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I know that in a quadratic field $K $ there is a splitting behavior for primes $p$.

$p$ splits in $K$ if $p\mathcal{O_K}=\mathfrak{p}\mathfrak{p'},$ for two prime ideals $\mathfrak{p}\neq\mathfrak{p'} $ of norm $p$.

$p$ is inert in $K$ if $p \mathcal{O_K}$ is a prime ideal in $\mathcal{O_K} $ of norm $p^2$.

$p$ is ramified in $K$ if $p \mathcal{O_K}=\mathfrak{p}^2$ for some ideal of norm $p$.

Is every prime ideal in $\mathcal{O_K}$ of this form?

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    $\begingroup$ Yes${}{}{}{}{}$. $\endgroup$ – Angina Seng May 1 '20 at 20:47
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Assuming it is question of a quadratic extension of number fields, $K/\Bbb Q$ is Galois, and a prime $p\in \Bbb Z=\mathcal O_{\Bbb Q}$ splits into $(P_1 P_2 \dots P_r )^e$ in $\mathcal O_K$ where the $P_i$ are distinct primes, all having the same inertial degree $f$ over $p$. Moreover $ref = [K : \Bbb Q ]=2$.

You listed all the possible splitting behaviours of $p\mathcal O_K$.

Now taking a random prime ideal $P\in \mathcal O_K$, it is not always of the form $p\mathcal O_K$ for a prime $p\in \Bbb Z$, as an example:

Take $K=\Bbb Q(i)$, and $P=(1-2i)$ is a prime of $\mathcal O_K$ lying over $5=(1-2i)(1+2i)$. But $P$ is not of the form $p\mathcal O_K$ for $p\in \Bbb Z$ prime.

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    $\begingroup$ thanks for the help. I just wanted to know whether each prime ideal in $\mathcal{O_K}$ is of the form $p\mathcal{O_K}$. $\endgroup$ – AnabolicHorse May 1 '20 at 21:15
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    $\begingroup$ You are welcome. Please check my edit. $\endgroup$ – PerelMan May 1 '20 at 21:51

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