---
tag: 9D7WZEN64H
kind: definition
name: Finite morphism of rings
msc: [13B10, 13B21]
defines: 1NBACAEC2V
uses: [0GGE0Z0JQ1, 0NZ8JA25WK, 05QY1VEBYD]
sources:
  - url: https://en.wikipedia.org/wiki/Finite_morphism
---

## Definition

Let $A$ and $B$ be commutative rings and let $f \colon A \to B$ be a [[ring homomorphism|0GGE0Z0JQ1]].
The ring $B$ is an [[$A$-module|0NZ8JA25WK]] by $a \cdot b = f(a) b$.
The homomorphism $f$ is *finite*, and $B$ is a *finite $A$-algebra*, if $B$ is a [[finitely generated|05QY1VEBYD]] $A$-module.
That is, there are $b_1, \ldots, b_n \in B$ such that every $b \in B$ can be written as $b = f(a_1) b_1 + \cdots + f(a_n) b_n$ with $a_i \in A$.

## Remarks

Example: $\Z \to \Z[\sqrt{2}]$ is finite, generated as a module by $1$ and $\sqrt{2}$.
Every surjective homomorphism $A \to A/I$ is finite, generated by $1$.

Non-example: $\Z \to \Q$ is not finite.
If $q_1, \ldots, q_n$ generated $\Q$ as a $\Z$-module, every rational number would have a denominator dividing the product of their denominators, which fails for $1/p$ with $p$ a large prime.

A homomorphism is finite if and only if it is [[integral|B7X0KFY1S1]] and of [[finite type|AC5TFFQT9M]].
A composition of finite homomorphisms is finite.