---
tag: QM5ZCBJYP1
kind: definition
name: Finite type algebra
msc: ["13E15", 13B02]
defines: AC5TFFQT9M
uses: [DPBJ04JMH8, 0SDC934R52]
sources:
  - url: https://en.wikipedia.org/wiki/Finitely_generated_algebra
---

## Definition

Let $A$ be a commutative ring and let $B$ be a commutative [[$A$-algebra|DPBJ04JMH8]] with structure map $f \colon A \to B$.
The algebra $B$ is *of finite type* over $A$, or *finitely generated* as an $A$-algebra, if it has a finite generating set.
That means there are $b_1, \ldots, b_n \in B$ such that every element of $B$ is $p(b_1, \ldots, b_n)$ for a polynomial $p$ with coefficients in $f(A)$.
Equivalently, there is a surjective $A$-algebra homomorphism from the [[polynomial ring|0SDC934R52]] $A[x_1, \ldots, x_n]$ onto $B$ for some $n \ge 0$.

## Remarks

Example: $k[x, x^{-1}]$ is of finite type over a field $k$.
It is generated by $x$ and $x^{-1}$, and it is isomorphic to $k[x, y]/(xy - 1)$.

Non-example: $\Q$ is not of finite type over $\Z$.
Suppose $\Q = \Z[q_1, \ldots, q_n]$, and let $d$ be the product of the denominators of the $q_i$.
Then every element of $\Q$ would have a power of $d$ as a denominator, which fails for $1/p$ with $p$ a prime not dividing $d$.

A [[finite|1NBACAEC2V]] algebra is of finite type, but $k[x]$ over $k$ shows that the converse fails.
A finitely generated algebra over a Noetherian ring is Noetherian, by [[Hilbert's basis theorem|0YKA9J3XEE]].