Going-down theorem - Imported Knowlpedia Content

For an integral extension $A\subseteq B$ , the going-up theorem says prime inclusions downstairs can be lifted to prime inclusions upstairs. The going-down phenomenon is the opposite direction: once you have fixed a prime upstairs lying over a larger prime downstairs, you can descend to primes lying over smaller primes.

Theorem (Going down). Let $A\subseteq B$ be an integral extension of integral domains. Assume that $A$ is an integrally closed domain. Let $\mathfrak p\subseteq \mathfrak p'$ be prime ideals of $A$ , and let $\mathfrak q'\in \operatorname{Spec}(B)$ satisfy $\mathfrak q'\cap A=\mathfrak p'$ . Then there exists a prime ideal $\mathfrak q\subseteq B$ such that

\mathfrak q\subseteq \mathfrak q' \qquad\text{and}\qquad \mathfrak q\cap A=\mathfrak p.

Equivalent characterizations

Equivalently, any chain of primes in $A$ can be realized as the contraction of a chain inside $\operatorname{Spec}(B)$ that ends at a prescribed prime lying over the top prime in the chain.

Remarks

The integrally closed hypothesis is essential: for general integral extensions, going-down can fail, even though lying over and going up always hold.

Examples

A Dedekind-domain example. The domain $\mathbb Z$ is integrally closed, and $\mathbb Z\subset\mathbb Z[i]$ is integral. Take the chain $(0)\subset (5)$ in $\mathbb Z$ and the prime $\mathfrak q'=(2+i)\subset \mathbb Z[i]$ lying over $(5)$ . Going-down provides a prime $\mathfrak q\subset (2+i)$ with $\mathfrak q\cap\mathbb Z=(0)$ ; necessarily $\mathfrak q=(0)$ .

From $k[t^2]$ to $k[t]$ . Let $A=k[t^2]$ and $B=k[t]$ for a field $k$ . Since $A\cong k[s]$ is a PID, it is integrally closed, and $A\subset B$ is integral. For the chain $(0)\subset (t^2)$ in $A$ and the prime $(t)\subset B$ lying over $(t^2)$ , going-down yields $(0)\subset (t)$ inside $B$ .

A quadratic integral extension of a PID. Let $A=k[x]$ and $B=k[x,y]/(y^2-x)$ . The ring $A$ is a PID, hence integrally closed, and $B$ is integral over $A$ . For the chain $(0)\subset (x)$ in $A$ and the prime $(x,y)\subset B$ lying over $(x)$ , going-down produces a prime inside $(x,y)$ contracting to $(0)$ ; again this is $(0)$ .