By Kummer-Dedekind theorem, to a prime $p$, we can find a product of $(p)$, s.t. we can find some prime ideals with norm $p$,however, how do we know these prime ideals are the all prime ideals with norm $p$? I tried to use unique factorisation, but I can't show why if $a$ is a prime ideal with norm $p$, then it has to be in the product of $(p)$.
Finding all the ideals of a given prime norm
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algebraic-number-theory
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1any ideal divides its norm ($N(I)\in I$ because, by definition, $N(I)=|O_K/I|$) – 2017-02-18
1 Answers
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If $\mathfrak{p}$ is a prime ideal of $\mathcal{O}_K$, then $\mathfrak{p}\cap \mathbb{Z}$ is a nonzero prime ideal of $\mathbb{Z}$, say $(p)$. Also $p$ is the only rational prime belonging to $\mathfrak{p}$ so $p$ is the only rational prime such that $\mathfrak{p}$ is over $p$.
The prime $p$ is the characteristic of the residue field $\mathcal{O}_K/\mathfrak{p}$ and therefore $\mathcal{O}_K/\mathfrak{p}$ is a field extension of $\mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$, the degree of this extension is by definition the inertia $f$ of $\mathfrak{p}$. Then by linear algebra over $\mathbb{F}_p$ you have $N(\mathfrak{p})=|\mathcal{O}_K/\mathfrak{p}|=p^f$.
In conclusion, every prime $\mathfrak{p}$ is above one and only one prime $p\in \mathbb{Z}$ and you can recognize this prime by computing the norm. Every prime with norm a power of $p$ need to be over $p$ and as the complete list of primes above $p$ are the primes that appears in the decomposition of $p$ you have that all the primes with norm a power of $p$ are the primes in the decomposition of $p$.
In particular the primes with norm exactly $p$ are the prime in the decomposition of $p$ with inertia 1.
As a remark. Using the unique factorization is easy to prove that an ideal $I$ have norm a power of prime iff $I$ is prime.