Question: Consider $A=\mathbb{Z}$ and $B=\mathbb{Z}[\sqrt{-3}]$. Let $I=7\mathbb{Z}$ be the prime ideal in $A$. I want to find a prime ideal in $B$ lying over $I$. That is, I need to find a prime ideal $J \subseteq B$ such that $J \cap A=I$.
I have tried the obvious choices: if I take $J=IB$, then $J \cap A=I$ but $IB$ is not prime. If I take $J$ to be the ideal generated by $\sqrt{-3}$ in $B$, then $J$ is prime but $J \cap A \neq I$. What should $J$ be ?
In general, is there any trick to do this type of problem?
Thank you !
So suppose our prime ideal was generated by some $a+\sqrt{-3}b$ with $a,b\in \mathbb Z$. Then, we would need $a^2 + 3b^2 = (a+b\sqrt{-3})(a-b\sqrt{-3})$ to be divisible by $7$ since it is in $\mathbb Z \cap J = I$. We can take $a=2,b=1$ to get $7$ exactly.
So you should guess that $J = (2+\sqrt{-3})$ (or $(2-\sqrt{-3}$) ). Now try and show that these are prime ideals using the norm.