---
tag: DRJPG78875
kind: definition
name: Integral closure
msc: [13B22, 13B21]
defines: G0V86SJBQW
uses: [1EHN4KPSFS, 0GNCRPYXYV, 0QVD7WBFGN, 0F2PTGX1A2]
sources:
  - url: https://en.wikipedia.org/wiki/Integral_element
---

## Definition

Let $B$ be a commutative ring and let $A$ be a [[subring|0GNCRPYXYV]] of $B$.
The *integral closure of $A$ in $B$* is the set
$$\overline{A} = \set{b \in B : b \text{ is integral over } A}$$
of the elements of $B$ that are [[integral|1EHN4KPSFS]] over $A$.
The ring $A$ is *integrally closed in $B$* if $\overline{A} = A$.
For an [[integral domain|0QVD7WBFGN]] $A$, the integral closure of $A$ in its [[field of fractions|0F2PTGX1A2]] is the *normalization* of $A$.

## Remarks

The set $\overline{A}$ is a subring of $B$ that contains $A$, and it is integrally closed in $B$ by [[transitivity of integrality|ZXZMQM818D]].

Example: the integral closure of $\Z$ in $\Q(i)$ is $\Z[i]$.
If $a + bi$ with $a, b \in \Q$ is integral, then so is its conjugate, so $2a$ and $a^2 + b^2$ are rational and integral over $\Z$, hence integers, and this forces $a, b \in \Z$.

Non-example: $\Z[\sqrt{5}]$ is not integrally closed in $\Q(\sqrt{5})$.
The element $(1 + \sqrt{5})/2$ is a root of $x^2 - x - 1$, so it is integral over $\Z$, but it is not in $\Z[\sqrt{5}]$.
The integral closure of $\Z$ in $\Q(\sqrt{5})$ is $\Z[(1 + \sqrt{5})/2]$.