Maximality Methods, Algebraic Dependence and Transcendence

If A is a set, let P(A)# denote the family of all its nonempty subsets. The axiom of choice states that if A is a nonempty set, then there exists a function β : P(A)# → A with β(S) ∈S for every nonempty subset S of A. Such a function β is called a choice function. Informally, the axiom of choice is a harmless looking statement; it says that we can simultaneously choose one element from each nonempty subset of a set. The axiom of choice is easy to accept, and it is one of the standard axioms of set theory. However, the axiom is not convenient to use as it stands. There are various equivalent forms of it that are more useful, the most popular of which are maximality principle and the well-ordering principle. Recall that a set X is a partially ordered set if there is a relation x ⪯y defined on X that is reflexive, antisymmetric, and transitive. We introduce some definitions to enable us to state the well-ordering principle.

A partially ordered set X is a chain if, for all x, y ∈X, either x ⪯y or y ⪯x. The set of real numbers R is a chain if one takes x ⪯y to be the usual inequality x ≤y. maximality principle. If X is a nonempty partially ordered set in which every chain has an upper bound in X, then X has a maximal element. Theorem. The following statements are equivalent: (i) maximality principle. (ii) The well-ordering principle. (iii) The axiom of choice.

A field K is algebraically closed if every nonconstant f (x) ∈K[x] has a root in K. An algebraic closure of a field k is an algebraic extension k of k that is algebraically closed. The algebraic closure of Q turns out to be the algebraic numbers: Q = A. The fundamental theorem of algebra says that C is algebraically closed; moreover, C is an algebraic closure of R. If f (x) ∈C[x] had no roots, then 1/f (x) would be a bounded entire function that is not constant. There are two main results here. First, every field has an algebraic closure; second, any two algebraic closures of a field are isomorphic. Our proof of existence will make use of a “big” polynomial ring: We assume that if k is a field and T is an infinite set, then there is a polynomial ring k[T ] having one variable for each t ∈T . (We have already constructed k[T ] when T is finite, and the infinite case is essentially a union of k[U], where U ranges over all the finite subsets of T .

Let E/k be a field extension. A subset U of E is algebraically dependent over k if there exists a finite subset {u1, . . . , un} ⊆U and a nonzero polynomial f (x1, . . . , xn) ∈k[x1, . . . , xn] with f (u1, . . . , un) = 0. A subset B of E is algebraically independent if it is not algebraically dependent. Let E/k be a field extension, let u1, . . . , un ∈E, and let ϕ : k[x1, . . . , xn] →E be the evaluation map; that is, ϕ is the homomorphism sending f (x1, . . . , xn) to f (u1, . . . , un) for all f (x1, . . . , xn) ∈k[x1, . . . , xn]. Now {u1, . . . , un} is algebraically dependent if and only if ker ϕ ̸= {0}. If {u1, . . . , un} is algebraically independent, then ϕ extends to an isomorphism k(x1, . . . , xn) ∼= k(u1, . . . , un) ⊆ E, where k(x1, . . . , xn) is the field of rational functions Frac(k[x1, . . . , xn]). In particular, {x1, . . . , xn} ⊆E = k(x1, . . . , xn) is algebraically independent, for ϕ is the identity map in this case. Since algebraically dependent subsets are necessarily nonempty, it follows that the empty subset ∅is algebraically independent. A singleton {e} ⊆E is algebraically dependent if e is algebraic over k; that is, e is a root of a nonconstant polynomial over k, and it is algebraically independent if e is transcendental over k, in which case k(e) ∼= k(x).

A dependency relation on a set $ is a relation ⪯from $ to P($) that satisfies the following axioms: (i) if x ∈S, then x ⪯S; (ii) if x ⪯S, then there exists a finite subset S′ ⊆S with x ⪯S′; (iii) (Transitivity) if x ⪯S and if, for some T ⊆$, we have s ⪯T for every s ∈S, then x ⪯T ; (iv) (Exchange Axiom) if x ⪯S and x ̸⪯S −{y}, then y ⪯(S −{y}) ∪{x}. The transitivity axiom says that if x is dependent on a set S, and if each element of S is dependent on another set T , then x is dependent on T .

Let k be a field of characteristic p > 0, and let f (x) ∈k[x]. If f (x) = g(x pe), where g(x) ∈k[x] but g(x) /∈k[x p], then deg( f ) = pe deg(g). We call pe the degree of inseparability of f (x), and we call deg(g) the reduced degree of f (x).

If E/k is a finite extension, define the separability degree by [E : k]s = [Es : k], and define the inseparability degree by [E : k]i = [E : Es]. Note that E/k is separable if and only if [E : k]i = 1. It is clear that [E : k] = [E : k]s[E : k]i.

