9$
- $2$ ramified primes.
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.
Best Answer
I'll be using stuffs mentioned in Marcus' Number Fields, Chapter 4. So it might be nice to read it before reading my proof.
For (1), let $P \subset \mathcal O_L$ be a prime lying over $\mathfrak p$. Suppose $K \subset L$ has intermediate fields. Now consider the inertia group $E = E(P|\mathfrak p)$ and the inertia field $L_E$. It is the largest subfield in which a ramification doesn't occur. As ramification occurs in each intermediate subfield we have that $L_E=L$ or $K$. However, the ramification is multiplicative, so there $e(P|\mathfrak p) \not = 1$. Thus $L_E=K$. However we also have:
$$e(P|\mathfrak p) = [L:L_E] = [L:K]$$
This implies that $\mathfrak p$ ramifies totally in $L$. Thus $K \subset L$ has not intermediate fields.
Similarly for (2), suppose $K \subset L$ has intermediate fields. Now consider the decomposition field $L_D$. It is the largest field in which there is only splitting of ideals. On the other side by the condition in each intermediate field there has to be a ramification or the inertial degree increases. Thus again $L_D = L$ or $K$. Once again, since both of these are multiplicative we must have that $L_D = K$. However, now $\mathfrak p$ splits into $[L_D:K] = 1$ ideals in $L$, a contradiction. Thus $K \subset L$ has not intermediate fields.
Finally, since the extension is Galois and it has no intermediate fields, we must have that $G$ has no nontrivial proper subgroups. The only such groups are the groups of prime order. (You can use Cauchy's Theorem to prove this claim).
For (1) of the second claim, consider the inertia field $L_E$. It is the largest subfield in which no ramification occurs. By the hypothesis it is an intermediate subfield, which contains every other proper subfield of $L$. Under Galois correspondence $E$ is a minimal subgroup of $G$ and is of order $[L:L_E]$. Note that it is the only subgroup of this order, as it is contained in all non-trivial subgroups of $G$. So it must be the unique smallest (non-trivial) subgroup of $G$.
For (2), you can use the same idea, as in the first claim. Consider the decomposition field $L_D$. As in (1) it must contain every other proper subfield of $L$. Hence $D$ is contained in every proepr subgroup of $G$, so it must be the unique smallest (non-trivial) subgroup of $G$.
However, we can derive more information for $H$, the smallest (non-trivial) subgroup of $G$. For example, $G$ must be a prime power group. Indeed, if $p$ and $q$ are two distinct prime divisors of $|G|$, by Cauchy's Theorem we have a subgroups of order $p$ and $q$. However, they don't have a common non-trivial subgroup, which contradicts the fact that the $H$ is contained in every proper subgroup of $G$. Thus $|G|=p^k$ and obviously $|H| = p$. Also $H \le Z(G)$, trivially.
For the third claim, consider the decomposition field $L_D$. It is the largest subfield in which $\mathfrak p$ splits completely. By the hypothesis and the fact that the inertia degree is multiplicative we must have that $L_D = K$. Also consider the inertia fields $L_E$. As in the second claim it must contain every proper subfield of $L$. Moreover, it can't be $L$, itself, as $\mathfrak p$ isn't inert in $L$, so it must ramify, since no splitting occurs, as $L_D=K$. So in the notation from the second claim $D=G$ and $E=H$.
Hence, $G/H = D/E$, which is a cyclic group. Now as $H \le Z(G)$, by the third isomorphism theorem, $G/Z(G)$ is isomorphic to a quotient of $D/E$ and in particular it is cyclic. Then, as in here we have that $G$ is abelian. Moreover, as in the second claim we have that $|G| = p^k$ and also it has a unique subgroup of order $p$. The last claim implies that $G$ is cyclic. Indeed, if $G = \mathbb{Z}_{p^{i_1}} \times \dots \times \mathbb{Z}_{p^{i_n}}$ it has $n$ subgroups of order $p$.
Hence the proof.