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College Algebra Exam Review 316

College Algebra Exam Review 316 - Furthermore if L D K.a...

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326 7. FIELD EXTENSIONS – FIRST LOOK The next proposition is a converse to Proposition 7.3.4 (b). Proposition 7.3.9. Let L D K.a 1 ;:::;a n / , where the a i are algebraic over K . (a) Then L is a finite extension of K , and, therefore, algebraic. (b) L D KŒa 1 ;:::;a n Ł , the set of polynomials in the a i with coeffi- cients in K . Proof. Let K 0 D K , and K i D K.a 1 ;:::;a i / , for 1 ± i ± n . Consider the tower of extensions: K ² K 1 ² ³³³ ² K n ± 1 ² L: We have K i C 1 D K i .a i C 1 / , where a i C 1 is algebraic over K , hence al- gebraic over K i . By Proposition 7.3.6 , dim K i .K i C 1 / is the degree of the minimal polynomial for a i C 1 over K i , and moreover K i C 1 D K i Œa i C 1 Ł . It follows by induction that K i D KŒa 1 ;:::;a i Ł for all i , and, in par- ticular, L D KŒa 1 ;:::;a n Ł . Moreover, dim K .L/ D dim K .K 1 / dim K 1 .K 2 / ³³³ dim K n ± 1 .L/ < 1 : By Proposition 7.3.4 , L is algebraic over K . n Combining Propositions 7.3.4 and 7.3.9 , we have the following propo- sition. Proposition 7.3.10. A field extension K ² L is finite if, and only if, it is generated by finitely many algebraic elements.
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Unformatted text preview: Furthermore, if L D K.a 1 ;:::;a n / , where the a i are algebraic over K , then L consists of polynomials in the a i with coefficients in K . Corollary 7.3.11. Let K ² L be a field extension. The set of elements of L that are algebraic over K form a subfield of L . In particular, the set of algebraic numbers (complex numbers that are algebraic over Q ) is a countable field. Proof. Let A denote the set of elements of L that are algebraic over K . It suffices to show that A is closed under the field operations; that is, for all a;b 2 A , the elements a C b , ab , ´ a , and b ± 1 (when b ¤ ) also are elements of A . For this, it certainly suffices that K.a;b/ ² A . But this follows from Proposition 7.3.9 ....
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