Page 254 - DMTH403_ABSTRACT_ALGEBRA

Unit 26: Splitting Fields, Existence and Uniqueness

4. The splitting field over the field K of a polynomial f(x)  K[x] is unique up to ............... Notes
(a) homomorphism (b) isomorphism

(c) automorphism (d) finite extension
26.2 Summary

If F is an extension field of K, and u  F is algebraic over K, then there exists a unique monic
irreducible polynomial p(x)  K[x] such that p(u) = 0. It is the monic polynomial of minimal
degree that has u as a root, and if f(x) is any polynomial in K[x] with f(u) = 0, then
p(x) | f(x).

Alternate Proof: The proof in the text uses some elementary ring theory. I’ve decided to include
a proof that depends only on basic facts about polynomials.
Let F be an extension field of K, and let u , u , ..., u  F. The smallest subfield of F that contains
2
n
1
K and u , u ,..., u will be denoted by K(u , u ,..., u ). It is called the extension field of K generated
1
2
n
2
n
1
by u , u ,...., u . If F = K(u) for a single element u  F, then F is said to be a simple extension of K.
n
1
2
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
2
a + a u + a u + ... + a u m ,
m
2
1
0
b + b u + b u + ... + b u n
2
1
0
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
j
i
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.
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
0 1 n-1 0 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.
The splitting field over the field K of a polynomial f(x)  K[x] is unique up to isomorphism.
26.3 Keywords
Splitting Field: Beginning with a field K, and a polynomial f(x)  K, we need to construct the
smallest possible extension field F of K that contains all of the roots of f(x). This will be called a
splitting field for f(x) over K.

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