Linear Algebra, Theory And Applications, 2012a

8.3. LOTS OF FIELDS 213
Then, since the l i are independent, it follows that
m∑
f ji k j =0
j=1
and since {k j } is independent, each f ji = 0 for each j for a given arbitrary i. Therefore,
{k j l i } is a basis.
Theorem 8.3.22 The set of all equivalence classes G ≡ F/ (p (x)) described above with
the multiplicative identity given by [1] and the additive identity given by [0] along with the
operations of Definition 8.3.17, is a field and p ([x]) = [0] . (Thus p has a root in this new
field.) In addition to this, [G : F] =n, thedegreeofp (x) .
Proof: Everything is obvious except for the existence of the multiplicative inverse and
the assertion that p ([x]) = 0. Suppose then that [a (x)] ≠[0]. That is, a (x) is not a multiple
of p (x). Why does [a (x)] −1 exist? By Theorem 8.3.6, a (x) ,p(x) are relatively prime and
so there exist polynomials ψ (x) ,φ(x) such that
1=ψ (x) p (x)+a (x) φ (x)
and so
which, by definition implies
1 − a (x) φ (x) =ψ (x) p (x)
[1 − a (x) φ (x)] = [1] − [a (x) φ (x)] = [1] − [a (x)] [φ (x)] = [0]
and so [φ (x)] = [a (x)] −1 . This shows G is a field.
Now if p (x) =a n x n + a n−1 x n−1 + ···+ a 1 x + a 0 ,p([x]) = 0 by (8.7) and the definition
which says [p (x)] = [0].
Consider the claim about the dimension. It was just shown that [1] , [x] , [ x 2] , ··· , [x n ]
is linearly dependent. Also [1] , [x] , [ x 2] , ··· , [ x n−1] is independent because if not, there
would exist a polynomial q (x) ofdegreen−1 which is a multiple of p (x) which is impossible.
Now for [q (x)] ∈ G, youcanwrite
q (x) =p (x) l (x)+r (x)
where the degree of r (x) is less than n or else it equals 0. Either way, [q (x)] = [r (x)] which
is a linear combination of [1] , [x] , [ x 2] , ··· , [ x n−1] . Thus [G : F] =n as claimed.
Note that if p (x) were not irreducible, then you could find a field extension G such that
[G : F] ≤ n. You could do this by working with an irreducible factor of p (x).
Usually, people simply write b rather than [b] ifb ∈ F. Then with this convention,
[bφ (x)] = [b][φ (x)] = b [φ (x)] .
This shows how to enlarge a field to get a new one in which the polynomial has a root.
By using a succession of such enlargements, called field extensions, there will exist a field
in which the given polynomial can be factored into a product of polynomials having degree
one. The field you obtain in this process of enlarging in which the given polynomial factors
in terms of linear factors is called a splitting field.