A prime ideal $\mathbf{p}$ "lying above" a prime p

Question

I am reading a book and we are considering an algebraic number field K, and its ring of algebraic integers O.

We know that O contains the (usual) integers and in particular it contains the prime number p.

The book speaks of a prime ideal $\mathbf{p}$ lying above p.

Im not exactly sure what they mean by this. Do they simply mean a prime ideal of O that contains p? My friend told me he thinks it means a prime ideal such that when intersected with the (usual) integers $\mathbf{Z}$ yields the ideal generated by p.

Answer 1

Both things you're saying are correct; that is, you can prove

If $K$ is a number field with ring of integers $\mathcal O_K$, and $\mathfrak p$ is a prime ideal of $\mathcal O_K$, then for a prime $p\in\mathbb Z$ the following are equivalent:

(a) $\mathfrak p$ contains $p$,

(b) $\mathfrak p\cap\Bbb Z=(p)$.

How to prove this? I think (b) implies (a) should be obvious, and on the other hand, if you want to prove (a) implies (b) then just notice that if $\mathfrak p$ contains $p$, then $\mathfrak p\cap\Bbb Z$ is an ideal of $\Bbb Z$ containing $(p)$, and then use the fact that $(p)$ is a maximal ideal in $\Bbb Z$.

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Another important thing I want to add is that every prime $\mathfrak p$ in $\mathcal O_K$ will lie above some prime $p$; this is because $\mathfrak p\cap\Bbb Z$ is always a prime ideal, and as you probably know any prime ideal of $\Bbb Z$ has the form $(p)$. The fact that $\mathfrak p\cap\Bbb Z$ is prime follows from the following more general fact

If $f:A\to B$ is a homomorphism of unital commutative rings and $\mathfrak q$ is a prime ideal of $B$, then $f^{-1}(\mathfrak q)$ is a prime ideal of $A$.

To prove this you just notice that $f^{-1}(\mathfrak q)$ is exactly the kernel of the composition $A\overset{f}\to B\to B/\mathfrak q$, and the latter is an integral domain because $\mathfrak q$ is prime, so you can use the (some number) isomorphism theorem to conclude that $A/f^{-1}(\mathfrak q)$ is an integral domain. To get the statement above you specialize to the case when $f$ is the inclusion $\Bbb Z\hookrightarrow \mathcal O_K$.