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Home , Algebraic extension, Field extension, Simple extension

10. extensions Notation: The letters F, K, L, M would usually denote fields. 10.1. Definition. Let F be a field. The of a field, denoted ch(F ) is the smallest natural n such that n.1F =0.Ifnosuchn exists, then one says that F has characteristic zero, otherwise one says that F has finite characteristic. If F is finite characteristic, then it is easy to see that ch(F ) is a . Clearly the intersection of a collection of subfields of F is again a subfield. If F is a field and S is some of F ,thenthesubfield of F generated by S is defined to be the intersection of all subfields of F that contain S. So the subfield of F generated by S is the smallest subfield of F that contains S. The prime subfield of F is defined to be the subfield generated by 1F .IfF has characteristic zero, then the prime subfield is isomorphic to Q.IfF has characteristic p, then the prime subfield is isomorphic to Fp. 10.2. Theorem. Let p(x) be a non-constant in F [x].Thenonecanconstructa field E such that F E and such that p(x) has a root in E. ⊆ 10.3.Fieldextensions: If K is a field containing a field F ,thenwesaythatK is an extension field of F .WeusetheshorthandK/F to denote a field extension. Two extensions K/F and K′/F are called isomorphic if there is a field φ : K K′ such that the restriction of φ to F is the identity. → If F K M is a “tower” of extensions, we say that K/F is a subextension of M/F. If K/F⊆ is⊆ an field extension, then K is a F -. The of this vector space is called the degree of the extension and denoted by [K : F ] = dimF (K). We say that K/F is a finite extension if [K : F ] is finite. An element a K called algebraic over F if there exists p(x) F [x]suchthatp(a)=0. Call K/F an algebraic∈ extension if each element of K is algebraic∈ over F . 10.4. Theorem. Suppose F K M is a tower of extensions. Then M/F is finite if and only if M/K and K/F are both⊆ finite.⊆ Suppose a1, ,am be a basis of K/F and let b1, ,bn be a basis of M/F.Thenthe a b :1 i ···m, 1 j n forms a basis of M/K.So···[M : F ]=[M : K][K : F ]. { i j ≤ ≤ ≤ ≤ } Proof. Exercise. ! 10.5. Theorem. Finite extensions are algebraic. Proof. Let K/F be a finite extension and a K. The elements 1,a,a2, cannot be all linearly independent over F , so some nontrivial∈ linear combination of these··· must be zero. In other words, a must satisfy of some polynomial with coefficients in F . ! 10.6.Minimalpolynomials: Suppose K/F is an and a K.Let F [a]andF (a) be respectively the and subfield of K generated by F and∈ a.Let φa : F [x] F [a] be the surjective homomorphism obtained by sending x to a.Then ker(φ )= →f F [x]: f(a)=0 is an in F [a]. Since a is algebraic over F , the ideal a { ∈ } ker(φa) is not zero. So there exists a unique pa(x)suchthatker(φa)= (pa(x)). Verify that pa(x) is the unique monic polynomial in F [x] of minimal degree such that pa(a) = 0. We call pa(x)theminimal polynomial of a (over F ). Thus we have

