In this section, we study some important properties of field extensions. We also give an introduction to Galois theory. Unless otherwise stated, the letters F, K and L stand for fields in this section.
We have seen that if F ⊆ K is a field extension, then K is a vector space over F. This observation leads to the following very useful definitions.
For a field extension F ⊆ K, the cardinality of any F-basis of K is called the degree of the extension F ⊆ K and is denoted by [K : F]. If [K : F] is finite, K is called a finite extension of F. Otherwise, K is called an infinite extension of F. |
Let F ⊆ K ⊆ L be a tower of field extensions. Then [L : F] = [L : K] [K : F]. In particular, the extension F ⊆ L is finite if and only if the extensions F ⊆ K and K ⊆ L are finite. In that case, [L : K] | [L : F] and [K : F] | [L : F]. Proof One can easily check that if S is an F-basis of K and S′ a K-basis of L, then the set is an F-basis of L. |
Recall the definitions of the rings F[X] of polynomials and F(X) of rational functions in one indeterminate X. These notations are now generalized. For a field extension F ⊆ K and for , we define:
and
Equation 2.3
It is easy to see that F[a] is the smallest (with respect to inclusion) of the integral domains that contain F and a. Similarly F(a) is the smallest of the fields that contain F and a. We also have F[a] ⊆ F(a). Now we state the following important characterization of algebraic elements.
For a field extension F ⊆ K and an element , the following conditions are equivalent:
Proof [(a)⇒(b)] Let be of degree d. Consider the ring homomorphism that takes f(X) ↦ f(a). From Proposition 2.28, Ker , and by the isomorphism theorem . Since h is irreducible over F, F[X]/〈h〉 and so Im are fields. Since Im contains F and a (note that ), we have , that is, . Finally, notice that [F[X]/〈h〉 : F] = d. [(b)⇒(c)] Let d := [F(a) : F]. Since the elements 1, a, a2, . . . , ad are linearly dependent over F, there exists , not all 0, such that α0 + α1a + · · · + αdad = 0. This, in turn, implies that there is an irreducible polynomial with h(a) = 0. Now consider . Clearly, h g (because otherwise g(a) = 0). Since h is irreducible, gcd(g, h) = 1, that is, there exist polynomials u(X), with u(X)g(X) + v(X)h(X) = 1, that is, with u(a)g(a) = 1. But then . [(c)⇒(a)] Clearly, the element 0 is algebraic over F. So assume a ≠ 0. Since , by hypothesis there is a polynomial such that 1/a = f(a). But then a is a root of the non-constant polynomial . |
For a field extension F ⊆ K, the set of elements in K that are algebraic over F is a field. Proof It is sufficient to show that if a, are algebraic over F, then the elements a ± b, ab and a/b (if b ≠ 0) are also algebraic over F. By Theorem 2.33, [F(a) : F] is finite. Since b is algebraic over F, it is also algebraic over F(a). In particular, [F(a)(b) : F(a)] is finite. But then the extension F(a)(b) is also finite over F and contains a ± b, ab and a/b (if b ≠ 0). |
The field F(a)(b) in the proof of the last corollary is also denoted as F(a, b). It is the smallest subfield of K that contains F, a and b, and it follows that F(a, b) = F(b, a). More generally, for a field extension F ⊆ K and for , each algebraic over F, the field F(a1, . . . , an) is defined as F(a1)(a2) . . . (an) and is independent of the order in which ai are adjoined.
Let F ⊆ K be a finite extension. Then K is algebraic over F. Proof For any , we have F ⊆ F(a) ⊆ K. Now use Proposition 2.30. |
The converse of the last corollary is not true, that is, it is possible that an algebraic extension has infinite extension degree. Exercise 2.59 gives an example.
