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Assume $K$ is a number field, $L$ is a Galois field extension of $K$ with Galois group $G$, $\mathfrak{p}$ is a fixed prime ideal of $\mathcal{O}_K$, where $\mathcal{O}_K$ is the ring of integers of $K$. I am trying to prove the following properties about ramification:

  1. In the following two cases, there are no intermediate fields and then $G$ is a cyclic group with prime order:

    (1) $\mathfrak{p}$ is totally ramified in each intermediate field but not in $L$;

    (2) Each intermediate field contains a unique prime lying over $\mathfrak{p}$ but not for $L$.

  2. In the following two cases, $G$ has a unique smallest (non-trivial) normal subgroup $H$, hence the orders of $G$ and $H$ are prime power and prime respectively, $H$ is in the center of $G$:

    (1) $\mathfrak{p}$ is unramified in each intermediate field but ramified in $L$;

    (2) $\mathfrak{p}$ completely splits in each intermediate field but not in $L$.

  3. $G$ is a cyclic group with prime power order if $\mathfrak{p}$ is inert in each intermediate field but not in $L$.

Notice every "intermediate field" we mentioned does not contain $K$ or $L$ themselves.

Any discussion about anyone of these properties will be helpful for me. Thanks!

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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.

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    $\begingroup$ I am really grateful for your answer! It indeed helps me a lot. Sorry for replying so late, since I was busy with other things a few days ago, also I read this answer carefully to make sure I understand all of it. So I have got everything except one: why cannot $L_D$ be $L$ in the third claim? I think we can't say because $L_D\subseteq L_E\subsetneq L$, since the proof of "$L_E\neq L$" uses the fact "$L_D=K$". Finally, thank you for helping me again. :) $\endgroup$ – Zigzag1263 Jun 15 at 14:42
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    $\begingroup$ @Zigzag1263 If L has no intermediate subfield, then it is possible to have $L=L_D$, however then the problem is trivial, as $G$ must be of prime order. On the other side if $L$ has an intermediate field, using the multiplicity of the inertia degree, we have that the inertia degree of $\mathfrak p$ in $L$ is greater than 1. Thus $L \not = L_D$, since in the latter only splitting of ideals occur. Hence $L_D = K$ $\endgroup$ – Stefan4024 Jun 16 at 9:58
  • $\begingroup$ Oh yes, no intermediate field implies $G$ is cyclic of prime order, I forgot it. Thanks. $\endgroup$ – Zigzag1263 Jun 16 at 10:24

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