文档介绍：page 1 of Chapter 3
CHAPTER 3 FIELD FUNDAMENTALS
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 and ). 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, X2 + 1 has no real roots, but if we consider the larger field plex 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.
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.
Lemma Let f : F → E be a homomorphism of fields, ., f(a + b)=f(a)+
f(b),f(ab)=f(a)f(b) (all a, b ∈ F ), and f(1F )=1E. Then f is a monomorphism.
Proof. First note that a field F has no ideals except {0} 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