Subject : MATHEMATICS Paper 1 : ABSTRACT ALGEBRA Chapter 8 : Field Extensions Module 3 : Algebraic extensions Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal 1 Algebraic extensions Learning Objectives: 1. Algebraic extensions. 2. Every finite extension is algebraic. 3. If F=L and L=K are algebraic then F=K is so. 4. Characterization of simple algebraic extensions. Every complex number a + ib is a root of the real polynomial (x − a)2 + b2. Thus in the extension C=R, every z 2 C is an algebraic element over R. But there are real numbers which are not algebraic elements over Q. In this module, we consider the field extensions F=K such that every element of F is algebraic over K. Such extensions are called algebraic extensions. Different characterizations of algebraic extensions have been done here. Definition 0.1. A field extension F=K is called an algebraic extension if every element of F is an algebraic element over K. Otherwise F=K is called transcendental extension. Thus an extension F=K is transcendental if and only if there is an element in F which is transcendental over K. Example 0.2. 1. C=R is an algebraic extension. 2. R=Q is an transcendental extension. (e; π 2 R are transcendental over Q) Consider a field extension F=K and an element c in F . In the previous module, we proved that K(c) is a finite extension of K if c is algebraic over K. Now we show that c is algebraic over K if K(c)=K is finite. In fact, we have the result in a more general form. Theorem 0.3. Every finite extension is algebraic. Proof. Let F=K be an extension of fields such that [F : K] = n. Consider c 2 F . Then 2 n f1; c; c ; ··· ; c g cannot be linearly independent. So there are k0; k1; ··· ; kn in K such that some of them are non zero and n k0 + k1c + ··· + knc = 0: n Then f(x) = k0 +k1x+···+knx 2 K[x] is a nonzero polynomial such that f(c) = 0, and so c is an algebraic element over K. Since c is an arbitrary element of F , F=K is an algebraic extension. 2 Now let us consider the converse of the above theorem. First note that if K(c)=K is algebraic then c is algebraic over K and hence K(c)=K is a finite extension. Thus a simple extension K(c)=K is algebraic if and only if it is finite. Following example shows that there are algebraic extensions which are not finite. Thus the converse of the above theorem is not true in general. Details of the following example can be found in \Fundamentals of Abstract Algebra" by Malik, Mordeson and Sen. p p p Example 0.4. Arrange primes 2; 3; 5; ··· in ascending order. Show that Q( 2; 3; 5; ··· )=Q is an infinite algebraic extension of fields. Now we prove a necessary and sufficient condition for an algebraic extension to be finite. Theorem 0.5. An algebraic extension F=K is finite if and only if F = K(c1; c2; ··· ; cn) where c1; c2; ··· ; cn 2 F . Proof. First assume that F=K is finite. Then F has a finite basis over K, say fc1; c2; ··· ; cng and it follows that F = K(c1; c2; ··· ; cn). Conversely, let F = K(c1; c2; ··· ; cn) for some c1; c2; ··· ; cn 2 F . Since F=K is an algebraic extension, c1; c2; ··· ; cn are algebraic over K. Denote the minimal polynomial of ci over K by mi(x) and di = deg mi(x) for all i = 1; 2; ··· ; n. Then the minimal polynomial of ci over K(c1; c2; ··· ; ci−1) is a factor of mi(x) and it follows that [K(c1; c2; ··· ; ci): K(c1; c2; ··· ; ci−1)] ≤ di. Thus, by the tower rule, [K(c1; c2; ··· ; cn): K] = [K(c1; c2; ··· ; cn): K(c1; c2; ··· ; cn−1)][K(c1; c2; ··· ; cn−1): K(c1; c2; ··· ; cn−2)] ··· [K(c1): K] ≤ d1d2 ··· dn which implies that F=K is finite. From the above proof it follows that K(c1; c2; ··· ; cn) is a finite extension of K if c1; c2; ··· ; cn are algebraic over K. Thus we have the following equivalent characterizations: Corollary 0.6. Let F=K be a field extension and c1; c2; ··· ; cn 2 F . Then the following conditions are equivalent: 1. K(c1; c2; ··· ; cn)=K is finite; 2. K(c1; c2; ··· ; cn)=K is algebraic; 3. c1; c2; ··· ; cn are algebraic over K. Also we have the following useful consequence of the above corollary. The following consequence is useful to characterize simplicity of an algebraic extension. Corollary 0.7. Let F=K be an algebraic extension. If it has only finite number of intermediate fields then F=K is a finite extension. 