This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Chapter 3 Field Fundamentals 3.1 Field Extensions If F is a field and F [ X ] is the set of all polynomials over F , that is, polynomials with coeﬃcients in F , we know that F [ X ] is a Euclidean domain, and therefore a principal ideal domain and a unique factorization domain (see Sections 2.6 and 2.7). Thus any nonzero polynomial f in F [ X ] can be factored uniquely as a product of irreducible polynomials. Any root of f must be a root of one of the irreducible factors, but at this point we have no concrete information about the existence of roots and how they might be found. For example, X 2 +1 has no real roots, but if we consider the larger field of complex numbers, we get two roots, + i and − i . It appears that the process of passing to a larger field may help produce roots, and this turns out to be correct. 3.1.1 Definitions If F and E are fields and F ⊆ E , we say that E is an extension of F , and we write F ≤ E , or sometimes E/F . If E is an extension of F , then in particular E is an abelian group under addition, and we may multiply the “vector” x ∈ E by the “scalar” λ ∈ F , and the axioms of a vector space are satisfied. Thus if F ≤ E , then E is a vector space over F . The dimension of this vector space is called the degree of the extension, written [ E : F ]. If [ E : F ] = n < ∞ , we say that E is a finite extension of F , or that the extension E/F is finite , or that E is of degree n over F . If f is a nonconstant polynomial over the field F , and f has no roots in F , we can always produce a root of f in an extension field of F . We do this after a preliminary result. 3.1.2 Lemma Let f : F → E be a homomorphism of fields, i.e., f ( a + b ) = f ( a )+ f ( b ) ,f ( ab ) = f ( a ) f ( b ) (all a,b ∈ F ), and f (1 F ) = 1 E . Then f is a monomorphism. 1 2 CHAPTER 3. FIELD FUNDAMENTALS Proof. First note that a field F has no ideals except { } and F . For if a is a nonzero member of the ideal I , then ab = 1 for some b ∈ F , hence 1 ∈ I , and therefore I = F . Taking I to be the kernel of f , we see that I cannot be all of F because f (1) 6 = 0. Thus I must be { } , so that f is injective. ♣ 3.1.3 Theorem Let f be a nonconstant polynomial over the field F . Then there is an extension E/F and an element α ∈ E such that f ( α ) = 0. Proof. Since f can be factored into irreducibles, we may assume without loss of generality that f itself is irreducible. The ideal I = h f ( X ) i in F [ X ] is prime (see (2.6.1)), in fact maximal (see (2.6.9)). Thus E = F [ X ] /I is a field by (2.4.3). We have a problem at this point because F need not be a subset of E , but we can place an isomorphic copy of F inside E via the homomorphism h : a → a + I ; by (3.1.2), h is a monomorphism, so we may identify F with a subfield of E . Now let α = X + I ; if f ( X ) = a + a 1 X + ··· + a n X n , then f ( α ) = ( a + I ) + a 1 ( X + I ) + ··· + a n ( X + I ) n = ( a + a 1 X + ··· + a n X n ) + I = f ( X ) + I...
View
Full
Document
This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.
 Spring '11
 sdd
 Polynomials

Click to edit the document details