In commutative algebra, an element b of a commutative ring B is said to be integral over a subring A of B if b is a root of some monic polynomial over A.^[1]

If A, B are fields, then the notions of "integral over" and of an "integral extension" are precisely "algebraic over" and "algebraic extensions" in field theory (since the root of any polynomial is the root of a monic polynomial).

The case of greatest interest in number theory is that of complex numbers integral over Z (e.g., ${\displaystyle {\sqrt {2}}}$ or ${\displaystyle 1+i}$); in this context, the integral elements are usually called algebraic integers. The algebraic integers in a finite extension field k of the rationals Q form a subring of k, called the ring of integers of k, a central object of study in algebraic number theory.

In this article, the term ring will be understood to mean commutative ring with a multiplicative identity.

Let ${\displaystyle B}$ be a ring and let ${\displaystyle A\subset B}$ be a subring of ${\displaystyle B.}$ An element ${\displaystyle b}$ of ${\displaystyle B}$ is said to be integral over ${\displaystyle A}$ if for some ${\displaystyle n\geq 1,}$ there exists ${\displaystyle a_{0},\ a_{1},\ \dots ,\ a_{n-1}}$ in ${\displaystyle A}$ such that

$${\displaystyle b^{n}+a_{n-1}b^{n-1}+\cdots +a_{1}b+a_{0}=0.}$$

The set of elements of ${\displaystyle B}$ that are integral over ${\displaystyle A}$ is called the integral closure of ${\displaystyle A}$ in ${\displaystyle B.}$ The integral closure of any subring ${\displaystyle A}$ in ${\displaystyle B}$ is, itself, a subring of ${\displaystyle B}$ and contains ${\displaystyle A.}$ If every element of ${\displaystyle B}$ is integral over ${\displaystyle A,}$ then we say that ${\displaystyle B}$ is integral over ${\displaystyle A}$, or equivalently ${\displaystyle B}$ is an integral extension of ${\displaystyle A.}$

Let B be a ring, and let A be a subring of B. Given an element b in B, the following conditions are equivalent:

(i) b is integral over A;

(ii) the subring A[b] of B generated by A and b is a finitely generated A-module;

(iii) there exists a subring C of B containing A[b] and which is a finitely generated A-module;

(iv) there exists a faithful A[b]-module M such that M is finitely generated as an A-module.

The usual proof of this uses the following variant of the Cayley–Hamilton theorem on determinants:

Theorem Let u be an endomorphism of an A-module M generated by n elements and I an ideal of A such that ${\displaystyle u(M)\subset IM}$. Then there is a relation:

${\displaystyle u^{n}+a_{1}u^{n-1}+\cdots +a_{n-1}u+a_{n}=0,\,a_{i}\in I^{i}.}$

This theorem (with I = A and u multiplication by b) gives (iv) ⇒ (i) and the rest is easy. Coincidentally, Nakayama's lemma is also an immediate consequence of this theorem.

Let A ⊂ B be rings and A' the integral closure of A in B. (See above for the definition.)

Integral closures behave nicely under various constructions. Specifically, for a multiplicatively closed subset S of A, the localization S⁻¹A' is the integral closure of S⁻¹A in S⁻¹B, and ${\displaystyle A'[t]}$ is the integral closure of ${\displaystyle A[t]}$ in ${\displaystyle B[t]}$.^[14] If ${\displaystyle A_{i}}$ are subrings of rings ${\displaystyle B_{i},1\leq i\leq n}$, then the integral closure of ${\displaystyle \prod A_{i}}$ in ${\displaystyle \prod B_{i}}$ is ${\displaystyle \prod {A_{i}}'}$ where ${\displaystyle {A_{i}}'}$ are the integral closures of ${\displaystyle A_{i}}$ in ${\displaystyle B_{i}}$.^[15]

The integral closure of a local ring A in, say, B, need not be local. (If this is the case, the ring is called unibranch.) This is the case for example when A is Henselian and B is a field extension of the field of fractions of A.

If A is a subring of a field K, then the integral closure of A in K is the intersection of all valuation rings of K containing A.

