College Algebra Exam Review 312

College Algebra Exam Review 312 - 322 7 FIELD EXTENSIONS...

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Unformatted text preview: 322 7. FIELD EXTENSIONS – FIRST LOOK 7.3. Adjoining Algebraic Elements to a Field The fields in this section are general, not necessarily subfields of the complex numbers. A field extension L of a field K is, in particular, a vector space over K . You are asked to check this in the Exercises. It will be helpful to review the definition of a vector space in Section 3.3 Since a field extension L of a field K is a K -vector space, in particular, L has a dimension over K , possibly infinite. The dimension of L as a K vector space is denoted dimK .L/ (or, sometimes, ŒL W K.) Dimensions of field extensions have the following multiplicative property: Proposition 7.3.1. If K  L  M are fields, then dimK .M / D dimK .L/ dimL .M /: Proof. Suppose that f 1 ; : : : ; r g is a subset of L that is linearly independent over K , and that f 1 ; : : : ; s g is a subset of M that is linearly independent over L. I claim that f i PW 1 Ä i Ä r; 1 ÄP Ä s g is linearly jP j independent over K . In fact, if 0 D i;j kij i j D j . i kij i / j , with kij 2 K , then linear independence of f j g over L implies that P i kij i D 0 for all j , and then linear independence of f i g over K implies that kij D 0 for all i; j , which proves the claim. In particular, if either dimK .L/ or dimL .M / is infinite, then there are arbitrarily large subsets of M that are linearly independent over K , so dimK .M / is also infinite. Suppose now that dimK .L/ and dimL .M / are finite, that the set f 1 ; : : : ; r g is a basis of L over K , and that the set f 1 ; : : : ; s g is a basis of M over L. The fact that f j g spans M over L and that f i g spans L over K implies that the set of products f i j g spans M over K (exercise). Hence, f i j g is a basis of M over K . I Definition 7.3.2. A field extension K  L is called finite if L is a finite– dimensional vector space over K . Now, consider a field extension K  L. According to Proposition 6.2.7, for any element ˛ 2 L there is a ring homomorphism (“evaluation” at ˛ ) from KŒx into L given by '˛ .f .x// D f .˛/. That is, '˛ .k0 C k1 x C C kn x n / D k0 C k1 ˛ C C kn ˛ n : (7.3.1) ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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