# A prime ideal $\mathbf{p}$ “lying above” a prime p

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.

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 $$\Bbb Z$$.

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 $$\Bbb Z\hookrightarrow \mathcal O_K$$.