1.15. FINITE FIELDS 27

and so, the left side is 0 since otherwise, you could multiply by the product of the pi (λ )k

and get equality of two polynomials of different degree. Hence you have

M

∑i=1

mi

∑k=1

nki (λ )

pi (λ )k =

M

∑i=1

mi

∑k=1

n̂ki (λ )

pi (λ )k (∗)

Now multiply both sides by ∏i̸= j pi (λ )mi p j (λ )

m j−1. Then

P(λ )+nm j j (λ )

p j (λ )= P̂(λ )+

n̂m j j (λ )

p j (λ )

then by the same argument, P(λ ) = P̂(λ ) and nownm j j(λ )

p j(λ )=

n̂m j j(λ )

p j(λ )and so

p j (λ )(nm j j (λ )− n̂m j j (λ )

)= 0

and from Lemma 1.13.2, since p j (λ ) ̸= 0, it follows that nm j j (λ )− n̂m j j (λ ) = 0. Thus in∗ the m j can be replaced with m j−1.

Continue this way, to show that nk j (λ )− n̂k j (λ ) = 0 for each k. Since j is arbitrary,this shows uniqueness. ■

1.15 Finite FieldsThe emphasis of the first part of this book will be on what can be done on the basis of alge-bra alone. Linear algebra only needs a field of scalars along with some axioms involving anAbelian group of vectors and there are infinitely many examples of fields, including somewhich are finite. Since it is good to have examples in mind, I will present the finite fieldsof residue classes modulo a prime number in this little section. Then, when linear algebrais developed in the first part of the book and reference is made to a field of scalars, youshould think that it is possible that the field might be this field of residue classes.

Here is the construction of the finite fields Zp for p a prime.

Definition 1.15.1 LetZ+ denote the set of nonnegative integers, Z+ = {0,1,2,3, · · ·}. Alsolet p be a prime number. We will say that two integers, a,b are equivalent and write a∼ bif a−b is divisible by p. Thus they are equivalent if a−b = px for some integer x.

Proposition 1.15.2 The relation ∼ is an equivalence relation. Denoting by n̄ the equiva-lence class determined by n ∈ N, the following are well defined operations.

n̄+ m̄≡ n+m

n̄m̄≡ nm

which makes the set Zp consisting of{

0̄, 1̄, · · · , p−1}

into a field.

Proof: First note that for n ∈ Z+ there always exists r ∈ {0,1, · · · , p−1} such thatn̄ = r̄. This is clearly true because if n ∈ Z+, then n = mp + r for r < p, this by theEuclidean algorithm. Thus r̄ = n̄. Now suppose that n̄1 = n̄ and m̄1 = m̄. Is it true thatn1 +m1 = n+m? Is it true that (n+m)− (n1 +m1) is a multiple of p? Of course since