Linear Algebra, Theory And Applications, 2012a

8.3. LOTS OF FIELDS 215
Proposition 8.3.25 Suppose p (x) ∈ F [x] is irreducible and has degree n. Then every
element of G = F [x] / (p (x)) is of the form [0] or [r (x)] where the degree of r (x) is less
than n.
Proof: This follows right away from the Euclidean algorithm for polynomials. If k (x)
has degree larger than n − 1, then
k (x) =q (x) p (x)+r (x)
where r (x) is either equal to 0 or has degree less than n. Hence
[k (x)] = [r (x)] .
Example 8.3.26 In the situation of the above example, find [ax + b] −1 assuming a 2 + b 2 ≠
0. Note this includes all cases of interest thanks to the above proposition.
and so
Thus
and so
You can do it with partial fractions as above.
1
(x 2 +1)(ax + b) = b − ax
(a 2 + b 2 )(x 2 +1) + a 2
(a 2 + b 2 )(ax + b)
1=
1
a 2 + b 2 (b − ax)(ax + b)+ a 2 (
x 2
(a 2 + b 2 +1 )
)
1
a 2 (b − ax)(ax + b) ∼ 1
+ b2 [ax + b] −1 =
[(b − ax)]
a 2 + b 2
= b − a [x]
a 2 + b 2
You might find it interesting to recall that (ai + b) −1 = b−ai
a 2 +b 2 .
8.3.3 The Algebraic Numbers
Each polynomial having coefficients in a field F has a splitting field. Consider the case of all
polynomials p (x) having coefficients in a field F ⊆ G and consider all roots which are also
in G. The theory of vector spaces is very useful in the study of these algebraic numbers.
Here is a definition.
Definition 8.3.27 The algebraic numbers A are those numbers which are in G and also
roots of some polynomial p (x) having coefficients in F.
Theorem 8.3.28 Let a ∈ A. Then there exists a unique monic irreducible polynomial p (x)
having coefficients in F such that p (a) =0. This is called the minimal polynomial for a.
Proof: By definition, there exists a polynomial q (x) having coefficients in F such that
q (a) =0. If q (x) is irreducible, divide by the leading coefficient and this proves the existence.
If q (x) is not irreducible, then there exist nonconstant polynomials r (x) andk (x) such that
q (x) =r (x) k (x). Then one of r (a) ,k(a) equals 0. Pick the one which equals zero and let it
playtheroleofq (x). Continuing this way, in finitely many steps one obtains an irreducible
polynomial p (x) such that p (a) = 0. Now divide by the leading coefficient and this proves
existence. Suppose p i ,i =1, 2 both work and they are not equal. Then by Lemma 8.3.7