Take the 2-minute tour ×
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.

I'm reading Algebraic Number Theory by Jurgen Neukirch. I have some problems with some of the exercises in Section 9 of Chapter 1.

They are:

1) If $L / K$ is a Galois extension of algebraic number fields with non-cyclic Galois group, then there are at most finitely many non-split prime ideals of $K$.

2) Let $L / K$ be a finite (not necessarily Galois) extension of algebraic number fields and $N / K$ the normal closure of $L / K$. Show that a prime ideal $p$ of $K$ is totally split in $L$ if and only if it is totally split in $N$.

I have worked on them for a long time but couldn't get any idea of it.

Can you please help? Thank you!

share|improve this question
    
For 1) think about what the Frobenius element at an unramified prime can be. –  Qiaochu Yuan May 16 '11 at 16:38
    
For (2), did you look at the bijection between the set of primes lying above $\mathfrak{p}$ in $L$ and the double cosets of $\mathrm{Gal}(N/K)$ modulo $\mathrm{Gal}(N/L)$ and $G_{\mathfrak{P}}$, where $\mathfrak{P}$ is a fixed prime of $L$ lying above $\mathfrak{p}$ (discussed at the top of page 55)? –  Arturo Magidin May 16 '11 at 17:23
    
can Mr Qiaochun Yuan share some more details for 1) than just a hint? thanks –  dust May 16 '11 at 18:47
    
@Qiaochu: Thank you very much but I can't follow your hint.(I'm a beginner.) Jeff has given an acceptable answer for question 1). But I'm also interested in your idea. Can you please give some more details? Thanks. –  Roun May 17 '11 at 5:52
    
@Arturo: Thank you for your comment. I think your advice is for the proof of the hint given by the book. I'm sorry for my unfamiliarity with Galois theory but I just don't know how to use this hint to prove the exercise. –  Roun May 17 '11 at 6:13

4 Answers 4

up vote 3 down vote accepted

For (2), I think the key is the discussion Neukirch has earlier in that section (top of page 55), namely:

Let $\mathfrak{p}$ be a prime in $K$, and let $P_{\mathfrak{p}}$ be the set of primes of $L$ lying over $\mathfrak{p}$. Let $G=\mathrm{Gal}(N/K)$, and let $H=\mathrm{Gal}(N/L)$ be the subgroup corresponding to $H$. If $\mathfrak{P}$ is a prime of $N$ lying over $\mathfrak{p}$, then $G_{\mathfrak{P}} = \{\sigma\in G\mid \sigma\mathfrak{P}=\mathfrak{P}\}$, the decomposition group of $\mathfrak{P}$, is also a subgroup of $G$.

Neukirch states (and leaves as an exercise) that the map from the double cosets of $G$ modulo $H$ and $G_{\mathfrak{P}}$, $H\setminus G/G_{\mathfrak{P}}$ to $P_{\mathfrak{p}}$ given by $$H\sigma G_{\mathfrak{P}} \longmapsto \sigma \mathfrak{P}\cap L$$ gives a bijection between the double cosets and $P_{\mathfrak{p}}$.


Assume that this is indeed the case (that is, the bijection is as given; I'll prove it below).

Showing that $\mathfrak{p}$ splits completely is equivalent to showing that $G_{\mathfrak{P}}$ is trivial (middle of page 54). So we prove that $G_{\mathfrak{P}}$ is trivial if and only if $\mathfrak{p}$ splits completely in $L$.

If $G_{\mathfrak{P}}$ is trivial (that is, if $\mathfrak{p}$ splits completely in $N$), then the double cosets are just the cosets of $H$ in $G$, and there are $[G:H] = [L:K]$ cosets (by the Fundamental Theorem of Galois Theory); that means that there are $[L:K]$ primes of $L$ lying over $\mathfrak{p}$, so $\mathfrak{p}$ splits completely in $L$ (this part can also be done simply by looking at the ramification and decomposition indices, which are multiplicative in towers).

Conversely, if $\mathfrak{p}$ splits completely in $L$, then the number of double cosets $H\setminus G/G_{\mathfrak{P}}$ equals $[L:K] = [G:H]$; this is the same as the number of right cosets of $H$; since each double coset decomposes as a disjoint union of right cosets of $H$, it follows that $H\sigma G_{\mathfrak{P}} = H\sigma$ for all $\sigma\in G$, and in particular all conjugates of $G_{\mathfrak{P}}$ are contained in $H$. That is, the normal subgroup generated by $G_{\mathfrak{P}}$ is contained in $H$.

But since $N$ is the normal closure of $L$, and $H$ corresponds to $L$, then by the Fundamental Theorem of Galois Theory we know that $H$ is core-free: the largest normal subgroup of $G$ contained in $H$ is the trivial group. That means that the normal subgroup generated by the decomposition group is trivial, hence the decomposition group $G_{\mathfrak{P}}$ itself is trivial. And this proves that $\mathfrak{p}$ splits completely in $N$, as desired.


So it all comes down to establishing the bijection mentioned above: the map takes the double coset $H\sigma G_{\mathfrak{P}}$ to $\sigma\mathfrak{P}\cap L$.

First, the map is well defined: if $\tau\in G_{\mathfrak{P}}$, then $\tau\mathfrak{P}=\mathfrak{P}$, so $\sigma\mathfrak{P}\cap L = \sigma\tau\mathfrak{P}\cap L$. And if $\rho\in H$, then $\rho$ fixes $L$ pointwise, so $\rho\sigma\mathfrak{P}\cap L = \rho(\sigma\mathfrak{P}\cap L) = \sigma\mathfrak{P}\cap L$. So $\rho\sigma\tau$ corresponds to the same prime of $L$ as $\sigma, and the map is well-defined.

