Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $K$.


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!

share|cite|improve this question
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. – Gerry Myerson Mar 27 '12 at 5:52
Thanks for responding! I still want to know if there are any properties for the normal subgroups of the simple, not cyclic group. – Qiang Zhang Mar 27 '12 at 6:05
@QiangZhang: As Gerry writes: "A simple group is a group which has no nontrivial proper normal subgroup." What properties are you asking for? – martini Mar 27 '12 at 6:09
@martini: I mean the properties concerning the non-cyclic restriction. – Qiang Zhang Mar 27 '12 at 6:17
up vote 1 down vote accepted

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.

share|cite|improve this answer
Where have you used the hypothesis that $G$ is simple? – Gerry Myerson Mar 27 '12 at 23:13
@GerryMyerson: I don't get it either. The restriction seemed unnecessary. – Qiang Zhang Mar 27 '12 at 23:50
Gerry, Qiang, simplicity is unnecessary and this is the correct general proof. See, e.g.,… – B R Mar 27 '12 at 23:56
@BR, a thousand thanks! It helps a lot. – Qiang Zhang Mar 28 '12 at 0:12
Qiang, you are welcome. I've always had a fondness for this exercise. – B R Mar 28 '12 at 0:16

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.

share|cite|improve this answer
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. – Qiang Zhang Mar 28 '12 at 1:57
@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. – Alex B. Mar 28 '12 at 4:45
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. – Qiang Zhang Mar 28 '12 at 11:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.