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Suppose that the extension $K/\mathbb{Q}$ is normal and has a Galois group which is simple, but not cyclic. Show that there is no rational prime $p$ such that $(p)$ remains prime in $K$.

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I am confused by the term simple but not cyclic. It would help me a lot to understand the concept more deeply if somenone is willing do this case. Thanks in advance!

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    Well, I can tell you "simple but not cyclic" means, if that's any help. A simple group is a group which has no nontrivial proper normal subgroup. If $p$ is prime then the (cyclic) group of order $p$ is simple, but there are (many) non-cyclic examples. The smallest is $A_5$, the alternating group on 5 symbols.2012-03-27
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    http://en.wikipedia.org/wiki/Simple_group2012-03-27
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    Thanks for responding! I still want to know if there are any properties for the normal subgroups of the simple, not cyclic group.2012-03-27
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    @QiangZhang: As Gerry writes: "A simple group is a group which has **no nontrivial proper normal subgroup**." What properties are you asking for?2012-03-27
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    @martini: I mean the properties concerning the non-cyclic restriction.2012-03-27
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    The non-cyclic restriction mainly says that the group is not abelian. Since the only abelian simple groups are the cyclic groups of prime order, which consequently are the only groups of prime order.2012-03-27
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    @JacobSchlather: Thanks! I've seen this as well. But I do not see anything to do with the problem itself. P.S. Why not give the non-abelian condition directly?2012-03-27
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    @GerryMyerson: Thanks a lot! But I am still confused with the relation between this restriction and the problem.2012-03-27
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    So you want to know why no rational prime remains prime. Wish I could help, but I don't see how to do that.2012-03-27
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    Here are some thoughts, but they don't get very far. The factorization of $(p)$ in $K$ is mirrored in the factorization of $f$ modulo $p$, where $f$ is the minimal polynomial for a generator of $K$ over the rationals. So we'd like to show that $f$ is never irreducible modulo $p$. Reducibility modulo $p$ is related to reducibility over the rationals, but the implications seem to go in the wrong direction, and anyway I don't see how to bring the Galois group into it.2012-03-27
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    @GerryMyerson: Big thanks! I have come up an idea, and I will post it later on. See if it is true.2012-03-27

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I have a proof like this:

We claim that there is no rational prime $p$ such that $(p)$ remains prime in $K$, i.e. $p$ is not inert in $K$. Otherwise, we have $$\left\{ \begin{array}{c} \mathfrak {P}_D=p, \\ \mathfrak{P}_I=\mathfrak{P}, \end{array}\right.\quad(*)$$ where $D\triangleq D_{\mathfrak{P}}$ and $I\triangleq I_{\mathfrak{P}}$ denotes the decomposition group and inertia group of $\mathfrak{P}$ respectively. $(*)$ is equivalent to $$\left\{ \begin{array}{c} G=D, \\ I=1_G. \end{array}\right.\quad(**)$$ From Galois theory, we know that $D/I$ is a cyclic group of order $f$. Therefore, we deduce from $(**)$ that $G$ and $D$ are cyclic, which is a contradiction.

See if it is true.

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    Where have you used the hypothesis that $G$ is simple?2012-03-27
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    @GerryMyerson: I don't get it either. The restriction seemed unnecessary.2012-03-27
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    Gerry, Qiang, simplicity is unnecessary and this is the correct general proof. See, e.g., http://mathoverflow.net/questions/12366/how-many-primes-stay-inert-in-a-finite-non-cyclic-extension-of-number-fields/12369#123692012-03-27
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    @BR, a thousand thanks! It helps a lot.2012-03-28
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    Qiang, you are welcome. I've always had a fondness for this exercise.2012-03-28
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    @BR, this one is interesting and confusing somehow since the "simplicity". It turns out I haven't done enough search work.2012-03-28
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If a prime $p$ remained prime in $K$, then $K_p/\mathbb{Q_p}$ would be an extension of local fields with Galois group $G$. But Galois groups of finite extensions of local fields are soluble, because they can be filtered by the ramification subgroups with subsequent quotients being abelian.

So more generally, the decomposition group at any prime in a Galois extension of number fields is always soluble. Further, the precise structure of the higher ramification groups gives more restrictions on the structure of these decomposition groups.

I don't know where this question was quoted from, so I don't know whether this is the intended solution. But it is the conceptual explanation.

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    Thanks a lot, but sorry to say, your answer is too deep for me, because I just began studying the a.n.t this semester, I haven't reached that far.2012-03-28
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    @Qiang What this answer is saying is that apart from the decomposition group and the inertia group, which you have already used, there is a deeper descending chain of subgroups of the inertia group with all successive quotients being abelian. In general, it is helpful for the answerer if you say where the problem comes from and what you have tried. Otherwise, it is not clear at what level to pitch the answer.2012-03-28
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    I really appreciated your comments and answer. I think I have learnt a lot from them with some re-consideration. Thanks for your patiently responding.2012-03-28