To see that the map is onto, given any prime $\mathfrak{q}$ of $L$ lying above $\mathfrak{p}$, there is a prime $\mathfrak{Q}$ of $N$ lying above $\mathfrak{q}$, and the transitive action of the Galois group guarantees the existence of $\sigma\in G$ such that $\sigma\mathfrak{P}=\mathfrak{Q}$. Thus, $H\sigma G_{\mathfrak{P}}$ maps to $\sigma\mathfrak{P}\cap L = \mathfrak{Q}\cap L = \mathfrak{q}$ (since $\mathfrak{Q}$ lies above $\mathfrak{q}$). This proves that the map is onto.

Finally, to show that the map is one to one, suppose that $\sigma\mathfrak{P}\cap L = \phi\mathfrak{P}\cap L = \mathfrak{q}$. Then $\sigma\mathfrak{P}$ and $\phi\mathfrak{P}$ both lie above $\mathfrak{q}$, so there exists $\rho\in \mathrm{Gal}(N/L) = H$ such that $\rho\sigma\mathfrak{P} = \phi\mathfrak{P}$. Therefore, $\phi^{-1}\rho\sigma\mathfrak{P} = \mathfrak{P}$, so $\phi^{-1}\tau\sigma\in G_{\mathfrak{P}}$, hence there exists $\rho\in G_{\mathfrak{P}}$ such that $\tau\sigma\rho^{-1} = \phi$. Hence, $\phi$ lies in the double coset $H\sigma G_{\mathfrak{P}}$, so $H\phi G_{\mathfrak{P}}=H\sigma G_{\mathfrak{P}}$, showing that the correspondence is one-to-one.

share|improve this answer

1) Let $G=$ Gal$(L/K)$, $p \in K$. Suppose $p$ is unramified and nonsplit. (Since only finitely many primes are ramified, it suffices to show that this cannot occur.) Since $p$ is unramified and nonsplit and $efg=|G|$, we see that $f=|G|$ and the decomposition group $D_p$ is isomorphic to $G$. But we also have that $D_p$ is isomorphic to the Galois group of the residue field of $L/K$ at $p$, which is cyclic of order $f$. This contradicts our hypothesis on $G$.

share|improve this answer

The following is a rather incomplete answer.

In the same section of Neukirch, there is the following exercise:

Exercise 3: If a prime ideal $\mathfrak{p}$ of $K$ is totally split in two separable extensions $L$ and $L'$ of $K$, then it also totally split in the compositum $LL'$.

It is clear that question 2) reduces to Exercise 3 since the Galois closure is the compositum of all "conjugates" of L. Each conjugate $L'$ is isomorphic to $L$, so $\mathfrak{p}$ splits completely in $L'$, too.

A reference for the proof of Exercise 3 is pp.40 (chapter II, section 1) of Lang's "algebraic number thoery", but it uses completion, which is yet to be introduced in Neukirch's book.

However, I had a hard time to prove Exercise 3 directly. I am also curious about the following exercise on the same page, which really puzzles me (I was hoping to reduce Ex3 to it).

Exerciese 2: For every integeral ideal $\mathfrak{a}$ of $\mathcal{O}_L$, there exists a $\theta \in \mathcal{O}_L$ such that the conductor $\mathfrak{F}=\{\alpha\in \mathcal{O}_L\mid \alpha \mathcal{O}_L \subseteq \mathcal{O}_K[\theta]\}$ is coprime to $\mathfrak{a}$ and $L=K(\theta)$.

share|improve this answer
    
Thank you for your answer. But Exercise 3 on my book is: Let $L\vert K$ be a solvable extension of prime degree $p$ (not necessarily Galois). If the unramified prime ideal $\mathfrak{p}$ in $L$ has two prime factors $\mathfrak{P}$ and $\mathfrak{P}'$ of degree 1, then it is already totally split (theorem of F.K. SCHMIDT). Do we mean the same book? Mine is Algebraic Number Theory –  Roun May 17 '11 at 6:09
    
Jiangwei, did you ever make any progress on Exercise 2 (concerning the conductor)? I've been stuck on it, and you might notice that I've posted a separate question on it. –  John M Jun 29 '11 at 18:48
    
Jiangwei, it does seem like you can use Exer 2 to show Exer 3 using elementary methods. If $L=K(\theta)$ then $LL'=L'(\theta)$. Now apply Prop 8.3 to $\mathfrak p$ in the smaller field, and then apply Prop 8.3 again to a prime $\mathfrak P$ over $\mathfrak p$ in $L'$. The minimal polynomial in the big field will divide the minimal polynomial in the small field, showing $\mathfrak P$ splits completely in $LL'$. –  John M Jun 29 '11 at 21:07

for question1,what about using the following lemma:

    If L / K is separable, then there are only finitely many prime
    ideals of K which are ramified in L.

we can suppose the extension L/K is finite.And we know a prime splits totally iff it is unramified,so we can say that when a prime ideal is nonsplit ,it must be ramified...so we can say 1) is right.

share|improve this answer
1  
A prime can be inert, i.e. unramified and not split. Or it can split, but not completely, and still be unramified. –  Alex B. May 17 '11 at 16:14
    
@Alex B oh....thanks –  dust May 17 '11 at 16:20

Your Answer

 
discard

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.