If a prime with prime norm is a split prime, in the number ring PID

Question

If a prime with prime norm is a split prime , in an number ring PID?

Example: $5-\sqrt{14}$ in $\mathbb{Z}[\sqrt{14}]$ has norm $11$, it is a split prime in $\mathbb{Z}[\sqrt{14}]$? Why?

Thanks

Answer 1

This is not quite true, I'm afraid.

I assume that the statement you want to prove is:

If $K$ is a number field, $\mathfrak{O}_K$ is its ring of integers, $x \in \mathfrak{O}_K$ is a prime element (that is, $\mathfrak{p} = x \mathfrak{O}_K$ is a prime ideal), and $|\operatorname{Norm}_{K / \mathbf{Q}}(x)|$ is a prime integer $p$, then $\mathfrak{p}$ is split (i.e. the completion of $K$ with respect to the $\mathfrak{p}$-adic valuation is isomorphic to $\mathbf{Q}_p$).

This is not quite true, because you could have ramification. For instance, if $K = \mathbf{Q}(\sqrt{-2})$ and $x = \sqrt{-2}$, then the norm of $x$ is 2 (which is prime!), but the completion of $K$ at the ideal generated by $x$ is a quadratic extension of $\mathbf{Q}_p$. But it's true if you assume that $x$ doesn't divide the discriminant of $K$, which only rules out finitely many primes (and works in the example you gave above). Moreover, you don't need to assume in advance that $x$ is prime; if the norm of $x$ is prime, then $x$ itself is prime.

Note that it makes no difference here whether $\mathfrak{O}_K$ is a PID or not.

(Edit: I should have said either "p doesn't divide the discriminant of K" - a statement about divisibility of integers -- or "x doesn't divide the different of K" -- a statement about divisibility of ideals of K. The different is an ideal of K, which provides slightly finer information about ramification than the discriminant, which is an integer, does.

There are numerous excellent books on this subject. Stewart and Tall's book "Algebraic number theory and Fermat's last theorem" is one of my favourites.)