Proof of Proposition 5.11 of David Cox's Primes of form $x^2+ny^2$? 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.
 A: 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 
$$
[L:K]=\sum_i e_{q_i/p}f_{q_i/p}.
$$
A: 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 


*

*$\frak{p}$ is unramified in $L$.

*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 $f$. 

*$\frak{p}$ 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.
