Kenneth Kuttler

16.4. VECTOR SPACES AND FIELDS∗ 389

1. p(x) and q(x) are relatively prime for any q(x) ∈ F [x] which is not a multiple ofp(x).

2. The definitions of addition and multiplication are well defined.

3. If a,b ∈ F and [a] = [b] , then a = b. Thus F is a subset of F [x]/(p(x)) .

4. F [x]/(p(x)) is a field in which the polynomial p(x) has a root.

5. F [x]/(p(x)) is a vector space with field of scalars F and its dimension is m where mis the degree of the irreducible polynomial p(x).

Proof: First consider the claim about p(x) ,q(x) being relatively prime. If ψ (x) is thegreatest common divisor, it follows ψ (x) is either equal to p(x) or 1. If it is p(x) , thenq(x) is a multiple of p(x) which does not happen. If it is 1, then by definition, the twopolynomials are relatively prime.

To show the operations are well defined, suppose

[a(x)] =[a′ (x)

], [b(x)] =

[b′ (x)

]It is necessary to show

[a(x)+b(x)] =[a′ (x)+b′ (x)

][a(x)b(x)] =

[a′ (x)b′ (x)

]Consider the second of the two.

a′ (x)b′ (x)−a(x)b(x)

= a′ (x)b′ (x)−a(x)b′ (x)+a(x)b′ (x)−a(x)b(x)

= b′ (x)(a′ (x)−a(x)

)+a(x)

(b′ (x)−b(x)

)Now by assumption (a′ (x)−a(x)) is a multiple of p(x) as is (b′ (x)−b(x)) , so the aboveis a multiple of p(x) and by definition this shows [a(x)b(x)] = [a′ (x)b′ (x)]. The case foraddition is similar.

Now suppose [a] = [b] . This means a−b = k (x) p(x) for some polynomial k (x) . Thenk (x) must equal 0 since otherwise the two polynomials a− b and k (x) p(x) could not beequal because they would have different degree.

It is clear that the axioms of a field are satisfied except for the one which says that nonzero elements of the field have a multiplicative inverse. Let [q(x)] ∈ F [x]/(p(x)) where[q(x)] ̸= [0] . Then q(x) is not a multiple of p(x) and so by the first part, q(x) , p(x) arerelatively prime. Thus there exist n(x) ,m(x) such that

1 = n(x)q(x)+m(x) p(x)

Hence[1] = [1−n(x) p(x)] = [n(x)q(x)] = [n(x)] [q(x)]

which shows that [q(x)]−1 = [n(x)] . Thus this is a field. The polynomial has a root in thisfield because if

p(x) = xm +am−1xm−1 + · · ·+a1x+a0,

[0] = [p(x)] = [x]m +[am−1] [x]m−1 + · · ·+[a1] [x]+ [a0]

16.4. VECTOR SPACES AND FIELDS* 389I. p(x) and q(x) are relatively prime for any q(x) € F [x] which is not a multiple ofP().The definitions of addition and multiplication are well defined.If a,b € F and {al = [b|, thena = b. Thus F is a subset of F |x] / (p(x)).F [x] / (p (x) is a field in which the polynomial p (x) has a root.Mw KR WNF [x] / (p (x)) is a vector space with field of scalars F and its dimension is m where mis the degree of the irreducible polynomial p (x).Proof: First consider the claim about p (x) ,q(x) being relatively prime. If y (x) is thegreatest common divisor, it follows y(x) is either equal to p(x) or 1. If it is p(x), thenq(x) is a multiple of p(x) which does not happen. If it is 1, then by definition, the twopolynomials are relatively prime.To show the operations are well defined, suppose[a(x)] = [a’(x)] [2 @)] = [0 )]It is necessary to showa (x) b! (x) — a(x) b(x)d' (x) b! (x) —a(x)b! (x) +.a(x)b' (x) — a(x) b(x)b' (x) (a’ (x) —a(x)) +(x) (6 (x) — B(x)Now by assumption (a’ (x) —a(x)) is a multiple of p(x) as is (b’ (x) — b(x)) , so the aboveis a multiple of p(x) and by definition this shows [a (x) b (x)] = [a’ (x) b’ (x)]. The case foraddition is similar.Now suppose [a] = [b]. This means a— b = k(x) p(x) for some polynomial k (x). Thenk(x) must equal 0 since otherwise the two polynomials a —b and k(x) p(x) could not beequal because they would have different degree.It is clear that the axioms of a field are satisfied except for the one which says that nonzero elements of the field have a multiplicative inverse. Let [g(x)] € F [x] /(p(x)) where[q(x)] 4 [0]. Then q(x) is not a multiple of p(x) and so by the first part, g(x), p(x) arerelatively prime. Thus there exist 1 (x) ,m(x) such that1 = n(x) q(x) +m(x) p(x)Hence[1] = [1 — n(x) p(~)] = [n (x) 4g (*)] = [n(®)] [2 @))which shows that [q(x)]~! = [n(x)]. Thus this is a field. The polynomial has a root in thisfield because ifp(x) =x" am x" | +++» Fayx+ao,[0] = [p ()] = ba" + [m1] fa"! +--+ [aa] fx] + [ao]