Kenneth Kuttler

7.3. LOTS OF FIELDS 201

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]

Thus [x] is a root of this polynomial in the field F [x] / (p (x)).Consider the last claim. Let f (x) ∈ F [x] / (p (x)) . Thus [f (x)] is a typical thing in

F [x] / (p (x)). Then from the division algorithm,

f (x) = p (x) q (x) + r (x)

where r (x) is either 0 or has degree less than the degree of p (x) . Thus

[r (x)] = [f (x)− p (x) q (x)] = [f (x)]

but clearly [r (x)] ∈ span([1] , · · · , [x]m−1

). Thus span

([1] , · · · , [x]m−1

)= F [x] / (p (x)).

Then{[1] , · · · , [x]m−1

}is a basis if these vectors are linearly independent. Suppose then

thatm−1∑i=0

ci [x]i=

[m−1∑i=0

cixi

]= 0

Then you would need to have p (x) /∑m−1

i=0 cixi which is impossible unless each ci = 0

because p (x) has degree m. ■From the above theorem, it makes perfect sense to write b rather than [b] if b ∈ 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 fieldin which the given polynomial can be factored into a product of polynomials having degreeone. The field you obtain in this process of enlarging in which the given polynomial factorsin terms of linear factors is called a splitting field.

7.3. LOTS OF FIELDS 201Consider the second of the two.— a(x) b(a)= a'(x)b' (x) — a(x) b' (x) +a (x) b' (x) — a(x) (2)= U(x) (a' (x) —a(2)) +a(2) (b (x) — b(2))Now by assumption (a’ (a) — a(«)) is a multiple of p (a) as is (b’ (2) — b(a)), 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(a) for some polynomial & (a). 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 [2] /(p(x)) where[q(x)] 4 [0]. Then q(x) is not a multiple of p(a) and so by the first part, q(x), p(x) arerelatively prime. Thus there exist n (a) ,m (a) such thatL=n(x)q(@) +m (2) p(2)Hence(l] = [L—n(@) p(@)] = [rn @) a (@)] = [rn @)] le)which shows that [q(x)] + = [n()]. Thus this is a field. The polynomial has a root in thisfield because ifp(2) =a" +am—12" | +++» + a2 +409,(0) = Co) = fa)” + [ana] fo] + +--+ faa fe] + [a]Thus [2] is a root of this polynomial in the field F [a] / (p (z)).Consider the last claim. Let f(a) € F[a]/(p(«)). Thus [f (a)] is a typical thing inF [x] / (p(a)). Then from the division algorithm,f(@) =p(2)q(@) +r)where r (x) is either 0 or has degree less than the degree of p(x). Thusbut clearly [r (x)] € span (1 yttty (2]”""*) . Thus span ((u yon .[x]"~*) = F(z] /(p(a)).Then {{1] titty ia} is a basis if these vectors are linearly independent. Suppose thenthatm-1 m-1Yee =|" ee! <01=0 1=0m-1Then you would need to have p(x) / 37/4 cix* which is impossible unless each c; = 0because p (a) has degree m. HlFrom the above theorem, it makes perfect sense to write b rather than [b] if b ¢ F. Thenwith this convention,[bd (x)] = [8] |b (x)] = b[¢ (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 fieldin which the given polynomial can be factored into a product of polynomials having degreeone. The field you obtain in this process of enlarging in which the given polynomial factorsin terms of linear factors is called a splitting field.