Let A be an ${\displaystyle \mathbb {N} }$-graded subring of an ${\displaystyle \mathbb {N} }$-graded ring B. Then the integral closure of A in B is an ${\displaystyle \mathbb {N} }$-graded subring of B.^[16]

There is also a concept of the integral closure of an ideal. The integral closure of an ideal ${\displaystyle I\subset R}$, usually denoted by ${\displaystyle {\overline {I}}}$, is the set of all elements ${\displaystyle r\in R}$ such that there exists a monic polynomial

${\displaystyle x^{n}+a_{1}x^{n-1}+\cdots +a_{n-1}x^{1}+a_{n}}$

with ${\displaystyle a_{i}\in I^{i}}$ with ${\displaystyle r}$ as a root.^[17][18] The radical of an ideal is integrally closed.^[19][20]

For noetherian rings, there are alternate definitions as well.

• ${\displaystyle r\in {\overline {I}}}$ if there exists a ${\displaystyle c\in R}$ not contained in any minimal prime, such that ${\displaystyle cr^{n}\in I^{n}}$ for all ${\displaystyle n\geq 1}$.
• ${\displaystyle r\in {\overline {I}}}$ if in the normalized blow-up of I, the pull back of r is contained in the inverse image of I. The blow-up of an ideal is an operation of schemes which replaces the given ideal with a principal ideal. The normalization of a scheme is simply the scheme corresponding to the integral closure of all of its rings.

The notion of integral closure of an ideal is used in some proofs of the going-down theorem.

Let B be a ring and A a subring of B such that B is integral over A. Then the annihilator of the A-module B/A is called the conductor of A in B. Because the notion has origin in algebraic number theory, the conductor is denoted by ${\displaystyle {\mathfrak {f}}={\mathfrak {f}}(B/A)}$. Explicitly, ${\displaystyle {\mathfrak {f}}}$ consists of elements a in A such that ${\displaystyle aB\subset A}$. (cf. idealizer in abstract algebra.) It is the largest ideal of A that is also an ideal of B.^[21] If S is a multiplicatively closed subset of A, then

${\displaystyle S^{-1}{\mathfrak {f}}(B/A)={\mathfrak {f}}(S^{-1}B/S^{-1}A)}$.

If B is a subring of the total ring of fractions of A, then we may identify

${\displaystyle {\mathfrak {f}}(B/A)=\operatorname {Hom} _{A}(B,A)}$.

Example: Let k be a field and let ${\displaystyle A=k[t^{2},t^{3}]\subset B=k[t]}$ (i.e., A is the coordinate ring of the affine curve ${\displaystyle x^{2}=y^{3}}$.) B is the integral closure of A in ${\displaystyle k(t)}$. The conductor of A in B is the ideal ${\displaystyle (t^{2},t^{3})A}$. More generally, the conductor of ${\displaystyle A=k[[t^{a},t^{b}]]}$, a, b relatively prime, is ${\displaystyle (t^{c},t^{c+1},\dots )A}$ with ${\displaystyle c=(a-1)(b-1)}$.^[22]

Suppose B is the integral closure of an integral domain A in the field of fractions of A such that the A-module ${\displaystyle B/A}$ is finitely generated. Then the conductor ${\displaystyle {\mathfrak {f}}}$ of A is an ideal defining the support of ${\displaystyle B/A}$; thus, A coincides with B in the complement of ${\displaystyle V({\mathfrak {f}})}$ in ${\displaystyle \operatorname {Spec} A}$. In particular, the set ${\displaystyle \{{\mathfrak {p}}\in \operatorname {Spec} A\mid A_{\mathfrak {p}}{\text{ is integrally closed}}\}}$, the complement of ${\displaystyle V({\mathfrak {f}})}$, is an open set.

An important but difficult question is on the finiteness of the integral closure of a finitely generated algebra. There are several known results.

