# Splitting of prime ideals in algebraic extensions

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.

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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

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}}$, showing that the correspondence is one-to-one. - 1) Let$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$. - The following is a rather incomplete answer. In the same section of Neukirch, there is the following exercise: Exercise 3: If a prime ideal$\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)$. - Thank you for your answer. But Exercise 3 on my book is: Let$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'$. – John M Jun 29 '11 at 21:07 @JohnM Prop 8.3 requires$\mathfrak{p}$be relative prime to the conductor, which can be done by using Exer 2. However, how to ensure$\mathfrak{P}\$ is again prime to the conductor of big field? –  X.G at Math Mar 23 at 9:00

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.

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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