A field extension F ⊆ K is called simple, if K = F(a) for some . |
Let F be a field of characteristic 0 and let a, b (belonging to some extension of F) be algebraic over F. Then the extension F(a, b) of F is simple. Proof Let p(X) and q(X) be the minimal polynomials (over F) of a and b respectively. Let d := deg p and d′ := deg q. The polynomials p and q are irreducible over F and hence by Exercise 2.61 have no multiple roots. Let a1, . . . , ad be the roots of p and b1, . . . , bd′ the roots of q with a = a1 and b = b1. For each i, j with j ≠ 1, the equation ai + λbj = a + λb has a unique solution for λ (not necessarily in F). Since F is infinite, we can choose which is not a solution of any of the equations just mentioned. Define c := a + μb, so that c ≠ ai + μbj for all i, j with j ≠ 1. Clearly, F(c) ⊆ F(a, b). To prove the reverse inclusion, note that by hypothesis q(b) = 0. Also if we define , we see that f(b) = p(a) = 0. By the choice of c, we have f(bj) ≠ 0 for j ≠ 1. Finally since q is square-free, we have . This implies that and so too. |
Let f(X) be a non-constant polynomial of degree d in F[X]. Assume that f does nor split over F. Consider an irreducible (in F[X]) factor f′ of f of degree d′ > 1. F′ := F[X]/〈f′〉 is a field extension of F. Furthermore, if , the elements 1, constitute a basis of F′ over F. In particular, [F′ : F] = d′ ≤ d. Now, one can write f(X) = (X – α1)g(X) for some . If g splits over F′, so does f too. Otherwise, choose any irreducible (in F′[X]) factor g′ of g with deg g′ > 1 and consider the field extension F″ := F′[X]/〈g′〉. Then [F″ : F′] = deg g′ ≤ deg g = d – 1, so that [F″ : F] ≤ d(d – 1). Moreover, if , then f(X) = (X –α1)(X –α2)h(X) for some . Proceeding in this way we get:
For a polynomial of degree d ≥ 1, there is a field extension K of F with [K : F] ≤ d!, such that f splits over K. |
We now establish the uniqueness of the splitting field of a polynomial . To start with, we set up certain notations. An isomorphism μ : F → F′ of fields induces an isomorphism μ* : F[X] → F′[Y] of polynomial rings, defined by adXd+ad–1Xd–1 + · · · + a0 ↦ μ(ad)Yd + μ(ad–1)Yd–1 + · · · + μ(a0). We have μ*(a) = μ(a) for all . Note also that is irreducible over F if and only if is irreducible over F′. With these notations we state the following important lemma.
Let the non-constant polynomial be irreducible over F. Let α and β be roots of f and μ*(f) respectively. Then there is an isomorphism τ : F (α) → F′(β) of fields such that τ(a) = μ(a) for all and τ(α) = β. Proof Since F(α) = F[α] and F′(β) = F′[β], we can define the map τ : F[α] → F′[β] by g(α) ↦ (μ*(g))(β) for each . It is now an easy check that τ is a well-defined isomorphism of fields with the desired properties. |
Roots of an irreducible polynomial are called conjugates (of each other). If α and β are two roots of an irreducible polynomial , the last lemma guarantees the existence of an isomorphism τ : F(α) → F(β) that fixes all the elements of F and that maps α ↦ β.
We use the maps μ : F → F′ and μ* : F[X] → F′[Y] as defined above. Let be a non-constant polynomial and let K and K′ be splitting fields of f and μ*(f) over F and F′ respectively. Then there is an isomorphism τ : K → K′ of fields, such that τ(a) = μ(a) for all . Proof We proceed by induction on n := [K : F]. (By Proposition 2.32 n is finite.) If n = 1, then K = F, that is, the polynomial f splits over F itself and so does μ*(f) over F′, that is, K′ = F′. Thus τ = μ is the desired isomorphism. Now assume that n > 1 and that the result holds for all fields L and for all polynomials in L[X] with splitting fields (over L) of extension degrees less than n. Consider an irreducible factor g of f with 1 < deg g ≤ deg f. Note that g also splits over K. We take any root of g and consider the tower of field extensions F ⊆ F(α) ⊆ K. Similarly, let be a root of μ*(g) and consider F′ ⊆ F′(β) ⊆ K′. By Lemma 2.5 there is an isomorphism ν : F(α) → F′(β) with ν(a) = μ(a) for all and ν(α) = β. Now [K : F(α)] = [K : F]/[F (α) : F] = [K : F ]/deg g < n. It is evident that K and K′ are splitting fields of f and μ*(f) over F(α) and F′(β) respectively. Hence by the induction hypothesis there is an isomorphism τ : K → K′ with τ(a) = ν(a) for all . In particular, τ(a) = μ(a) for all . |
The results pertaining to the splitting field of a polynomial can be generalized in the following way. Let S be a non-empty subset of F[X]. A splitting field of S over F is a minimal field K containing F such that each polynomial splits in K. If S = {f1, . . . , fr} is a finite set, the splitting field of S is the same as the splitting field of f = f1 · · · fr (Exercise 2.57). But the situation is different, if S is infinite. Of particular interest is the set S consisting of all irreducible polynomials in F[X]. In this case, the splitting field of S is an algebraic closure of F.
