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

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

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Could you clarify your question? It isn't very clear to me what you're asking. Are you asking why certain primes split in quadratic fields? Or do you want to deduce something about the class number from this information...? – Dylan Moreland Sep 18 '11 at 16:34
Also there is no need to write so many ?s. – Ted Sep 18 '11 at 17:11

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.)

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many thanks David when you say "x doesn't divide the discriminant of K" , in my example : x=5-sqrt{14} doesn't divide the discriminant of Q[sqrt{14}] is 4*14=56, but the division is in the number ring or in Z. Can you recommend some book that comes it explained thanks again PD: Sorry for my level of english – Sidi Sep 18 '11 at 22:25