A ring extension A→B is integral if every element of B is integral over A.
Let A→B be a homomorphism of commutative rings. The map (or the extension) is called integral if every element b∈B is integral over A; equivalently, for each b∈B there is a monic polynomial in A[T] having b as a root.
A frequently used sufficient condition is:
Finite algebra implies integral. If B is finitely generated as an A-module, then A→B is integral.
Integral extensions behave well with localization: localizing an integral extension at any multiplicative set preserves integrality (compare localization). They also satisfy strong prime ideal behavior, formalized by results such as lying over and going up.
Examples
Adjoining a root of a monic polynomial. For any ring A and monic f(T)∈A[T], the quotient
B:=A[T]/(f)
is integral over A: the class of T in B satisfies the monic equation f(T)=0.
Classical quadratic extensions. The inclusion Z⊆Z[i] is integral since i satisfies T2+1=0 over Z, and every element of Z[i] is a polynomial in i with integer coefficients.
A cusp subring inside a polynomial ring. Let k be a field, A=k[x2,x3]⊆B=k[x]. Then B is integral over A because x∈B satisfies the monic equation T2−x2=0 with x2∈A, and hence every polynomial in x is integral over A.
Non-example: polynomial extensions are not integral. The inclusion A⊆A[x] is typically not integral: the element x does not satisfy any monic polynomial with coefficients in A.
A commutative ring is a ringR such that ab=ba for all a,b∈R.
Let A→B be a homomorphism of commutative rings (often viewed as an inclusion A⊆B). An element b∈B is integral over A if there exists a monic polynomial
Tn+an−1Tn−1+⋯+a1T+a0∈A[T]
such that b is a root, i.e.
bn+an−1bn−1+⋯+a1b+a0=0in B.
A fundamental equivalent criterion is:
Finite-module criterion. An element b∈B is integral over A if and only if the A-subalgebra A[b]⊆B is finitely generated as an A-module.
Quadratic integers. In B=Z[2], the element 2 is integral over A=Z because it satisfies the monic polynomial T2−2.
A non-example: a localization element. In B=Z[1/2], the element 1/2 is not integral over Z. (Intuitively, integrality would force Z[1/2] to be a finite Z-module via the finite-module criterion, which it is not.)
Integral over a subring generated by squares. Let A=k[x2]⊆B=k[x] for a field k. Then x∈B is integral over A, since it satisfies the monic equation T2−x2=0 with coefficient x2∈A.
As a set, S−1R can be constructed from pairs (r,s)∈R×S modulo the equivalence relation
(r,s)∼(r′,s′)⟺∃t∈S such that t(rs′−r′s)=0 in R.
Write the class of (r,s) as sr. Addition and multiplication are defined by
sr+s′r′=ss′rs′+r′s,sr⋅s′r′=ss′rr′.
The canonical map is ι(r)=1r.
If 0∈S, then ι(0) is invertible, hence 1=0 in S−1R; in this case S−1R is the zero ring.
Universal property
The localization is characterized by the following universal mapping property:
If A is any commutative ring and φ:R→A is a ring homomorphism such that φ(s) is a unit of A for every s∈S, then there exists a unique ring homomorphism φ:S−1R→A with φ∘ι=φ. Explicitly,
Theorem (Going up). Assume A⊆B is an integral extension. Let p⊆p′ be prime ideals of A, and let q∈Spec(B) satisfy q∩A=p. Then there exists a prime ideal q′⊆B such that
q⊆q′andq′∩A=p′.
More generally, for any chain of prime ideals p0⊆⋯⊆pn in A and any prime q0 of B lying over p0, there is a chain q0⊆⋯⊆qn in B with qi∩A=pi for all i.
In terms of the prime spectrum, going-up says the contraction map Spec(B)→Spec(A) has the property that prime inclusions downstairs can be realized upstairs, provided one starts with a prime lying over the smaller one.
Examples
A chain in Z lifted to Z[i]. The extension Z⊂Z[i] is integral. Consider the chain (0)⊂(5) in Z. The prime (0)⊂Z[i] lies over (0). Going-up produces a prime q′⊂Z[i] with (0)⊂q′ and q′∩Z=(5); one choice is q′=(2+i).
From k[t2] to k[t]. With A=k[t2]⊂B=k[t] integral, the chain (0)⊂(t2) in A lifts starting from (0)⊂B: going-up gives the chain (0)⊂(t) in B, where (t)∩A=(t2).
Adjoining a square root of x. Let A=k[x]⊂B=k[x,y]/(y2−x), which is integral. The chain (0)⊂(x) in A lifts starting from the prime (0)⊂B to the chain (0)⊂(x,y) in B, since (x,y)∩A=(x).