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I'm asked to computing the higher ramification group for quadratic extensions $K=\mathbb{Q}(\sqrt{d})|\mathbb{Q}$. They are defined as follows, for a prime ideal $\mathfrak{p}$, $$G_{\mathfrak{p}}^{(i)}:=\{\sigma \in Gal(\mathbb{Q}(\sqrt{d})|\mathbb{Q})\mid \forall \alpha \in \mathcal{O}_K, \sigma(\alpha)=\alpha \pmod{\mathfrak{p}^{i+1}}\}$$

clearly for $i=0$ we have that the ramification group is the inertia group. By the fact that the extension is clearly Galois, the definition of inertia group depends only on the prime number contained in the prime ideal.

Then we have clearly that the cardinality of the inertia group at a prime is the ramification index of such prime, and therefore $$ p \nmid \delta_K \Leftrightarrow G_{\mathfrak{p}}^{(0)}\cong \{1\}$$ and therefore all the higher ramification groups are trivial.

Is there a clever way to deal with the case $p \mid \delta_K$? More importantly, what properties do the higher ramification groups have in order to compute them in this easy case?

ADDENDUM Brute computations shows that for $p$ odd and $d\neq 1 \pmod{4}$ $G_p^{(1)}$ is trivial. I think one can do other cases by brute computations as well. Is there a smarter way?

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  • $\begingroup$ “Brute computation”, as you’ve done, is the best way to get a feel for what is really happening, but as often happens, a few well-chosen theorems can shorten the process immensely. In the cases you’ve calculated, since the degree is $2$, all your ramification will be tame, except at the prime above $2$, and tame ramification has only one break, at $0$. The whole story is more interesting, but I don’t have time to go into it now. Maybe in a few hours’ time. $\endgroup$ – Lubin Feb 16 '16 at 23:29
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Here’s a method you can use for the higher ramification in a ramified quadratic extension. If you look at the definition of the ramification groups, you see that everything comes down to the distance between a prime element $\pi$ and its conjugate $\bar\pi$.

The computation is purely local, which means you can do it over $\Bbb Q_p$ if you like, and it boils down to this: take a prime element $\pi$, then the unique break number is $v_\pi(\bar\pi-\pi)-1$. I won’t justify that decrease by $1$, but you’ll see it when you trace through the standard definition.

Let’s look at examples: first, $\Bbb Q(\sqrt d\,)$ at odd $p|d$: Here, locally at $p$, $\sqrt d$ is a prime, and you take $v_{\sqrt d}(2\sqrt d\,)=1$ and subtract $1$ to get the result I stated in my comment, that the break is at $0$, since you have tame ramification. So $G_0$ is the whole group, $G_1$ is trivial.

All remaining cases are with $p=2$, first for $d\equiv3\pmod4$. Then $\sqrt d-1=\pi$ is a good prime, locally at $2$ remember, and its minimal polynomial is $X^2+2X+1-d$, Eisenstein because $d\equiv3$ modulo $4$. Now, $\bar\pi-\pi=-2-2\pi$, which has $v_\pi$-value $2$, so the break-number is $1$.

Finally, for $d\equiv2$ modulo $4$, $\sqrt d=\pi$ is your local prime, and $v_\pi(\bar\pi-\pi)=3$, so that the break-number is $2$.

Maybe I should add a slightly philosophical note: what ramification theory tells you is how far the various conjugates of a prime element are from each other (whether or not your extension is normal). In a tame extension, all the conjugates are equally far from each other, like the $n$ vertices of an $(n-1)$-simplex. In a wild extension, they may be at various distances from each other. Consider the primitive $16$-th roots of unity $\zeta$ and the corresponding prime elements $\pi=\zeta-1$. There are eight of them in all and if you fix one of them, you will find one other that’s moderately close to it; two others that are somewhat farther away, and the remaining four are at a greater distance yet. The break numbers tell you exactly what these distances are.

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