Lying-over theorem - Imported Knowlpedia Content

Let $A\subseteq B$ be an extension of commutative rings. The extension is an integral extension if every element of $B$ is integral over $A$ .

Theorem (Lying over). Assume $A\subseteq B$ is an integral extension. Then for every prime ideal $\mathfrak p\subseteq A$ there exists a prime ideal $\mathfrak q\subseteq B$ such that

\mathfrak q\cap A=\mathfrak p.

Equivalent characterizations

Equivalently, the natural map of prime spectra

\operatorname{Spec}(B)\longrightarrow \operatorname{Spec}(A),\qquad \mathfrak q\mapsto \mathfrak q\cap A

is surjective. In particular, every maximal ideal of $A$ is the contraction of some maximal ideal of $B$ , so the induced map on maximal spectra is also surjective.

Remarks

Lying-over is frequently used as the existence input for going up and, under extra hypotheses, for going down.

Examples

Gaussian integers over the integers. The inclusion $\mathbb Z\subset \mathbb Z[i]$ is integral (since $i$ satisfies $T^2+1=0$ ). For the prime ideal $(5)\subset\mathbb Z$ , lying-over guarantees a prime $\mathfrak q\subset \mathbb Z[i]$ with $\mathfrak q\cap\mathbb Z=(5)$ . Concretely, one can take $\mathfrak q=(2+i)$ (or $\mathfrak q=(2-i)$ ), both lying over $(5)$ .

A simple subring of a polynomial ring. Let $k$ be a field and set $A=k[t^2]\subset B=k[t]$ . The element $t$ is integral over $A$ (it satisfies $T^2-t^2=0$ ), so $A\subset B$ is integral. The prime ideal $(t^2)\subset A$ is the contraction of the prime ideal $(t)\subset B$ .

Adjoining a square root. Let $A=k[x]$ and $B=k[x,y]/(y^2-x)$ . The image of $y$ is integral over $A$ (it satisfies $T^2-x=0$ ), hence $A\subset B$ is integral. The prime ideal $(x)\subset A$ is the contraction of the prime ideal $(x,y)\subset B$ .