# Definition of split prime in quadratic fields

I have the following definition of split prime number $p \in \mathbb{Z}$ in my lecture notes that I don't understand. Let $K$ be a quadratic field, the definition I have says:
$p$ is called split in $K$ if $\mathcal{O}_K/(p)\cong \mathbb{F}^2_p$
What I don't like is that $\mathbb{F}^2_p$ is a field which implies that $p$ remains prime in $\mathcal{O}_k$. However I though that $p$ remains prime if and only if it is inert. So what am I missing here? Is $\mathbb{F}^2_p$ not a field ? The definition I have for $\mathbb{F}^2_p$ in my head is the direct product $\mathbb{F}_p\times\mathbb{F}_p$ which is a direct product of fields so it must be a field.

Thanks in advance

• A direct product of fields is NEVER a field! – Ferra Apr 20 '16 at 13:44

## 1 Answer

Its the most unintuitive definition I have seen. And also $F\times K$ is not a field if $F$ and $K$ are fields. What for example is the inverse of $(1,0)$. A prime splits if $(p)=\mathfrak{p}_1\mathfrak{p}_2$ where $\mathfrak{p}_1$ and $\mathfrak{p}_2$ are distinct prime ideals. So $\mathcal{O}/(p)=\mathcal{O}/\mathfrak{p}_1 \times \mathcal{O}/\mathfrak{p}_2$ gives you the connection with the given definition.

• Thanks a lot that makes sense – TheGeometer Apr 20 '16 at 13:49
• You are welcome, I am glad. – Rene Schipperus Apr 20 '16 at 14:14