Let $L/K$ be a finite Galois extension where $K= \operatorname{Frac}(O)$ is complete and $O$ is a Dedekind domain. Let $\Delta=Gal(L/K)$. Since $K$ is complete, there is a unique prime $Q$ associated to $O_L$ where $O_L$ is the integral closure of $O_K$ in $L$. Then let $\Delta_i=\{\delta\in\Delta\vert\forall a\in O_L,\delta(a)-a\in Q^i\}$.

$\Delta_1$ is unique $p$-Sylow subgroup of $\Delta_0$. It is clear that $\Delta_0$ is the inertia group associated to prime $Q$ of $O_L$. If $\Delta_1$ is unique $p$-Sylow of $\Delta_0$, then $\Delta_1$ is normal by all $p$-Sylow being conjugates.

$\Delta_0$ is normal due to exact sequence $1\to\Delta_0\to\Delta\to \operatorname{Gal}(k_L/k)\to 1$ where $k$ is residue field of $K$ and $k_L$ is the residue field of $L$.

$\textbf{Q:}$ The book says it is follows easily $\Delta_1\leq\Delta$ is normal. Why $\Delta_1$ is normal here? Normality is non-transitive. I knew $\Delta_1\leq\Delta_0$ normal and $\Delta_0\leq\Delta$ normal. $ef=[L:K]$. I knew $|\Delta_0|=e$. However there is no reason to expect $p\nmid f$.

Ref:Algebraic Number Theory by Taylor, Frohlich, Chpter 4 section 4.

  • $\begingroup$ The definition of $\Delta_i$ uses $\forall a\in O_L$. $Q$ is fixed by $\Delta$ (as is $O_L$). Hence, one can replace $a$ by $ga$, and $Q$ by $gQ$ in the definition, for any fixed $g\in \Delta$. OK? $\endgroup$ – peter a g Jul 26 '18 at 0:33
  • $\begingroup$ @peterag Then you know $p\nmid f$ by direct consequence. Why this is the case then? What do you mean by $gQ$ here? There is only 1 non trivial prime in $O_L$ by $K$ complete implying $L$ complete.($O_L$ is DVR.) $gQ=Q$ by $g$ is ring homomorphism. $\endgroup$ – user45765 Jul 26 '18 at 1:22
  • $\begingroup$ @peterag I do not see that is the author's intention. It should be deduced from $p-$sylow part. I do not see any particular reason making this clear. $p\nmid f$ is equivalent to $\Delta_1$ normal. $\endgroup$ – user45765 Jul 26 '18 at 1:46
  • 1
    $\begingroup$ (before your new comment - I'm sorry I didn't read your question carefully! but I think the following still worthwhile. )As you say$gQ =Q$, because there is one prime. So if $\delta g a - ga \in gQ^{i+1}$, for all $a$, we have that $ (g^{-1} \delta g) a - a \in Q^{i+1}$ for all $a$, and conversely. (I think you are missing a $+1$ in your definition, btw, if you are using the usual numbering.) Hence $\Delta_i$ is normal. For your question about $p\not | f$: certainly there are unramified extensions of degree $p$. For instance $p=2$, the spl field of $x^2+x+1=0$ over $\mathbb Q_2$. $\endgroup$ – peter a g Jul 26 '18 at 1:50
  • 1
    $\begingroup$ Because $\Delta_0 = g\Delta_0g^{-1}$. OK? $\endgroup$ – peter a g Jul 26 '18 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.