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In the concrete case that $\mathcal{O}_K = \mathbb{Z}[i]$, and $\mathfrak{p} = (1+i)$, how to make sense of $\mathfrak{p}\cap \mathbb{Z}$?

I want to know if (1+i) lies above/over 2.

sycs
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  • $(1+i)$ lies over $2$, since $\langle 1+i \rangle^{2} = \langle 2 \rangle$ in $\mathbb{Z}[i]$. Do you want proof of the general fact in the title? If so, I'll post an answer in a bit. – Alex Wertheim Jul 24 '16 at 04:31
  • Is $\langle 1+i \rangle^{2} $ meant as the product of the ideal (1+i) with itself? If so, how to see that it equals (2)? – sycs Jul 24 '16 at 15:08

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Let $i : \mathbb{Z}\hookrightarrow\mathbb{Z}[i]$ be the inclusion. Then $\mathfrak{p}\cap\mathbb{Z}$ is just the preimage $i^{-1}(\mathfrak{p})$. It's easy to prove that the preimage of a prime ideal is also a prime ideal, so in this case the preimage is $(p)$ for some prime $p$. Since $(1+i)(1-i) = 2$, we find that $2\in\mathfrak{p}$, and hence $2\in\mathfrak{p}\cap\mathbb{Z} = i^{-1}(\mathfrak{p})$. Since $(2)$ is a maximal ideal of $\mathbb{Z}$, we see that the preimage must be precisely $(2)$.

oxeimon
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  • So as long as we find a prime $p\in\mathfrak{p}\cap\mathbb{Z}$, then $\mathfrak{p}$ lies over $p$? – sycs Jul 24 '16 at 15:14
  • The definition of $\mathfrak{p}$ lying over $(p)$ is that $\mathfrak{p}\cap\mathbb{Z} = (p)$ – oxeimon Jul 24 '16 at 18:11