I have a question about the notion of algebraic integer, and especially the prime ideal of $O_K$ where $K$ is an algebraic number field. Actually, we know that any ideal $I \subset O_K$ has the form $P_{1}^{e_1}…P_{r}^{e_r}$ where the $P_i$ are some prime ideal of $O_K$, and we have the uniqueness of such a decomposition. In the same way, if we have $P$ a prime ideal of $O_K$, then $P \cap \mathbb{Z} = p\mathbb{Z}$, with $p$ a prime number, which is also the characteristic of the field $O_K / P$.
In the same way, if $p$ is a prime number, we have : $pO_K = P_{1}^{e_1}…P_{r}^{e_r}$, we have $N(pO_K) = p^{e_1f_1+…+e_nf_n}$, where $p^{f_i} = N(P_i)$.
We have also by the chinese reminder theorem : $O_K/pO_K \cong (O_K/P_{1}^{e_1})
\times … \times (O_K/P_{1}^{e_n})$.
Then, and this is the point where I'm stuck, it's written that :
Describing the prime ideal of K is equivalent to describe the decomposition of the prime numbers in $\mathbb{Z}$ in $O_K$.
Actually, I don't understand why it is equivalent. For example, when $K=\mathbb{Q}(\sqrt{d})$, if $p$ is prime we have that :
- $pO_K = P_1P_2$, with $N(P_i) = p$ $(1)$
- $pO_K = P_1$, with $N(P_1) = p^2$ $(2)$
- $pO_K = P_1^{2}$, with $N(P_1) = p$ $(3)$
and then, it's written that :
$p$ a prime number in $\mathbb{Z}$ is a norm of an ideal in $O_K$ $\iff$ $p$ is in the case $(1)$ or $(3)$
I don't understand how to prove the direction $\Leftarrow$. The other one is a consequence of what we have done previously, but I don't figure out why we have the equivalence.
Actually, I think my problem is this one : I have understand how to describe $pO_K$ where $p$ is a prime number, but given a prime ideal of $O_K$, what is the link with some $pO_K$ ?
Thank you for enlightening me !
Best Answer
What you need to show is that if $p$ is inert, with $pO_K$ prime, then no ideal of $K$ has norm $p$. But if it did, then by what you've already stated, it would lie over $p$. That is impossible since the only ideal over $p$ has norm $p^2$.