Integral Extensions, Algebraic Integers and One-Dimensional Domains

A ring extension R∗/R is a commutative ring R∗containing R as a subring. If R∗/R is a ring extension, then an element a ∈R∗is integral over R if it is a root of a monic polynomial in R[x]. A ring extension R∗/R is an integral extension if every a ∈R∗is integral over R.

An algebraic number field is a finite field extension of Q. If E is an algebraic number field, then OE/Z is usually denoted by OE instead of by OE/Z, and it is called the ring of integers in E. Because of this new use of the word integers, algebraic number theorists often speak of the ring of rational integers when referring to Z.

If R is a domain with F = Frac(R), then R is a G-domain if F/R is a finitely generated ring extension. An ideal I in a commutative ring R is a G-ideal 4 if R/I is a G-domain. Every field is a G-domain, and so every maximal ideal in a commutative ring is a Gideal. If I is a G-ideal, then R/I is a G-domain, hence a domain; therefore, every G-ideal is a prime ideal. Corollary 11.61 says that Z is not a G-domain; it follows that the prime ideal (x) in Z[x] is not a G-ideal.

A quadratic field is an algebraic number field E with [E : Q] = 2. one-dimensional integrally closed ACC domains

If E is an algebraic number field, then an integral basis for OE is a list β1, . . . , βn in OE such that every α ∈OE has a unique expression α = c1β1 + · · · + cnβn, where ci ∈Z for all i. We now prove that integral bases always exist.

If OE is the ring of integers in an algebraic number field E, then a discriminant of OE is Δ(OE) = → i< j (αi −α j)2, where α1, . . . , αn is an integral basis of OE. 11.43 Let d be a squarefree integer, and let E = Q (√ d ) . (i) If d ≡2 mod 4 or d ≡3 mod 4, prove that 1, √ d is an integral basis of OE, and prove that a discriminant of OE is 4d. (ii) If d ≡1 mod 4, prove that 1, 1 ( 1 + √ d ) is an integral basis of OE, and prove that a discriminant of OE is d. 11.44 Let p be an odd prime, and let E = Q ( ζp ) be the cyclotomic field. (i) Show that 1, 1 −ζp, (1 −ζp)2, . . . , (1 −ζp)p−2 is an integral basis for OE. (ii) Prove that a discriminant of OE is (−1) 2 (p−1) p p−2. Hint. 11.45 (i) If A is the field of all algebraic numbers, prove that OA is not ascending-chain-finite. (ii) Prove that every nonzero prime ideal in OA is a maximal ideal. Hint. Use the proof of Corollary 11.53. Characterizations of one-dimensional integrally closed ACC domains The following definition involves some of the ring-theoretic properties enjoyed by the ring of integers OE in an algebraic number field E.

If R is a one-dimensional integrally closed ACC domains, then its class group C(R) is defined by C(R) = F(R)/P(R), where P(R) is the subgroup of all nonzero principal ideals. The usual proof of finiteness of the class number uses a geometric theorem of H.

Let R be a ring and let A be a left R-module. If there is a finite projective resolution 0 →Pn →· · · →P1 →P0 →A →0, then we write pd(A) ≤n. If n ≥0 is the smallest integer such that pd(A) ≤n, then we say that A has projective dimension n; if there is no finite projective resolution of A, then pd(A) = ∞.

If R∗/R is a ring extension, then the following conditions on a nonzero element u ∈R∗are equivalent. (i) u is integral over R. (ii) There is a finitely generated R-submodule B of R∗with uB ⊆B. (iii) There is a finitely generated faithful R-submodule B of R∗with uB ⊆B; that is, if d B = {0} for some d ∈R, then d = 0.

Let R∗/R be a ring extension of domains with R∗integral over R. Then R∗is a field if and only if R is a field.

Let R∗/R be a ring extension with R∗integral over R. If p ⊆q are prime ideals in R, and if p∗is a prime ideal in R∗lying over p, then there exists a prime ideal q∗lying over q with p∗⊆q∗.

If R is a commutative ring, then nil(R) = " p=prime ideal p = " p=G-ideal p. Remark. If R is a domain, then {0} is a prime ideal, and so nil(R) = {0} (alternatively, there are no nonzero nilpotent elements in a domain). The intersection of all the nonzero prime ideals in a commutative ring R may be larger than nil(R); this happens, for example, when R is a DVR. ◀

If m is a maximal ideal in k[x1, . . . , xn], where k is an algebraically closed field, then there are a1, . . . , an ∈k such that m = (x1 −a1, . . . , xn −an).

Let E/k be a field extension of finite degree n, and let u ∈E. If u = u1, . . . , us are the roots, with multiplicity, of irr(u, k) (in some extension field of E), that is, irr(u, k) = s i=1(x −ui), then tr(u) = [E : k(u)] s i=1 ui and N(u) = ( s→ i=1 ui )[E:k(u)]. Remark. Of course, if u is separable over k, then irr(u, k) has no repeated roots and each ui occurs exactly once in the formulas. ◀

But tr(eiα) = tr ( j c jei f j ) = j c j tr(ei f j) = c jδi j = ci. Therefore, ci ∈R for all i, and so α = i ci fi lies in the free R-module with basis f1, . . . , fn. • one-dimensional integrally closed ACC domains

A domain R is a DVR if and only if it is ascending-chain-finite, integrally closed, and has a unique nonzero prime ideal.

(i) If R is a UFD, then a nonzero ideal I in R is invertible if and only if it is principal. (ii) A one-dimensional integrally closed ACC domains R is a UFD if and only if it is a PID.

The following conditions are equivalent for a domain R. (i) R is a one-dimensional integrally closed ACC domains. (ii) Every submodule of a projective R-module is projective. (iii) Every quotient of an injective R-module is injective. one-dimensional integrally closed ACC domains

If R is a one-dimensional integrally closed ACC domains and M is a finitely generated torsion-free R-module, then M ∼= I1 ⊕· · · ⊕In, where Ii is an ideal in R.

The following statements are equivalent for a domain R. (i) R is a one-dimensional integrally closed ACC domains. (ii) An R-module E is injective if and only if it is divisible.

Let R be a one-dimensional integrally closed ACC domains. If R ⊕G ∼= R ⊕H, where G and H are R-modules, then G ∼= H.

If R is a one-dimensional integrally closed ACC domains, then K0(R) ∼= C(R) ⊕Z, where C(R) is the class group of R.

It states that if k is a field and A is a finitely generated k-algebra, then there exist algebraically independent elements a1, . . . , an in A so that A is integral over k[a1, . . . , an]. ◀ Recall that a complex number is an algebraic integer if it is a root of a monic polynomial in Z[x], so that algebraic integers are integral over Z. The reader should compare the next lemma with Proposition 7.24.

We have seen that R = Z √ −5 is a one-dimensional integrally closed ACC domains that is not a PID. Any non-principal ideal gives an example of a projective R-module that is not free. ◀ Remark. A not necessarily commutative ring R is called left hereditary if every left ideal is a projective R-module (there exist rings that are left hereditary but not right hereditary). Some examples of left hereditary rings aside from one-dimensional integrally closed ACC domains are semisimple rings, noncommutative principal ideal rings, and FIRs (free ideal rings—all left ideals are free Rmodules). Cohn proved that polynomial rings over a field in noncommuting variables are FIRs, and so there exist left hereditary rings that are not left ascending-chain-finite. ◀ The projective and injective modules over a one-dimensional integrally closed ACC domains are well-behaved.