Integral element

There are many examples of integral closure which can be found in algebraic number theory since it is fundamental for defining the ring of integers for an algebraic field extension ${\displaystyle K/\mathbb {Q} }$ (or ${\displaystyle L/\mathbb {Q} _{p}}$).

Integers are the only elements of Q that are integral over Z. In other words, Z is the integral closure of Z in Q.

The Gaussian integers are the complex numbers of the form ${\displaystyle a+b{\sqrt {-1}},\,a,b\in \mathbf {Z} }$, and are integral over Z. ${\displaystyle \mathbf {Z} [{\sqrt {-1}}]}$ is then the integral closure of Z in ${\displaystyle \mathbf {Q} ({\sqrt {-1}})}$. Typically this ring is denoted ${\displaystyle {\mathcal {O}}_{\mathbb {Q} [i]}}$.

The integral closure of Z in ${\displaystyle \mathbf {Q} ({\sqrt {5}})}$ is the ring

${\displaystyle {\mathcal {O}}_{\mathbb {Q} [{\sqrt {5}}]}=\mathbb {Z} \!\left[{\frac {1+{\sqrt {5}}}{2}}\right]}$

This example and the previous one are examples of quadratic integers. The integral closure of a quadratic extension ${\displaystyle \mathbb {Q} ({\sqrt {d}})}$ can be found by constructing the minimal polynomial of an arbitrary element ${\displaystyle a+b{\sqrt {d}}}$ and finding number-theoretic criterion for the polynomial to have integral coefficients. This analysis can be found in the quadratic extensions article.

Let ζ be a root of unity. Then the integral closure of Z in the cyclotomic field Q(ζ) is Z[ζ].[2] This can be found by using the minimal polynomial and using Eisenstein's criterion.

The integral closure of Z in the field of complex numbers C, or the algebraic closure ${\displaystyle {\overline {\mathbb {Q} }}}$ is called the ring of algebraic integers.

The roots of unity, nilpotent elements and idempotent elements in any ring are integral over Z.

In geometry, integral closure is closely related with normalization and normal schemes. It is the first step in resolution of singularities since it gives a process for resolving singularities of codimension 1.

• For example, the integral closure of ${\displaystyle \mathbb {C} [x,y,z]/(xy)}$ is the ring ${\displaystyle \mathbb {C} [x,z]\times \mathbb {C} [y,z]}$ since geometrically, the first ring corresponds to the ${\displaystyle xz}$-plane unioned with the ${\displaystyle yz}$-plane. They have a codimension 1 singularity along the ${\displaystyle z}$-axis where they intersect.
• Let a finite group G act on a ring A. Then A is integral over A^G, the set of elements fixed by G; see Ring of invariants.
• Let R be a ring and u a unit in a ring containing R. Then[3]

1. u⁻¹ is integral over R if and only if u⁻¹ ∈ R[u].
2. ${\displaystyle R[u]\cap R[u^{-1}]}$ is integral over R.
3. The integral closure of the homogeneous coordinate ring of a normal projective variety X is the ring of sections[4]

${\displaystyle \bigoplus _{n\geq 0}\operatorname {H} ^{0}(X,{\mathcal {O}}_{X}(n)).}$

• If ${\displaystyle {\overline {k}}}$ is an algebraic closure of a field k, then ${\displaystyle {\overline {k}}[x_{1},\dots ,x_{n}]}$ is integral over ${\displaystyle k[x_{1},\dots ,x_{n}].}$
• The integral closure of C[[x]] in a finite extension of C((x)) is of the form ${\displaystyle \mathbf {C} [[x^{1/n}]]}$ (cf. Puiseux series)

It follows from the above four equivalent statements that the set of elements of ${\displaystyle B}$ that are integral over ${\displaystyle A}$ forms a subring of ${\displaystyle B}$ containing ${\displaystyle A}$. (Proof: If x, y are elements of ${\displaystyle B}$ that are integral over ${\displaystyle A}$, then ${\displaystyle x+y,xy,-x}$ are integral over ${\displaystyle A}$ since they stabilize ${\displaystyle A[x]A[y]}$, which is a finitely generated module over ${\displaystyle A}$ and is annihilated only by zero.)[5] This ring is called the integral closure of ${\displaystyle A}$ in ${\displaystyle B}$.

Another consequence of the above equivalence is that "integrality" is transitive, in the following sense. Let ${\displaystyle C}$ be a ring containing ${\displaystyle B}$ and ${\displaystyle c\in C}$. If ${\displaystyle c}$ is integral over ${\displaystyle B}$ and ${\displaystyle B}$ integral over ${\displaystyle A}$, then ${\displaystyle c}$ is integral over ${\displaystyle A}$. In particular, if ${\displaystyle C}$ is itself integral over ${\displaystyle B}$ and ${\displaystyle B}$ is integral over ${\displaystyle A}$, then ${\displaystyle C}$ is also integral over ${\displaystyle A}$.

