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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.

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    I would like to know how to use "Let L / K be a solvable extension of prime degree p"?2011-05-17
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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.2011-05-17
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    What is the definition of a solvable non-Galois extension?2011-05-17
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    Your link is broken.2011-05-17
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    @Arturo: Thank you for the advice. I've edited the post.2011-05-17
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    @Arturo: Sorry for that. It is fixed now.2011-05-17
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    @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?2011-05-17
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    @Alex: it means that the Galois group of this extension is solvable but it is not Galois.2011-05-17
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    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."2011-05-17
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    @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.2011-05-17
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    @dust: Actually a solvable non-Galois separable extension means that Galois group of its Galois closure is solvable.2011-07-06

1 Answers 1

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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 $\mathfrak p$.