---
tag: Y2XHZKR0TX
kind: definition
name: Finitely presented algebra
msc: ["13E15", 13B02]
defines: 764GZHTGV8
uses: [DPBJ04JMH8, 0SDC934R52, 0GSW76Q14N, 0076HWHGHQ, AC5TFFQT9M]
sources:
  - url: https://en.wikipedia.org/wiki/Glossary_of_ring_theory
---

## Definition

Let $A$ be a commutative ring and let $B$ be a commutative [[$A$-algebra|DPBJ04JMH8]].
The algebra $B$ is *finitely presented* over $A$ if it has a presentation with finitely many generators and relations.
That means there are an integer $n \ge 0$ and a finitely generated ideal $I = (g_1, \ldots, g_m)$ of the [[polynomial ring|0SDC934R52]] $A[x_1, \ldots, x_n]$ with $B \cong A[x_1, \ldots, x_n]/I$ as $A$-algebras.
Here $A[x_1, \ldots, x_n]/I$ is the [[quotient ring|0076HWHGHQ]].

## Remarks

Example: $\Z[i]$ is finitely presented over $\Z$, since $\Z[i] \cong \Z[x]/(x^2 + 1)$ and the ideal $(x^2 + 1)$ is generated by one element.

Non-example: let $A = k[y_1, y_2, \ldots]$ be a polynomial ring over a field $k$ in countably many variables, and let $B = A/(y_1, y_2, \ldots) \cong k$.
Then $B$ is of [[finite type|AC5TFFQT9M]] over $A$, generated by no elements at all, but it is not finitely presented.
If an algebra is finitely presented, then the kernel of every surjection $A[x_1, \ldots, x_n] \to B$ is finitely generated, and the kernel of $A \to B$ is $(y_1, y_2, \ldots)$, which is not.

If $A$ is Noetherian, every finite type $A$-algebra is finitely presented, because every ideal of $A[x_1, \ldots, x_n]$ is finitely generated by [[Hilbert's basis theorem|0YKA9J3XEE]].