Page 252 - DMTH403_ABSTRACT_ALGEBRA

Unit 26: Splitting Fields, Existence and Uniqueness

In the above proof, the monic polynomial p(x) of minimal degree in K[x] such that p(u) = 0 is Notes
called the minimal polynomial of u over K, and its degree is called the degree of u over K.
Let F be an extension field of K, and let u , u , ..., u  F. The smallest subfield of F that contains
2
1
n
K and u , u ,..., u will be denoted by K(u , u ,..., u ). It is called the extension field of K generated
1
2
2
n
1
n
by u1, u ,...., u . If F = K(u) for a single element u  F, then F is said to be a simple extension of K.
2
n
Let F be an extension field of K, and let u  F. Since K(u) is a field, it must contain all elements of
the form
a + a u + a u + ... + a u m ,
2
m
2
0
1
b + b u + b u + ... + b u n
2
0
1
2
n
where a , b  K for i = 1,..., m and j = 1,... n. In fact, this set describes K(u), and if u is transcendental
i
j
over K, this description cannot be simplified. On the other hand, if u is algebraic over K, then the
denominator in the above expression is unnecessary, and the degree of the numerator can be
assumed to be less than the degree of the minimal polynomial of u over K.
If F is an extension field of K, then the multiplication of F defines a scalar multiplication,
considering the elements of K as scalars and the elements of F as vectors. This gives F the
structure of a vector space over K, and allows us to make use of the concept of the dimension of
a vector space. The next result describes the structure of the extension field obtained by adjoining
an algebraic element.
Definition: Let F be an extension field of K and let u  F be an element algebraic over K.
(a) K(u)  K[x]= hp(x)i, where p(x) is the minimal polynomial of u over K.
(b) If the minimal polynomial of u over K has degree n, then K(u) is an n-dimensional vector
space over K.
Alternate proof: The standard proof uses the ring homomorphism  : K[x]  F defined by
evaluation at u. Then the image of  is K(u), and the kernel is the ideal of K[x] generated by the
minimal polynomial p(x) of u over K. Since p(x) is irreducible, ker() is a prime ideal, and so
K[x] = ker() is a field because every nonzero prime ideal of a principal ideal domain is maximal.
Thus K(u) is a field since K(u) 245= K[x]= ker().
The usual proof involves some ring theory, but the actual ideas of the proof are much simpler.
To give an elementary proof, define  : K[x]= {p(x)}  K(u) by ([f(x)]) = f(u), for all congruence
classes [f(x)] of polynomials (modulo p(x)). This mapping makes sense because K(u) contains u,
together with all of the elements of K, and so it must contain any expression of the form a +a u+
1
0
... + a u , where a  K, for each subscript i. The function  is well-defined, since it is also
m
m
i
independent of the choice of a representative of [f(x)]. In fact, if g(x)  K[x] and f(x) is equivalent
to g(x), then f(x) – g(x) = q(x)p(x) for some q(x)  K[x], and so f(u) – g(u) = q(u)p(u) = 0, showing
that ([f(x)]) = ([g(x)]).
Since the function  simply substitutes u into the polynomial f(x), and it is not difficult to show
that it preserves addition and multiplication. It follows from the definition of p(x) that 
is one-to-one. Suppose that f(x) represents a nonzero congruence class in K[x]= {p(x)}. Then
p(x) | f(x), and so f(x) is relatively prime to p(x) since it is irreducible. Therefore, there exist
polynomials a(x) and b(x) in K[x] such that a(x)f(x) + b(x)p(x) = 1. It follows that [a(x)][f(x)] = [1]
for the corresponding equivalence classes, and this shows that K[x] /{p(x)} is a field. Thus the
image E of  in F must be subfield of F. On the one hand, E contains u and K, and on the other
hand, we have already shown that E must contain any expression of the form a + a u + ... + a u ,
m
1
0
m
where a  K. It follows that E = K(u), and we have the desired isomorphism.
i
(b) It follows from the description of K(u) in part (a) that if p(x) has degree n, then the set
B = {1, u, u ,..., u } is a basis for K(u) over K.
2
n-1
LOVELY PROFESSIONAL UNIVERSITY 245