Proof of equivalence of definitions of split primes etc.

I think my definitions of a prime being ramified, split and inert are non-standard. Also I do not see how my definitions are equivalent to (what appear to be) the standard ones.

My definition: Consider $\mathcal O_{K}$, where $K=\mathbb Q(\sqrt d)$ and let $\omega=\sqrt d$ if $d \equiv 2,3 \mod 4$ and $\omega=\frac{1+\sqrt d}{2}$ if $d \equiv 1 \mod 4$. Let $f(x)$ be the minimal polynomial of $\omega$ over $\mathbb Z$ (the monic polynomial with coefficients in $\mathbb Z$ of least degree with $\omega$ as a root). A prime $p \in \mathbb Z$ is ramified, split and inert if $f$ has a repeated root, two distinct roots or no roots in $\mathbb F_p$ respectively. Another, more useful, definition that I have is that $\mathcal O_K/(p) \cong \mathbb F_p[x]/(x^2), \, \mathbb F_p^2$ and $\mathbb F_{p^2}$ respectively, but this can be deduced from the first definition (see my previous question).

Other definition: Wikipedia says that $p$ is inert if $(p)$ is a prime ideal, $p$ splits if $(p)$ is a product of two distinct prime ideals of $\mathcal O_K$ and $p$ is ramified if $(p)$ is the square of a prime ideal of $\mathcal O_K$. (Can this be generalised to non-quadratic fields $K$?)

How could I prove that these definitions are equivalent?

• This can help you maybe.(1) RAMIFIED: all odd prime divisor of d and 2 if d ≡ 2 or 3 (mod4); (2) SPLIT: odd primes p such that d is square in Zp and 2 if d ≡ 1 (mod8); (3) INERT: odd prime such that d is not square in Zp and 2 if d ≡ 5 (mod8) – Piquito Apr 26 '15 at 14:44

A prime ideal $I$ in $\mathcal O_K$ of norm $p^n$ will contain $p$, since $I\bar I= (Norm(I))=(p^n)\implies p \in I$ ($I$ is prime), and so will correspond to an prime ideal of $\mathcal O_K/(p)$ by the correspondence theorem.

Ramified: $\mathcal O_K/(p) \cong \mathbb F_p[x]/(x^2)$ and the only prime ideal of this is $(x)$, by here. So there is only one prime ideal of $p$-power norm and this ideal has norm $p$ (e.g. there are no prime ideals of norm $p^2$). Hence $(p)=(x)(x)=(x)^2$ in the prime ideal factorisation in $\mathcal O_K.$

Split: $\mathcal O_K/(p) \cong \mathbb F_p \times \mathbb F_p$ and this has two prime ideals: $\mathbb F_p \times \{0\}$ and $\{0\} \times \mathbb F_p$, since any element of $\mathbb F_p$ is a unit or $0$. So there are two ideals of $p$-power norm, and both of these have norm $p$. These two ideals are also conjugate, I claim. Studying this in terms of $\bar f$ (notation as here) we find \begin{align*} \frac{\mathcal O_K}{(p) } & \cong \frac{\mathbb F_p[x]}{((\overline {f(x)}))} \\ & \cong \frac{\mathbb F_p[x]}{((x-\alpha)(x-\beta))} \\ & \cong \frac{\mathbb F_p[x]}{(x-\alpha)} \times \frac{\mathbb F_p[x]}{(x-\beta)}\\ & \cong\mathbb F_p^2. \end{align*} So conjugation will send one root to the other, i.e. $\alpha \mapsto \beta$, $\beta \mapsto \alpha$. Hence it will also map the corresponding $\mathbb F_p \times \{0\}$ to $\{0\} \times \mathbb F_p$ and vice versa. So the claim is proved. Now, by a theorem, $I\bar I=(Norm(I)),$ so $(p)=I\bar I$ where $I$ and $\bar I$ are the two ideals of norm $p$.

Inert: $\mathcal O_K/(p) \cong \mathbb F_{p^2}$ and the prime ideal of this is $\{0\}$ since we already have a field and the only proper ideal of a field is $\{0\}.$ This ideal has norm $p^2$ and cprresponds to $(p)$. So $(p)$ is already prime in $\mathcal O_K$.

By the previous question each of these cases correspond to when $\bar f$ (which is $f$ reduced modulo $p$) has a repeated root, two distinct roots and no roots. Hence one of these cases must always occur. This proves the converse. E.g. if $(p)=I^2$ for some $I\triangleleft \mathcal O_K$ then $p$ can't be split or inert, so it must be ramified.

Ramified, split and inert are adjectives for the ideals p*$O_K$ really (the analogues of pZ with rational numbers) in search of unique factorization which is reach with ideals but in general not with $O_K$ It is usual to say "the prime" and not "the ideal" for commodity (and by abuse of language).From the standard definition I feel it is easy to deduce your non-standard one. You have to prove the reciprocal and if it is easy it would be just un exercise. If difficult it will be valuable maybe. On the other hand it is maybe false, I hope it is true.

$\mathcal O_K = \Bbb Z[\omega]$ and so $\mathcal O_K/(p) = \Bbb Z/(p)[\omega] = \Bbb F[x]/f(x)$. This needs a few more details but the basic idea is alright. This is a standard result that goes by some name I can't quite remember...