We give a sketch of the proof that even when S is infinite, a splitting field for S can be constructed. This, in particular, establishes the existence of an algebraic closure of any field. We may assume that S comprises non-constant polynomials only. For each , we define an indeterminate Xf and consider the ring and the ideal of A generated by f(Xf) for all . We have and, therefore, there is a maximal ideal of A containing (Exercise 2.23). Consider the field F1 := A/m containing F. Every polynomial contains at least one root in F1. Now we replace F by F1 and as above get another field F2 containing F1 (and hence F), such that every polynomial in S (of degree ≥ 2) has at least two roots in F2. We continue this procedure (infinitely often, if necessary) and obtain a sequence of fields F ⊆ F1 ⊆ F2 ⊆ F3 ⊆ · · ·. Define K to be the field consisting of all elements of , that are algebraic over F. Each polynomial in S splits in K, but in no proper subfield of K, that is, K is a splitting field of S.
It turns out that the splitting field of S is unique up to isomorphisms that fix elements of F. In particular, the algebraic closure of F is unique up to isomorphisms that fix elements of F, and is denoted by .
For a field K, the set Aut K of all automorphisms of K is a group under (functional) composition. We extend this concept now. Let F ⊆ K be an extension of fields.
For every intermediate field L (that is, a field L with F ⊆ L ⊆ K), we have a subgroup AutL K of AutF K. Conversely, given a subgroup H of AutF K we have the intermediate fixed field FixF H. It is a relevant question to ask if there is any relationship between the subgroups of AutF K and the intermediate fields. A nice correspondence exists for a particular type of extensions that we define now.
A field extension F ⊆ K is said to be a Galois extension (or K is said to be a Galois extension over F), if FixF (AutF K) = F. Thus K is Galois over F if and only if for every there is a with . |
Let K be the splitting field of a non-constant polynomial . By Exercise 2.77, the extension F ⊆ K is normal. Assume that F ⊆ K is a separable extension (Exercise 2.75). Consider an element and let g be the minimal polynomial of α over F. Then deg g > 1 and g splits in K[X]. By assumption (of separability), there is a root of g with β ≠ α. Lemma 2.5 shows that there is a such that τ(α) = β. Thus, K is Galois over F. In particular, if char F = 0 or if , then F ⊆ K is separable and so Galois. For example, is a Galois extension of . |
The following theorem establishes the correspondence we are looking for.
For a finite Galois extension F ⊆ K, there is a bijective correspondence between the set of all intermediate fields and the set of all subgroups of AutF K (given by L ↦ AutL K and H ↦ FixF H) such that the following assertions hold:
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A proof of this theorem is rather long and uses many auxiliary results which we would not need otherwise. We, therefore, choose to omit the proof here.
2.73 | Let α be transcendental over F. Show that the domain F[α] and the field F(α) are respectively isomorphic to the polynomial ring F[X] and the field F(X) of rational functions in one indeterminate X. Generalize the result for an arbitrary family αi, , of elements each of which is transcendental over F. |
2.74 | Let F ⊆ K be a field extension and let be an endomorphism of K with for every .
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2.75 | Let F ⊆ K be a field extension.
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2.76 | F is called a perfect field, if every irreducible polynomial in F[X] is separable over F. |
2.77 | A field extension F ⊆ K is called normal, if every irreducible polynomial in F[X], that has a root in K, splits in K[X].
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2.78 | Prove the following assertions:
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2.79 | Let F ⊆ K be a field extension and let L be the fixed field of AutF K over F. Show that K is a Galois extension of L. |
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