If ${\displaystyle A}$ happens to be the integral closure of ${\displaystyle A}$ in ${\displaystyle B}$, then A is said to be integrally closed in ${\displaystyle B}$. If ${\displaystyle B}$ is the total ring of fractions of ${\displaystyle A}$, (e.g., the field of fractions when ${\displaystyle A}$ is an integral domain), then one sometimes drops the qualification "in ${\displaystyle B}$ " and simply says "integral closure of ${\displaystyle A}$" and "${\displaystyle A}$ is integrally closed."[6] For example, the ring of integers ${\displaystyle {\mathcal {O}}_{K}}$ is integrally closed in the field ${\displaystyle K}$.

Let A be an integral domain with the field of fractions K and A' the integral closure of A in an algebraic field extension L of K. Then the field of fractions of A' is L. In particular, A' is an integrally closed domain.

This situation is applicable in algebraic number theory when relating the ring of integers and a field extension. In particular, given a field extension ${\displaystyle L/K}$ the integral closure of ${\displaystyle {\mathcal {O}}_{K}}$ in ${\displaystyle L}$ is the ring of integers ${\displaystyle {\mathcal {O}}_{L}}$.

Note that transitivity of integrality above implies that if ${\displaystyle B}$ is integral over ${\displaystyle A}$, then ${\displaystyle B}$ is a union (equivalently an inductive limit) of subrings that are finitely generated ${\displaystyle A}$-modules.

If ${\displaystyle A}$ is noetherian, transitivity of integrality can be weakened to the statement:

There exists a finitely generated ${\displaystyle A}$-submodule of ${\displaystyle B}$ that contains ${\displaystyle A[b]}$.

Finally, the assumption that ${\displaystyle A}$ be a subring of ${\displaystyle B}$ can be modified a bit. If ${\displaystyle f:A\to B}$ is a ring homomorphism, then one says ${\displaystyle f}$ is integral if ${\displaystyle B}$ is integral over ${\displaystyle f(A)}$. In the same way one says ${\displaystyle f}$ is finite (${\displaystyle B}$ finitely generated ${\displaystyle A}$-module) or of finite type (${\displaystyle B}$ finitely generated ${\displaystyle A}$-algebra). In this viewpoint, one has that

${\displaystyle f}$ is finite if and only if ${\displaystyle f}$ is integral and of finite type.

Or more explicitly,

${\displaystyle B}$ is a finitely generated ${\displaystyle A}$-module if and only if ${\displaystyle B}$ is generated as an ${\displaystyle A}$-algebra by a finite number of elements integral over ${\displaystyle A}$.

An integral extension A ⊆ B has the going-up property, the lying over property, and the incomparability property (Cohen–Seidenberg theorems). Explicitly, given a chain of prime ideals ${\displaystyle {\mathfrak {p}}_{1}\subset \cdots \subset {\mathfrak {p}}_{n}}$ in A there exists a ${\displaystyle {\mathfrak {p}}'_{1}\subset \cdots \subset {\mathfrak {p}}'_{n}}$ in B with ${\displaystyle {\mathfrak {p}}_{i}={\mathfrak {p}}'_{i}\cap A}$ (going-up and lying over) and two distinct prime ideals with inclusion relation cannot contract to the same prime ideal (incomparability). In particular, the Krull dimensions of A and B are the same. Furthermore, if A is an integrally closed domain, then the going-down holds (see below).

In general, the going-up implies the lying-over.[7] Thus, in the below, we simply say the "going-up" to mean "going-up" and "lying-over".

When A, B are domains such that B is integral over A, A is a field if and only if B is a field. As a corollary, one has: given a prime ideal ${\displaystyle {\mathfrak {q}}}$ of B, ${\displaystyle {\mathfrak {q}}}$ is a maximal ideal of B if and only if ${\displaystyle {\mathfrak {q}}\cap A}$ is a maximal ideal of A. Another corollary: if L/K is an algebraic extension, then any subring of L containing K is a field.

Let B be a ring that is integral over a subring A and k an algebraically closed field. If ${\displaystyle f:A\to k}$ is a homomorphism, then f extends to a homomorphism B → k.[8] This follows from the going-up.

Let ${\displaystyle f:A\to B}$ be an integral extension of rings. Then the induced map

${\displaystyle {\begin{cases}f^{\#}:\operatorname {Spec} B\to \operatorname {Spec} A\\p\mapsto f^{-1}(p)\end{cases}}}$

is a closed map; in fact, ${\displaystyle f^{\#}(V(I))=V(f^{-1}(I))}$ for any ideal I and ${\displaystyle f^{\#}}$ is surjective if f is injective. This is a geometric interpretation of the going-up.