φa 0 (pa(x)) F [x] F [a] 0, where φa(x)=a. → → −→ 29 → Recall that in a PID every prime ideal is maximal. Since F [a] F [x]/(pa) is an , (p ) is a prime ideal, hence maximal. So F [a] F [x]/(p≃) is actually a field. So a ≃ a F [x]/(p ) F [a]=F (a). a ≃ 2 d−1 If the polynomial pa has degree d then one verifies that 1,a,a , ,a forms a basis of F (a)asaK vector space. Hence [F (a):F ]=d. Conversely we have··· the following: 10.7. Theorem. Let p be an irreducible monic polynomial of degree d in F [x].Thenthere exists a extension K/F of degree d and a K such that p is the minimal polynomial of a and K = F [a] F [x]/(p(x)). ∈ Suppose K and≃ K′ are two extensions of F and a K, a′ K′ such that p(x) is the minimal polynomial of a and a′ over F .Thentheextensions∈ K/F∈ and K′/F are isomorphic via an isomorphism that sends a to a′. Proof. Since p(x) is irreducible, it is a prime. In a PID every prime ideal is maximal, so K = F [x]/(p) is a field. Let φ : F [x] F [x]/(p) be the surjection. The composition → φ F F [x] K = F [x]/(p) → −→ is injective (since F is a field and the of the composite is not the whole of F ,hence must be zero). Identify F with the image of of this injection. In other words, we make the identification φ(α)=α for α F .ThenK/F is a field extension and we can consider p(x) to be a polynomial in K[x]. For∈ f(x)= α xi F [x], note that i i ∈ i ! i i φ(f(x)) = φ( αix )= φ(αi)φ(x) = αiφ(x) = f(φ(x)). "i "i "i where the second equality follows since φ is a homomorphism and the third one uses the identification φ(F )=F . Let a = φ(x). Then K = F [a]andp(a)=p(φ(x)) = φ(p(x)) = 0 holds in K. Since p(x) is irreducible, it must be the minimal polynomial of a. Finally, 1,a,a2, ,ad−1 is a basis of K as a F -vector space. ··· ! 10.8.Generatorsforafieldextension: Let F [x , ,x ] denote the ring of 1 ··· n in n variables x1, ,xn, with coefficients in the field F . The fraction field of F [x1, ,xn] is the field of all rational··· functions, denoted by F (x , ,x ). ··· 1 ··· n Let K/F be a field extension and a1, ,an K. Then one verifies that the smallest subfield of K generated by F and a , ···,a consists∈ of all rational expressions of the form 1 ··· n f(a1, ,an)/g(a1, ,an)wheref,g F [x1, ,xn]andg(a1, ,an) = 0. This field is denoted··· by F (a , ···,a ). Say that the∈ extension··· K/F is generated··· by a̸ , ,a K if 1 ··· n 1 ··· n ∈ K = F (a1, ,an). Say that an extension K/F is finitely generated if K = F (a1, ,an) for some a ···, ,a .Observethat ··· 1 ··· n F (a , ,a )=F (a , ,a )(a ). 1 ··· n 1 ··· n−1 n Say that K/F is a if K = F (a)forsomea K. ∈ 10.9. Theorem. Let K/F be an extension. Then the following are equivalent: (a) K/F is a finite extension. par (b) There exists finitely many elements a , ,a K 1 ··· n ∈ such that each aj is algebraic over F and K = F (a1, ,an). (c) There exists a , ,a K such that K = F···(a , ,a ) and a is algebraic over 1 ··· n ∈ 1 ··· n i F (a1, ,ai−1) for each i. ··· 30 Proof. Suppose K/F is finite. Let a1, ,an be a basis of K as a F -vector space. Since finite extensions are algebraic, each a ···is algebraic over F and a , ,a generate K as a j 1 ··· n F -vector space, hence also as a field, that is, F (a1, ,an)=K. Thus (a) implies (b). Suppose a , ,a K such that each a is algebraic··· over F and K = F (a , ,a ). 1 ··· n ∈ j 1 ··· n Let mj be the degree of the minimal polynomial of aj over F . A fortiori, aj

satisfies a polynomial of degree m which has coefficients in F (a , ,a 1 ). So j 1 ··· j [F (a1, ,aj−1)(aj),F(a1, ,aj−1] mj. Thus (b) implies (c). Now··· assume (c). Then every··· step in≤ the tower F F (a ) F (a ,a ) F (a , ,a ) ⊆ 1 ⊆ 1 2 ⊆ ···⊆ 1 ··· n is finite and 10.4 implies that F (a , ,a )/F is finite too. ! 1 ··· n 10.10. Corollary. (a) Let F K M be fields. If K/F and M/K are algebraic then so is M/F. ⊆ ⊆ (b) Let K/F be an extension and a, b K are algebraic over F .Then(a + b) and ab are also algebraic over F .Sothesetofelementsin∈ K that are algebraic over F form a subextension of K/F. Proof. (a) Let a M. Then there is a polynomial f(x)= n a xi K[x]suchthat ∈ i=0 i ∈ f(a)=0.Soa is algebraic over L = F (a0, ,an), so L(a)/L is! finite. Since each aj K is algebraic over F , 10.9 implies that the extension··· L/F is finite. So L(a)/F is also finite∈ and hence a is algebraic over F . (b) Since a and b are algebraic over F , the extension F (a, b)/F is finite. Since (a + b) belong to F (a, b), the extensions F (a + b)/F is also finite. Hence a + b is also algebraic over F . Similar argument for ab. ! 10.11. Definition. Let K, K ,K ,F be fields such that F K K for j =1, 2. Then we 1 2 ⊆ j ⊆ define K1K2 to be the subfield generated by K1 and K2, that is, the smallest subfield of K that contains both K1 and K2.WesaythatK1K2 is the compositum of K1 and K2. Exercise: If [K ,F] and [K : F ] are finite, then [K K : F ] [K : F ][K : F ]. 1 2 1 2 ≤ 1 2 10.12. Definition. An injective homomorphism of fields is called an of fields. Let σ : F L be a homomorphism of fields. Since F does not have any nonzero ideal, ker(σ) is either→ 0 or is equal to F . So either σ =0orσ is an embedding. Let K/F be an extension and σ : F L be an embedding. We say that σ extends to K if there exists a →′ ′ field homomorphism σ : K L such that σ F = σ. Let R, S be rings. Any ring→ homomorphism| σ : R S induces a R[x] S[x] by defining σ( a xi)= σ(a )xi. This→ homomorphism R[x] S[x] will → i i i i → again be denoted by σ. ! !

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