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Abstract Algebra
Notes Let F be an extension field of K. The dimension of F as a vector space over K is called the degree
of F over K, denoted by [F : K]. If the dimension of F over K is finite, then F is said to be a finite
extension of K. Let F be an extension field of K and let u F. The following conditions are
equivalent: (1) u is algebraic over K; (2) K(u) is a finite extension of K; (3) u belongs to a finite
extension of K.
Never underestimate the power of counting: the next result is crucial. If we have a tower of
extensions K E F, where E is finite over K and F is finite over E, then F is finite over K, and
[F : K] = [F : E][E : K]. This has a useful corollary, which states that the degree of any element of
F is a divisor of [F : K].
Let K be a field and let f(x) = a + a x + ... + a x be a polynomial in K[x] of degree n > 0. An
n
n
1
0
extension field F of K is called a splitting field for f(x) over K if there exist elements r , r ,..., r n
2
1
F such that
(i) f(x) = a (x r )(x r ) ... (x rn) and
1
n
2
(ii) F = K(r , r ..., r ).
1
2,
n
In this situation we usually say that f(x) splits over the field F. The elements r , r ,..., r are roots
2
n
1
of f(x), and so F is obtained by adjoining to K a complete set of roots of f(x). An induction
argument (on the degree of f(x)) can be given to show that splitting fields always exist. Theorem
states that if f(x) K[x] is a polynomial of degree n > 0, then there exists a splitting field F for f(x)
over K, with [F : K] n!.
The uniqueness of splitting fields follows from two lemmas. Let : K L be an isomorphism of
fields. Let F be an extension field of K such that F = K(u) for an algebraic element u F. Let p(x)
be the minimal polynomial of u over K. If v is any root of the image q(x) of p(x) under , and
E = L(v), then there is a unique way to extend to an isomorphism : F E such that (u) = v and
(a) = (a) for all a K. The required isomorphism : K(u) L(v) must have the form
(a + a u + ... + a u ) = (a ) + (a )v + ... + (a )v n-1
n-1
1
0
0
1
n-1
n-1
The second lemma is stated as follows. Let F be a splitting field for the polynomial f(x) K[x]. If
: K L is a field isomorphism that maps f(x) to g(x) L[x] and E is a splitting field for g(x) over
L, then there exists an isomorphism : F E such that (a) = (a) for all a K. The proof uses
induction on the degree of f(x), together with the previous lemma.
Theorem states that the splitting field over the field K of a polynomial f(x) K[x] is unique up
to isomorphism. Among other things, this result has important consequences for finite fields.
Self Assessment
1. If F is an extension field k and u F is algebraic over K, then their exists a ...............
(a) different (b) finite
(c) infinite (d) unique
2. The monic polynomial P(x) of minimal degree in K[x] such that P(u) = 0 is called is
............... of r over K and its degree is called the degree of u over K.
(a) maximal polynomial (b) minimal polynomial
(c) finite polynomial (d) infinite polynomial
3. The dimension of F as a vector space K is called the ............... of F over K, denoted by [f : k]
(a) range (b) domain
(c) degree (d) field
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