Going-Down Theorem

Statement

Let $A \subseteq B$ be an integral extension of domains, and assume $A$ is an integrally closed domain .

Let $\mathfrak p \subseteq \mathfrak q$ be prime ideals in $A$, and let $\mathfrak Q$ be a prime ideal of $B$ such that

Then there exists a prime ideal $\mathfrak P \subseteq B$ with

Informally: under integrality + normality of the base, inclusions of primes in $A$ can be lifted downward inside a chosen prime of $B$.

Cross-links

• Integrality and lifting upward: integral extension , going-up theorem
• Base condition: integrally closed domain
• Existence of primes above: lying-over theorem

Examples

1. $\mathbb Z \subset \mathbb Z[i]$.
   $\mathbb Z$ is integrally closed, and $\mathbb Z[i]$ is integral over $\mathbb Z$.
   Take $(0)\subset (5)$ in $\mathbb Z$ and choose $\mathfrak Q=(2+i)\subset \mathbb Z[i]$ lying over $(5)$.
   Going-down produces a prime $\mathfrak P\subset (2+i)$ lying over $(0)$; here $\mathfrak P=(0)$ works.

2. A quadratic integral extension of a PID.
   Let $A=k[x]$ and $B=k[x,y]/(y^2-x)\cong k[y]$. The inclusion $k[x]\hookrightarrow k[y]$ via $x\mapsto y^2$ is integral, and $k[x]$ is integrally closed.
   For the chain $(0)\subset (x)$ in $A$, pick $\mathfrak Q=(y)\subset B$ lying over $(x)$.
   Going-down gives $\mathfrak P=(0)\subset (y)$ lying over $(0)$.

3. Warning (no example needed to use it).
   If $A$ is not integrally closed, going-down can fail for some integral extensions; this is one reason normality (integral closedness) is a standard hypothesis in dimension theory.