2.8. Fields

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.

2.8.1. Properties of Field Extensions

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.

Definition 2.54.

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.

Proposition 2.30.

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.

Theorem 2.33.

For a field extension F ⊆ K and an element , the following conditions are equivalent:

  1. The element a is algebraic over F.

  2. The extension F(a) is finite over F.

  3. F(a) = F[a].

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, hg (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 .

Corollary 2.10.

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.

Corollary 2.11.

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.

Corollary 2.12.

If F ⊆ K and K ⊆ L are algebraic field extensions, then F ⊆ L is also algebraic.

Proof

Take an arbitrary . Since K ⊆ L is algebraic, there is a non-zero polynomial such that f(a) = 0. It then follows that a is algebraic over F(α0, . . . , αn). Since each αi is algebraic over F, the degree [F(α0, . . . , αn) : F] is finite. Therefore, [F(α0, . . . , αn)(a) : F] = [F(α0, . . . , αn)(a) : F(α0, . . . , αn)] [F(α0, . . . , αn) : F] is also finite and hence F(α0, . . . , αn)(a) and, in particular, a are algebraic over F.

Definition 2.55.

A field extension F ⊆ K is called simple, if K = F(a) for some .

Proposition 2.31.

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.

Corollary 2.13.

A finite extension F ⊆ K of fields of characteristic 0 is simple.

Proof

We proceed by induction on d := [K : F]. The result vacuously holds for d = 1. So let us assume that d > 1 and that the result holds for all smaller values of d. Choose an element . Then [F(a) : F] > 1 and divides d. If [F(a) : F] = d, we are done. So assume [F(a) : F] < d. Since [K : F(a)] < d, by the induction hypothesis the extension F(a) ⊆ K is simple, say K = F(a)(b) = F(a, b). The result now follows immediately from the previous proposition.

2.8.2. Splitting Fields and Algebraic Closure

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:

Proposition 2.32.

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.

Lemma 2.5.

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.

Proposition 2.33.

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 .

*2.8.3. Elements of Galois Theory

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.

Definition 2.56.

An automorphism is called an F-automorphism of K, if fixes all the elements of F(which means that for all ). The set of all F-automorphisms of K is denoted by AutF K or by Gal(K|F) and is a subgroup of Aut K. The Galois group of a polynomial is defined to be the group AutF K, where K is the splitting field of f over F.

Conversely, for a subgroup H of AutF K the set of elements of K that are fixed by all the automorphisms of H, that is, the set of all with for every , is a subfield of K, called the fixed field of H (over F) and denoted as FixF H. Clearly, F ⊆ FixF H ⊆ K.

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.

Definition 2.57.

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 .

Example 2.18.

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.

Theorem 2.34. Fundamental theorem of Galois theory

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:

  1. AutFixF H K = H for every subgroup H of AutF K.

  2. FixF (AutL K) = L for every field L with F ⊆ L ⊆ K.

  3. For field extensions F ⊆ L ⊆ L′ ⊆ K, the extension degree [L′ : L] is the same as the index [AutL K : AutL′ K]. In particular, the order of AutF K is [K : F].

  4. For every intermediate field L, one has:

  1. K is Galois over L.

  2. L is Galois over F if and only if AutL K is a normal subgroup of AutF K. In this case, AutF L ≅ AutF K/AutL K.

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.

Exercise Set 2.8

2.73Let 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.
2.74Let F ⊆ K be a field extension and let be an endomorphism of K with for every .

  1. If a non-constant polynomial has a root , show that is also a root of f. For example, if , and is the automorphism mapping z to its (complex) conjugate , then we conclude that if a complex number z is a root of , then is also a root of f. A similar result holds for the extension , where m is a non-square rational number.

  2. If K is algebraic over F, show that is an automorphism. [H]

2.75Let F ⊆ K be a field extension.

  1. An irreducible polynomial is said to be separable over F, if f has no multiple roots. An algebraic element is said to be separable over F, if the minimal polynomial of α over F is separable. K is called a separable extension of F, if every element of K is (algebraic and) separable over F. Show that if char F = 0 or if , and if K is an algebraic extension of F, then K is separable over F · [H]

  2. An algebraic element is called purely inseparable over F, if the minimal polynomial of α over F factors in K[X] as (X – α)n for some . If every element of K is (algebraic and) purely inseparable over F, then K is called a purely inseparable extension of F. Show that is both separable and purely inseparable if and only if . Thus, if char F = 0 or , then F has no purely inseparable extension other than itself.

  3. If p := char F > 0, then an element is purely inseparable over K if and only if minpolyα,F(X) = Xpr + a for some r ≥ 0 and . In particular, show that if K is a finite purely inseparable extension of F, then [K : F ] = ps for some s ≥ 0.

2.76F is called a perfect field, if every irreducible polynomial in F[X] is separable over F.

  1. Show that F is a perfect field if and only if every algebraic extension of F is separable over F. In particular, the fields of characteristic 0 and the fields , , are perfect.

  2. Let p := char F > 0. Show that F is perfect if and only if every element of F has a p-th root in F. [H]

2.77A field extension F ⊆ K is called normal, if every irreducible polynomial in F[X], that has a root in K, splits in K[X].

  1. If K is the splitting field of a polynomial over F, show that K is a normal extension of F. [H]

  2. If [K : F] = 2, show that F ⊆ K is a normal extension.

  3. Consider the tower of field extensions to conclude that if F ⊆ K and K ⊆ L are normal extensions, then F ⊆ L need not be normal.

2.78Prove the following assertions:

  1. is an infinite extension of . [H]

  2. . [H]

2.79Let 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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