3 Proof. First we show that there are n 2 N and c1; c2; ··· ; cn 2 F such that F = K(c1; c2; ··· ; cn). If not, then we have an infinite strictly ascending sequence of intermediate fields K ⊆ K(c1) ( K(c1; c2) ( ··· of F=K. This contradicts that the extension F=K has only finite number of intermediate fields. Thus F = K(c1; c2; ··· ; cn) where n 2 N and c1; c2; ··· ; cn 2 F . Since c1; c2; ··· ; cn are all algebraic over K it follows that F=K is finite. Now we show that if F=K is a field extension, then the set of all algebraic elements of F over K forms a field. Theorem 0.8. Let F=K be a field extension. Then the set L of all algebraic elements of F over K is an intermediate field of F=K Proof. Every k 2 K is a root of the polynomial x − k 2 K[x] and so is algebraic over K. Thus K ⊆ L. Let a; b 2 L; b 6= 0. Then K(a; b)=K is an algebraic extension and hence a − b; ab−1 2 L showing that L is a subfield. Recall that the complex numbers which are algebraic elements over the field Q of all rationals, are called algebraic numbers. Denote the set of all algebraic numbers by A. Then from the above result, we have: Corollary 0.9. The set A of all algebraic numbers is a subfield of C. Thus the sum, products, quotients of algebraic numbers are again algebraic numbers. Now we show that algebraic extension of an algebraic extension is algebraic. Theorem 0.10. Let F=K be an extension and L an intermediate field of F=K. Then both F=L and L=K are algebraic if and only if F=K is so. Proof. First assume that both F=L and L=K are algebraic extensions. Consider a 2 F . Then a is n n−1 algebraic over L. Let m(x) = x + l1x + ··· + ln 2 L[x] be the minimal polynomial of a over L. n n−1 Then m(x) = x +l1x +···+ln 2 K(l1; l2; ··· ; ln) implies that a is algebraic over K(l1; l2; ··· ; ln) and so K(l1; l2; ··· ; ln; a)=K is an algebraic extension. Now l1; l2; ··· ; ln are algebraic over K implies that K(l1; l2; ··· ; ln)=K is an algebraic extension. Thus K(l1; l2; ··· ; ln; a)=K is an algebraic extension showing that a is algebraic over K. Hence F=K is an algebraic extension. Conversely, let F=K be an algebraic extension. Consider a 2 L. Since every element of F is algebraic over K, a is so. Thus L=K is an algebraic extension. Now consider a 2 F . Let m(x) 2 K[x] be the minimal polynomial of a over K. Then K ⊆ L shows that m(x) 2 L[x] and hence a is algebraic over L. Thus F=L is an algebraic extension. 4 Recall that an extension F=K is called simple if there is c 2 F such that F = K(c). A simple extension is algebraic if and only if it is finite. Also, if an algebraic extension F=K has only finitely many intermediate fields then it is finite. In the following we characterize simple algebraic extensions. Theorem 0.11. Let F be an algebraic extension of an infinite field K. Then F=K is a simple extension if and only if it has only finite number of intermediate fields. Proof. First assume that F = K(c) for some c 2 F . Let m(x) be the minimal polynomial of c over K. Let L be an intermediate field of the extension F=K. Then c is algebraic over L and the minimal polynomial f(x) 2 L[x] of c over L divides m(x) in L[x] and so in F [x]. Let n n−1 0 0 0 f(x) = x + l1x + ··· + ln and L = K(l1; l2; ··· ; ln). Then f(x) 2 L [x]. Since L ⊆ L, irreducibility of f(x) over L implies that f(x) is irreducible over L0. Hence f(x) is the minimal polynomial of c over L0. This implies that [L0(c): L0] = [L(c): L]. Now F = K(c) implies that F = L(c) = L0(c) and hence [F : L] = [F : L0] = [F : L][L : L0]. Then [L : L0] = 1 forcing L = L0. Thus the intermediate field L is determined uniquely by the coefficients of f(x) which is a monic factor of m(x) in F [x]. Since m(x) has only a finite number of factors in F [x], it follows that F=K has only finite number of intermediate fields.