I'm trying to make sense of all these theorems related to ramification. I was hoping someone would summarize these results. Assume we have:

  1. An extension $L/K$ and a some subextensions $E_1,\ldots,E_k$ sitting in between (without necessarily having $E_i\leq E_j$ or vice versa for pairs).

  2. What about if some of these extensions are Galois?

I was wondering what ramification of a prime in $L/K$ tells us about ramification for the $E_i/K$ or $L/E_i$ and vice versa? How much can we pin down where the ramification has to be in these subextensions or do we always have to just compute discriminants?

I'm also trying to figure out how to compute Hilbert class fields. This is why I'm interested in how given some field $K/\mathbb{Q}$ I should go about trying to find a field $L/K$ that kills all ramification? I should be able to somehow figure out what's allowed to ramify in $L/\mathbb{Q}$?

EDIT: This is one argument that I just don't get. The Hilbert Class Field of $\mathbb{Q}(\sqrt{-5})$ is $\mathbb{Q}(\sqrt{-5},\sqrt{-1})$. Here's the standard argument:

$\mathbb{Q}(\sqrt{5})/\mathbb{Q}$ is unramified outside of 5 and $\mathbb{Q}(\sqrt{-1})/\mathbb{Q}$ is unramified outside 2. Hence, $\mathbb{Q}(\sqrt{-5},\sqrt{-1})/\mathbb{Q}(\sqrt{-5})$ is unramified everywhere.

Why can't there be some prime of $\mathbb{Q}(\sqrt{-5})$ that ramifies and ramifies above the other two extensions?

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    $\begingroup$ I can actually answer my edit myself. We should know that the discriminant of $\mathbb{Q}(\sqrt{-5},\sqrt{-1})$ is $2^a5^b$, so only 2 and 5 ramify. Their ramification degrees in $\mathbb{Q}(\sqrt{-5},\sqrt{-1})/\mathbb{Q}$ is at most 2 and both ramify in $\mathbb{Q}(\sqrt{-5})/\mathbb{Q}$. This means that nothing can ramify in $\mathbb{Q}(\sqrt{-5},\sqrt{-1})/\mathbb{Q}(\sqrt{-5})$. $\endgroup$ – dstt Apr 18 '11 at 3:49
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    $\begingroup$ Hi dstt, thank you for your question! could you tell me how you got that the absolute discriminant of $\Bbb Q(\sqrt{-5},\sqrt{-1})$ is $2^a5^b$? Could you also say something more generally about the discriminant of $\Bbb Q(\sqrt{-p_1,...p_n},\sqrt{p_1},\sqrt(p_2)...\sqrt{p_i}), i<n$? $\endgroup$ – Rodrigo May 3 '14 at 21:06
  • $\begingroup$ @Rodrigo : see math.stackexchange.com/questions/81994 $\endgroup$ – Watson Jan 8 '17 at 17:48

$ \newcommand{\fp}{{\mathfrak p}} \newcommand{\fP}{{\mathfrak P}} $ The ramification in intermediate fields is a slightly delicate (purely group theoretic) issue. First, assume without loss of generality that $L/K$ is Galois with Galois group $G$, otherwise pass to the Galois closure and repeat the following argument for $L$ being the intermediate field.

Suppose that you have a prime $\fp$ of $K$ with prime $\fP$ of $L$ lying above $\fp$. Let $D=D_{\fP/\fp}$ be the decomposition group, i.e. the subgroup of $G$ consisting of elements that fix the place $\fP$. Inside that, you have the inertia subgroup $I=I_{\fP/\fp}$. Recall that the index of $D$ in $G$ is the number of primes in $L$ lying above $\fp$, and $|I|$ is the ramification index of $\fP/\fp$. The residue field degree is the index of $I$ in $D$. Now, let $E$ be an intermediate field, corresponding to a subgroup $H$ of $G$, so that $H=\text{Gal}(L/E)$. Then

Exercise 1: There is a natural bijection between double cosets $H\backslash G/D$ and the primes of $E$ lying above $\fp$. (You might want to begin by checking that this is independent of the choice of $\fP$. Another prime above $\fp$ would have given a conjugate decomposition group.)