Let B be a ring and A a subring that is a noetherian integrally closed domain (i.e., ${\displaystyle \operatorname {Spec} A}$ is a normal scheme.) If B is integral over A, then ${\displaystyle \operatorname {Spec} B\to \operatorname {Spec} A}$ is submersive; i.e., the topology of ${\displaystyle \operatorname {Spec} A}$ is the quotient topology.[9] The proof uses the notion of constructible sets. (See also: Torsor (algebraic geometry).)

If ${\displaystyle B}$ is integral over ${\displaystyle A}$, then ${\displaystyle B\otimes _{A}R}$ is integral over R for any A-algebra R.[10] In particular, ${\displaystyle \operatorname {Spec} (B\otimes _{A}R)\to \operatorname {Spec} R}$ is closed; i.e., the integral extension induces a "universally closed" map. This leads to a geometric characterization of integral extension. Namely, let B be a ring with only finitely many minimal prime ideals (e.g., integral domain or noetherian ring). Then B is integral over a (subring) A if and only if ${\displaystyle \operatorname {Spec} (B\otimes _{A}R)\to \operatorname {Spec} R}$ is closed for any A-algebra R.[11] In particular, every proper map is universally closed.[12]

Proposition. Let A be an integrally closed domain with the field of fractions K, L a finite normal extension of K, B the integral closure of A in L. Then the group ${\displaystyle G=\operatorname {Gal} (L/K)}$ acts transitively on each fiber of ${\displaystyle \operatorname {Spec} B\to \operatorname {Spec} A}$.

Proof. Suppose ${\displaystyle {\mathfrak {p}}_{2}\neq \sigma ({\mathfrak {p}}_{1})}$ for any ${\displaystyle \sigma }$ in G. Then, by prime avoidance, there is an element x in ${\displaystyle {\mathfrak {p}}_{2}}$ such that ${\displaystyle \sigma (x)\not \in {\mathfrak {p}}_{1}}$ for any ${\displaystyle \sigma }$. G fixes the element ${\displaystyle y=\prod \nolimits _{\sigma }\sigma (x)}$ and thus y is purely inseparable over K. Then some power ${\displaystyle y^{e}}$ belongs to K; since A is integrally closed we have: ${\displaystyle y^{e}\in A.}$ Thus, we found ${\displaystyle y^{e}}$ is in ${\displaystyle {\mathfrak {p}}_{2}\cap A}$ but not in ${\displaystyle {\mathfrak {p}}_{1}\cap A}$; i.e., ${\displaystyle {\mathfrak {p}}_{1}\cap A\neq {\mathfrak {p}}_{2}\cap A}$.

The Galois group ${\displaystyle \operatorname {Gal} (L/K)}$ then acts on all of the prime ideals ${\displaystyle {\mathfrak {q}}_{1},\ldots ,{\mathfrak {q}}_{k}\in {\text{Spec}}({\mathcal {O}}_{L})}$ lying over a fixed prime ideal ${\displaystyle {\mathfrak {p}}\in {\text{Spec}}({\mathcal {O}}_{K})}$.[13] That is, if

${\displaystyle {\mathfrak {p}}={\mathfrak {q}}_{1}^{e_{1}}\cdots {\mathfrak {q}}_{k}^{e_{k}}\subset {\mathcal {O}}_{L}}$

then there is a Galois action on the set ${\displaystyle S_{\mathfrak {p}}=\{{\mathfrak {q}}_{1},\ldots ,{\mathfrak {q}}_{k}\}}$. This is called the Splitting of prime ideals in Galois extensions.

The same idea in the proof shows that if ${\displaystyle L/K}$ is a purely inseparable extension (need not be normal), then ${\displaystyle \operatorname {Spec} B\to \operatorname {Spec} A}$ is bijective.

Let A, K, etc. as before but assume L is only a finite field extension of K. Then

(i) ${\displaystyle \operatorname {Spec} B\to \operatorname {Spec} A}$ has finite fibers.

(ii) the going-down holds between A and B: given ${\displaystyle {\mathfrak {p}}_{1}\subset \cdots \subset {\mathfrak {p}}_{n}={\mathfrak {p}}'_{n}\cap A}$, there exists ${\displaystyle {\mathfrak {p}}'_{1}\subset \cdots \subset {\mathfrak {p}}'_{n}}$ that contracts to it.

Indeed, in both statements, by enlarging L, we can assume L is a normal extension. Then (i) is immediate. As for (ii), by the going-up, we can find a chain ${\displaystyle {\mathfrak {p}}''_{i}}$ that contracts to ${\displaystyle {\mathfrak {p}}'_{i}}$. By transitivity, there is ${\displaystyle \sigma \in G}$ such that ${\displaystyle \sigma ({\mathfrak {p}}''_{n})={\mathfrak {p}}'_{n}}$ and then ${\displaystyle {\mathfrak {p}}'_{i}=\sigma ({\mathfrak {p}}''_{i})}$ are the desired chain.