# Is each prime ideal in a quadratic field of norm $p$ or $p^2$?

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?

• Yes${}{}{}{}{}$. – Angina Seng May 1 '20 at 20:47

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.
• thanks for the help. I just wanted to know whether each prime ideal in $\mathcal{O_K}$ is of the form $p\mathcal{O_K}$. – AnabolicHorse May 1 '20 at 21:15