A separating transcendence basis of a field extension E/k is a transcendence basis B with E/k(B) a separable extension. Not every extension E/k has a separating transcendence basis. For example, if E/k is an inseparable algebraic extension, then the only transcendence basis is ∅; but k(∅) = k, and E/k(∅) is inseparable. If a field extension E/k has a separating transcendence basis, then E and k1/p are linearly disjoint intermediate fields of E, the algebraic closure of E. Conversely, if E and k1/p are linearly disjoint and E/k is finitely generated, that is, E = k(u1, . . . , un), then E/k has a separating transcendence basis.

If C is a chain and S = {x1, . . . , xn} ⊆C, then there exists some xi, for 1 ≤i ≤n, with x j ⪯xi for all x j ∈S.

Every vector space V over a field F has a basis. Indeed, every linearly independent subset B of V is contained in a basis of V ; that is, there is a subset B′ so that B ∪B′ is a basis of V .

A commutative ring R is ascending-chain-finite if and only if every prime ideal in R is finitely generated.

Let k be a field, and let k[T ] be the polynomial ring in a set T of variables. If t1, . . . , tn ∈T are distinct and if fi(ti) ∈k[ti] ⊆k[T ] are nonconstant polynomials, then the ideal I = ( f1(t1), . . . , fn(tn)) in k[T ] is a proper ideal. Remark. If n = 2, then f1(t1) and f2(t2) are relatively prime, and this lemma says that 1 is not a linear combination of them. ◀

, E/k is an algebraic extension. Hence, p(x) = irr(α1, k) ∈k[x]. By item (ii), p(x) splits over K, so that {α1, . . . , αm} ⊆K; that is, E ⊂K. Therefore, g(x) splits in K[x], and so K is algebraically closed. •

If k/k is an algebraic closure, and if F/k is an algebraic extension, then there is an injective k-map ψ : F →k.

Let ϕ ∈k(x), where k(x) is the field of rational functions over a field k. Then k(ϕ) = k(x) if and only if ϕ is a linear fractional transformation.

If k(x) is a simple transcendental extension, then every intermediate field B is also a simple transcendental extension of k: There is ϕ ∈B with B = k(ϕ).

Let ⪯be a dependency relation on a set $. If T ⊆$ is independent and z ̸⪯T for some z ∈$, then T ∪{z} ⊋T is a strictly larger independent subset.

If E/k is a field extension, then there exists a transcendence basis B. If F = k(B), then F/k is purely transcendental and E/F is algebraic. Moreover, if B and C are maximal algebraically independent subsets, then |B| = |C|.

If k is a field of characteristic p > 0 and f (x) ∈k[x], then there exists e ≥0 and a polynomial g(x) ∈k[x] with g(x) /∈k[x p] and f (x) = g(x pe). Moreover, if f (x) is irreducible, then g(x) is separable.

(i) Let k ⊆B ⊆E be a tower of fields with E/k algebraic. If E/k is separable, then E/B is separable. (ii) Let E/k be an algebraic field extension, where k has characteristic p > 0. If E/k is a separable extension, then E = k(E p). Conversely, if E/k is finite and E = k(E p), then E/k is separable.

If E/K is an algebraic extension, define Es = { α ∈E : α is separable over k } ; then Es is an intermediate field that is the unique maximal separable extension of k contained in E.

If k ⊆B ⊆E is a tower of finite extensions, where k is a field of characteristic p > 0, then [E : k]s = [E : B]s[B : k]s and [E : k]i = [E : B]i[B : k]i.

(i) A partially ordered set may have no maximal elements. For example, R, with its usual ordering, has no maximal elements. (ii) A partially ordered set may have many maximal elements. For example, if X is the partially ordered set of all the proper subsets of a set U, then a subset S is a maximal element if and only if S = U −{u} for some u ∈U; that is, S is the complement of a point. (iii) If X is the family of all the proper ideals in a commutative ring R, partially ordered by inclusion, then a maximal element in X is a maximal ideal. ◀ maximality principle gives a condition that guarantees the existence of maximal elements.

Let k be a perfect field of characteristic p, let k(x) be the function field, and define E = k({un, for n ≥1 : u pn n = x}). Since k is perfect, every extension of k is separable, and so E ∩k1/p = k. However, we claim that E/k does not have a separating transcendence basis. deg(E/k) = 1, because any pair x1/pn and x1/pm are algebraically dependent; let {β} be a transcendence basis. Now k(β) ̸= E, and so there exists some un with un /∈k(β); choose n minimal. Consider the tower k(β) ⊆k(β, un) ⊆E. If {β} were a separating transcendence basis, then E/k(β, un) would be separable, by Proposition 6.81(i). But irr(un, k(β)) is a nonlinear divisor of y pn −x pn, because un /∈k(β), and hence it has repeated roots; therefore, E/k(β, un) is inseparable, a contradiction. ◀