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I'm just beginning to read the paper Finding Eisenstein Elements in Cyclic Number Fields of Odd Prime Degree.*

On the third page, in Lemma 2, the author references Proposition 5.11 of David Cox's Primes of the form $x^2+ny^2$, (John Wiley and Sons, New York, 1989), which is apparently found on page 102. It deals with inert primes.

Unfortunately, I don't have access to this book, and it is checked out at my local library for another 3 weeks. Could someone with access perhaps provide the proof of the proposition in question? I'd be very grateful.

*For some reason the link I copy from my browser doesn't seem to work as a link. Googling brings up the paper as the first result though.

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2 Answers 2

Here are lecture notes of a number theory course in Cambridge written by Markus Schepke. Theorem 20 and Proposition 21 are essentially proposition 5.11 in Cox's book. However the author of your paper only uses a small corollary of the proposition, which is easy to prove on its own:

Let $L=K(\alpha)/K$ with $\alpha\in\mathcal{O}_L$ be an extension of number fields, $p$ a prime of $K$. If the minimal polynomial of $\alpha,~m_\alpha(x)$ is irreducible over $\mathcal{O}_K/p$, then $p$ is inert.

Proof: Let $q$ be a prime of $L$ above $p$. Then $\mathcal{O}_L/q$ contains the image of $\alpha$, a root of $m_\alpha(x)\mod p$ and hence $[L:K]\geq f_{q/p}=[\mathcal{O}_L/q:\mathcal{O}_K/p]\geq\deg m_\alpha(x)=[L:K]$. So $q$ is the only prime above $p$.

Remark: For the first inequality above and the conclusion you can use Theorem 19 in the lecture notes I sent you. If $p\mathcal{O}_L=\prod_i q_i^{e_i}$, then $$ [L:K]=\sum_i e_{q_i/p}f_{q_i/p}. $$

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Thanks Michalis. For some reason the lecture notes give me a 404 error. Have they been taken down? –  Cenk May 2 '12 at 18:55
    
@Cenk: That's strange. They were up yesterday and now I can't even access his site anymore. Maybe it's just temporary. In any case, as I said, I don't think you don't need the full theorem, but just the corollary that I prove. –  Michalis May 2 '12 at 20:07
    
Oh well, hopefully it will come back online soon. Thanks for this simplified version, I think it's clearer than the prop in Cox. Just to be clear, what does the notation $f_{q/p}$ mean here? –  Cenk May 3 '12 at 3:21
    
@Cenk: It's the residue degree of $q$ over $p$. It is defined as the degree of the field extension $\mathcal{O}_L/q$ of $\mathcal{O}_K/p$. –  Michalis May 3 '12 at 6:33

Just for reference, here is Cox's Proposition 5.11 and proof from the book.


Proposition 5.11. Let $K\subset L$ be a Galois extension, where $L=K(\alpha)$ for some $\alpha\in\mathcal{O}_L$. Let $f(x)$ be the monic minimal polynomial of $\alpha$ over $K$, so that $f(x)\in \mathcal{O}_K[x]$. If $\frak{p}$ is prime in $\mathcal{O}_K$ and $f(x)$ is separable modulo $\frak{p}$, then

  1. $\frak{p}$ is unramified in $L$.
  2. If $f(x)\equiv f_1(x)\cdots f_g(x)\mod \frak{p}$, where the $f_i(x)$ are distinct and irreducible modulo $\frak{p}$, then ${\frak{P}}_i={\frak{p}}\mathcal{O}_L+f_i(\alpha)\mathcal{O}_L$ is a prime ideal of $\mathcal{O}_L$, ${\frak{P}}_i\neq{\frak{P}}_j$ for $i\neq j$, and $${\frak{p}}\mathcal{O}_L= {\frak{P}}_1\cdots {\frak{P}}_g.$$ Furthermore, all of the $f_i(x)$ have the same degree, which is the inertial degree $f$.
  3. $\frak{p}$ splits completely in $L$ if and only if $f(x)\equiv 0\mod \frak{p}$ has a solution in $\mathcal{O}_K$.

Proof. Note that 1. and 3. are immediate consequences of 2. (see Exercise 5.5). To prove 2., note that $f(x)$ separable modulo $\frak{p}$ implies that $$f(x)\equiv f_1(x)\cdots f_g(x)\mod\frak{p},$$ where the $f_i(x)$ are distinct and irreducible modulo $\frak{p}$. The fact that the above congruence governs the splitting of $\frak{p}$ in $\mathcal{O}_L$ is a general fact that holds for arbitrary finite extensions (see Marcus [77, Theorem 27]). However, the decomposition group from Proposition 5.10 makes the proof in the Galois case especially easy. See Exercise 5.6.$\qquad$ Q.E.D.

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Thanks for typing this out. –  Cenk May 2 '12 at 18:49
    
@Cenk Well I happen to have the book handy, though I've never read this far into it. Except for the first chapters, this material is beyond me. –  Byron Schmuland May 2 '12 at 18:53

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