Exercise 2: If, under the above correspondence, the double coset $HgD$ corresponds to the prime $\fp'$ of $L$, then $$ D_{\fP/\fp'} = H\cap D^g\leq H\text{ and } I_{\fP/\fp'} = H\cap I^g\leq H. $$ This allows you to compute the ramification index and the residue field degree of $\fP/\fp'$ as above.

I hope that that answers your questions. As for explicit computation of Hilbert class fields, you might want to consult Sections 3 and 4 of Cohen's Advanced Topics in Computational Number Theory - a wonderful book that treats your question in great detail.


The ramification index is multiplicative in towers. That is:

If $K\subseteq E\subseteq L$ is a tower of extensions, $P$ is a prime in $K$, $Q$ is a prime in $E$ with $Q\cap K=P$, and $Q'$ is a prime in $L$ with $Q'\cap E=Q$, then letting $e(Q|P)$ be the ramification index of $Q$ over $P$, $e(Q'|Q)$ the ramification index of $Q'$ over $Q$, and $e(Q'|P)$ the ramification index of $Q'$ over $P$, we have have $$e(Q'|P) = e(Q'|Q)e(Q|P).$$ To see this, you do the factoring: if we factor $P$ into primes in $E$, then $Q$ occurs with exponent $e(Q|P)$; then we factor in $L$, the exponent of $Q'$ will be its exponent in the factorization of $Q$, $e(Q'|Q)$, times the exponent to which $Q$ occurs in that of $P$, $e(Q|P)$, since $(\mathfrak{P}^a)^b = \mathfrak{P}^{ab}$.

If all extensions are Galois (that is, $L$ and $E$ are both Galois over $K$), then the ramification index depends only on $P$ and not on $Q$ or $Q'$, but it's still multiplicative, but here you get the further information that the ramification index also divides the order of the extension.

All of that information it may give you enough leverage to determine the ramification index in $E_i/K$ in some circumstances (if you know the ramification in $L/K$, then the multiplicativity means that knowing the index in $E_i/K$ is equivalent to knowing it in $L/E_i$). But it may not. For example, if you have a biquadratic extension $L/K$ whose Galois group is the Klein $4$-group, with a prime $P$ that has ramification index $2$, then for the two intermediate fields $E_1$ and $E_2$ you could have that $P$ already ramifies in $E_1$, but the ramification index in $L/E_1$ is $1$; or the other way around.

Added. To answer your final question without reference to discriminants:

No rational prime can have ramification degree $4$ in $K=\mathbb{Q}(\sqrt{-5},\sqrt{-1})$: that would require it to ramify in both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$ by the multiplicativity of the ramification degree, but no prime ramifies in both. So the ramification degree of any ramified rational prime is $2$.

Let $L_1 = \mathbb{Q}(\sqrt{5})$, $L_2=\mathbb{Q}(\sqrt{-5})$, $L_3=\mathbb{Q}(\sqrt{-1})$. These are all the intermediate fields of the extension.

Let $p$ be a prime that ramifies in $K$. Let $Q$ be a prime of $K$ lying over $p$, and let $P_i$ be the prime of $L_i$ lying under $Q$. If $I_p$ is the inertia group of $p$, and $L_i$ is the fixed field of $I_p$, then $e(Q|P_i) = e(Q|p) = 2$, and $e(Q|P_j)=1$ with $j\neq i$. Again, multiplicativity of the ramification tells you that $p$ would then need to ramify in both the "other two" subextensions.

In particular, if $L_i = L_2 = \mathbb{Q}(\sqrt{-5})$ (so that there is a prime of $\mathbb{Q}(\sqrt{-5})$ that ramifies in $K$), then $p$ would have to ramify in both $\mathbb{Q}(\sqrt{5})$ and in $\mathbb{Q}(\sqrt{-1})$. But no rational prime ramifies in both; so we can never have the fixed field of $I_p$ equal to $\mathbb{Q}(\sqrt{-5})$, which means $K/\mathbb{Q}(\sqrt{-5})$ must be unramified everywhere, as claimed.


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