The integral closure of a Dedekind domain in a finite extension of the field of fractions is a Dedekind domain; in particular, a noetherian ring. This is a consequence of the Krull–Akizuki theorem. In general, the integral closure of a noetherian domain of dimension at most 2 is noetherian; Nagata gave an example of dimension 3 noetherian domain whose integral closure is not noetherian.^[23] A nicer statement is this: the integral closure of a noetherian domain is a Krull domain (Mori–Nagata theorem). Nagata also gave an example of dimension 1 noetherian local domain such that the integral closure is not finite over that domain.^{[citation needed]}

Let A be a noetherian integrally closed domain with field of fractions K. If L/K is a finite separable extension, then the integral closure ${\displaystyle A'}$ of A in L is a finitely generated A-module.^[24] This is easy and standard (uses the fact that the trace defines a non-degenerate bilinear form.)

Let A be a finitely generated algebra over a field k that is an integral domain with field of fractions K. If L is a finite extension of K, then the integral closure ${\displaystyle A'}$ of A in L is a finitely generated A-module and is also a finitely generated k-algebra.^[25] The result is due to Noether and can be shown using the Noether normalization lemma as follows. It is clear that it is enough to show the assertion when L/K is either separable or purely inseparable. The separable case is noted above, so assume L/K is purely inseparable. By the normalization lemma, A is integral over the polynomial ring ${\displaystyle S=k[x_{1},...,x_{d}]}$. Since L/K is a finite purely inseparable extension, there is a power q of a prime number such that every element of L is a q-th root of an element in K. Let ${\displaystyle k'}$ be a finite extension of k containing all q-th roots of coefficients of finitely many rational functions that generate L. Then we have: ${\displaystyle L\subset k'(x_{1}^{1/q},...,x_{d}^{1/q}).}$ The ring on the right is the field of fractions of ${\displaystyle k'[x_{1}^{1/q},...,x_{d}^{1/q}]}$, which is the integral closure of S; thus, contains ${\displaystyle A'}$. Hence, ${\displaystyle A'}$ is finite over S; a fortiori, over A. The result remains true if we replace k by Z.

The integral closure of a complete local noetherian domain A in a finite extension of the field of fractions of A is finite over A.^[26] More precisely, for a local noetherian ring R, we have the following chains of implications:^[27]

(i) A complete ${\displaystyle \Rightarrow }$ A is a Nagata ring

(ii) A is a Nagata domain ${\displaystyle \Rightarrow }$ A analytically unramified ${\displaystyle \Rightarrow }$ the integral closure of the completion ${\displaystyle {\widehat {A}}}$ is finite over ${\displaystyle {\widehat {A}}}$ ${\displaystyle \Rightarrow }$ the integral closure of A is finite over A.

Noether's normalisation lemma is a theorem in commutative algebra. Given a field K and a finitely generated K-algebra A, the theorem says it is possible to find elements y₁, y₂, ..., y_m in A that are algebraically independent over K such that A is finite (and hence integral) over B = K[y₁,..., y_m]. Thus the extension K ⊂ A can be written as a composite K ⊂ B ⊂ A where K ⊂ B is a purely transcendental extension and B ⊂ A is finite.^[28]

In algebraic geometry, a morphism ${\displaystyle f:X\to Y}$ of schemes is integral if it is affine and if for some (equivalently, every) affine open cover ${\displaystyle U_{i}}$ of Y, every map ${\displaystyle f^{-1}(U_{i})\to U_{i}}$ is of the form ${\displaystyle \operatorname {Spec} (A)\to \operatorname {Spec} (B)}$ where A is an integral B-algebra. The class of integral morphisms is more general than the class of finite morphisms because there are integral extensions that are not finite, such as, in many cases, the algebraic closure of a field over the field.

Let A be an integral domain and L (some) algebraic closure of the field of fractions of A. Then the integral closure ${\displaystyle A^{+}}$ of A in L is called the absolute integral closure of A.^[29] It is unique up to a non-canonical isomorphism. The ring of all algebraic integers is an example (and thus ${\displaystyle A^{+}}$ is typically not noetherian).

• Normal scheme
• Noether normalization lemma
• Algebraic integer
• Splitting of prime ideals in Galois extensions
• Torsor (algebraic geometry)