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The following exercise and hint appear in Neukirch's Algebraic Number Theory (Section 9, Exercise 3, page 58)

Let $L/K$ 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).

Neukirch also gives this hint:

If $G$ is a transitive solvable permutation group of prime degree $p$, then there is no nontrivial permutation $\sigma \in G$ which fixes two distinct letters.

Can we use the conclusion of my former question to solve this one? Thanks!

I would like to know how to use "Let $L/K$ be a solvable extension of prime degree $p$".

P.S. I'm new here and I've asked my friend Roun to ask the "former question" for me.

Community
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dust
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  • I would like to know how to use "Let L / K be a solvable extension of prime degree p"? – dust May 17 '11 at 16:39
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    (i) If you are going to refer to previous question, you need to link to them, not just say "my former question". (ii) You should write the important information in the question, not in comments. In particular, the note that "[you] would like to know how to use" the separability hypothesis should be in the body of the question, not in a comment. – Arturo Magidin May 17 '11 at 16:42
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    What is the definition of a solvable non-Galois extension? – Alex B. May 17 '11 at 16:48
  • Your link is broken. – Arturo Magidin May 17 '11 at 16:52
  • @Arturo: Thank you for the advice. I've edited the post. – dust May 17 '11 at 16:53
  • @Arturo: Sorry for that. It is fixed now. – dust May 17 '11 at 16:55
  • @dust: When citing a book, it's good form to give the citation; note also that this is exercise 3 in the book, while the previous post are (1) Exercise 1 and (2) Exercise 4; I doubt that you are expected to use Exercise 4 to solve Exercise 3. Also, there is an extensive hint in the book, why not copy that too? – Arturo Magidin May 17 '11 at 16:57
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    @Alex: it means that the Galois group of this extension is solvable but it is not Galois. – dust May 17 '11 at 16:57
  • Hint from Neukirch: "Use the following result of Galois: if $G$ is a transitive solvable permutation group of prime degree $p$, then there is no nontrivial permutation $\sigma\in G$ which fixes two distinct letters." – Arturo Magidin May 17 '11 at 16:59
  • @Arturo:thank you very much for telling me this,I find some difficulty in using this.....so I try not to type to many words....I am sorry. – dust May 17 '11 at 17:01
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    @dust: Actually a solvable non-Galois separable extension means that Galois group of its Galois closure is solvable. – John M Jul 06 '11 at 01:23

1 Answers1

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Let $\mathcal O_K$ be a Dedekind domain, $L$ the fraction field of $\mathcal O_K$. Let $L/K$ be a finite, separable extension, not necessarily Galois, of degree $p$. Let $N$ be the normal closure of $L/K$. Let $G=\text{Gal}(N/K)$ and $H=\text{Gal}(N/L)$.

Let $\mathfrak p$ be a prime of $K$ (i.e. of $\mathcal O_K$). Let $Q$ be a prime of $N$ above $\mathfrak p$. Let $G_Q$ denote the decomposition group of $Q$ over $K$. Then (as discussed in Neukirch pg. 55) there is a bijection from the set of double cosets $H\backslash G/G_Q$ to the set $P_\mathfrak p$ of primes of $L$ above $\mathfrak p$, given by: $$H\backslash G/G_Q\rightarrow P_\mathfrak p, \quad H\sigma G_Q\mapsto\sigma Q\cap L$$

Now suppose the prime $\mathfrak p$ is unramified in $L$. Then $\mathfrak p$ is also unramified in $N$.

Note that there are $p$ cosets $H\sigma_1,\dots,H\sigma_p$ of $H\backslash G$ where $p=[L:K]$. There is an action of $G$ that permutes the cosets $H\sigma_i$ by right multiplication. They key observation is this:

The size of the orbit of the coset $H\sigma_i$ under the right action of the decomposition group $G_Q$ equals the inertia degree of the prime $\sigma_iQ\cap L$ over $\mathfrak p$.

To show this, first observe that, for $\rho\in G_Q$ and $\sigma_i\in G$, $$H\sigma_i\rho=H\sigma_i \Longleftrightarrow \rho\in\sigma_i^{-1}H_{\sigma_iQ}\sigma_i$$ where $H_{\sigma_iQ}$ is the decomposition group of $\sigma_iQ$ over $L$.

Thus the size of the orbit of $H\sigma_i$ is $$[G_Q:\text{stab}(H\sigma_i)]=[G_Q:\sigma_i^{-1}H_{\sigma_iQ}\sigma_i]=[\sigma_iG_Q\sigma_i^{-1}:H_{\sigma_iQ}]=[G_{\sigma_iQ}:H_{\sigma_iQ}]$$ $[G_{\sigma_iQ}:H_{\sigma_iQ}]$ equals the inertia degree of $\sigma_iQ\cap L$ over $\mathfrak p$, proving the highlighted claim above.

Now assume the degree $p$ of $L/K$ is prime, and assume that $\mathfrak p$ has two prime factors $\mathfrak P_1$ and $\mathfrak P_2$ in $L$ of degree 1. This implies we have two cosets $H\sigma_1$ and $H\sigma_2$ who orbits under the action of $G_Q$ are of size 1. $G$ is a solvable group with a transitive action on the $p$ cosets $H\sigma_1,\dots,H\sigma_p$. Thus each element of $G_Q$ fixes the two cosets $H\sigma_1$ and $H\sigma_2$, so by the theorem given in the Hint, each element of $G_Q$ must fix all the cosets, so $G_Q$ partitions the $H\sigma_i$ into $p$ distinct orbits of one element each. Thus, every prime factor of $\mathfrak p$ in $L$ is of degree 1 over $\mathfrak p$.

John M
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  • may I ask you something? How can I derive that $[G_{\sigma_iQ} : H_{\sigma_iQ} ]$ equals the inertia degree? – hew Jan 14 '21 at 11:56