Lying-Over Theorem

Statement

Let $A \subseteq B$ be an integral extension .

For every prime ideal $\mathfrak p \subseteq A$, there exists a prime ideal $\mathfrak P \subseteq B$ such that

Equivalently, the induced map on prime spectra

is surjective.

Cross-links

• Integrality: integral extension
• Lifting chains after choosing a prime above: going-up theorem
• Prime spectra viewpoint: prime spectrum

Examples

1. Gaussian integers again.
   $\mathbb Z \subseteq \mathbb Z[i]$ is integral.
   For the prime $(5)\subset \mathbb Z$, there are primes in $\mathbb Z[i]$ lying over it, e.g. $(2+i)$ and $(2-i)$, each contracting to $(5)$.

2. Square map subring.
   With $A=k[t^2]\subseteq B=k[t]$, every prime of $A$ has a prime above it in $B$.
   For instance, $(t^2)\subset A$ has the lying-over prime $(t)\subset B$, since $(t)\cap k[t^2]=(t^2)$.

3. Adjoining an integral element.
   Let $B=A[\alpha]$ where $\alpha$ is integral over $A$ (so $A\subseteq B$ is integral).
   Then every $\mathfrak p\in Spec (A)$ arises as the contraction of some $\mathfrak P\in Spec (B)$.