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Linear Algebra, Theory And Applications, 2012a

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210 VECTOR SPACES AND FIELDS<br />

Corollary 8.3.11 Let q (λ) = ∏ p<br />

i=1 φ i (λ) ki where the k i are positive integers and the φ i (λ)<br />

are irreducible monic polynomials. Suppose also that p (λ) is a monic polynomial which<br />

divides q (λ) . Then<br />

p∏<br />

p (λ) = φ i (λ) r i<br />

i=1<br />

where r i is a nonnegative integer no larger than k i .<br />

Proof: Using Theorem 8.3.10, let p (λ) =b ∏ s<br />

i=1 ψ i (λ) r i<br />

where the ψ i (λ) areeach<br />

irreducible and monic and b ∈ F. Sincep (λ) is monic, b =1. Then there exists a polynomial<br />

g (λ) such that<br />

s∏<br />

p∏<br />

p (λ) g (λ) =g (λ) ψ i (λ) r i<br />

= φ i (λ) k i<br />

Hence g (λ) must be monic. Therefore,<br />

p (λ) g (λ) =<br />

i=1<br />

{<br />

p(λ)<br />

}} {<br />

s∏<br />

ψ i (λ) ri<br />

i=1<br />

i=1<br />

l∏<br />

η j (λ) =<br />

j=1<br />

p∏<br />

φ i (λ) ki<br />

for η j monic and irreducible. By uniqueness, each ψ i equals one of the φ j (λ) and the same<br />

holding true of the η i (λ). Therefore, p (λ) is of the desired form. <br />

8.3.2 Polynomials <strong>And</strong> Fields<br />

When you have a polynomial like x 2 − 3 which has no rational roots, it turns out you can<br />

enlarge the field of rational numbers to obtain a larger field such that this polynomial does<br />

have roots in this larger field. I am going to discuss a systematic way to do this. It will<br />

turn out that for any polynomial with coefficients in any field, there always exists a possibly<br />

larger field such that the polynomial has roots in this larger field. This book has mainly<br />

featured the field of real or complex numbers but this procedure will show how to obtain<br />

many other fields which could be used in most of what was presented earlier in the book.<br />

Here is an important idea concerning equivalence relations which I hope is familiar.<br />

Definition 8.3.12 Let S be a set. The symbol, ∼ is called an equivalence relation on S if<br />

it satisfies the following axioms.<br />

1. x ∼ x for all x ∈ S. (Reflexive)<br />

2. If x ∼ y then y ∼ x. (Symmetric)<br />

3. If x ∼ y and y ∼ z, thenx ∼ z. (Transitive)<br />

Definition 8.3.13 [x] denotes the set of all elements of S which are equivalent to x and<br />

[x] is called the equivalence class determined by x or just the equivalence class of x.<br />

Also recall the notion of equivalence classes.<br />

Theorem 8.3.14 Let ∼ be an equivalence class defined on a set, S and let H denote the<br />

set of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either<br />

x ∼ y and [x] =[y] or it is not true that x ∼ y and [x] ∩ [y] =∅.<br />

i=1

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