Lying-over theorem

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

Equivalently, the natural map of prime spectra

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.

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

Examples

1. 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 $(5)$.

2. 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 $(t)\subset B$.

3. 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 $(x,y)\subset B$.