Kenneth Kuttler

3.4. POLYNOMIALS AND FIELDS 67

Then you would need to have p(x)/∑m−1i=0 cixi which is impossible unless each ci = 0 be-

cause p(x) has degree m. ■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.

Definition 3.4.10 A commutative ring is just a field in which the assumption that multi-plicative inverses for nonzero elements may not exist. An ideal I in a commutative ringR is a subset of R closed with respect to addition and additive inverses such that rI ⊆ Imeaning that something in I multiplied by r ∈ R will yield something in I. Then F [x] is acommutative ring and (p(x)) is an example of an ideal. A maximal ideal is an ideal forwhich the only ideal containing it is itself or the entire ring.

Example 3.4.11 The polynomial x2 − 2 is irreducible in Q(x) . This is because if x2 −2 = p(x)q(x) where p(x) ,q(x) both have degree less than 2, then they both have degree1. Hence you would have x2− 2 = (x+a)(x+b) which requires that a+ b = 0 so thisfactorization is of the form (x−a)(x+a) and now you need to have a =

√2 /∈ Q. Now

Q(x)/(x2−2

)is of the form a+b [x] where a,b∈Q and [x]2−2 = 0. Thus one can regard

[x] as√

2. Q(x)/(x2−2

)is of the form a+b

√2.

In the above example,[x2 + x

]is not zero because it is not a multiple of x2− 2. What

is[x2 + x

]−1? You know that the two polynomials are relatively prime and so there existsn(x) ,m(x) such that

1 = n(x)(x2−2

)+m(x)

(x2 + x

)Thus [m(x)] =

[x2 + x

]−1. How could you find these polynomials? First of all, it sufficesto consider only n(x) and m(x) having degree less than 2. Otherwise, reiterating the above,m(x) = p(x) l (x)+ r (x) where r (x) has degree smaller than the degree of p(x) and youcould simply use r (x) in place of m(x).

1 = (ax+b)(x2−2

)+(cx+d)

(x2 + x

)1 = ax3−2b+bx2 + cx2 + cx3 +dx2−2ax+dx

Now you solve the resulting system of equations.

a =12,b =−1

2,c =−1

2,d = 1

Then the desired inverse is[− 1

2 x+1]. To check,(

−12

x+1)(

x2 + x)−1 =−1

2(x−1)

(x2−2

)Thus

[− 1

2 x+1][

x2 + x]− [1] = [0].

The above is an example of something general described in the following definition.

Definition 3.4.12 Let F ⊆ K be two fields. Then clearly K is also a vector space over F.Then also, K is called a finite field extension of F if the dimension of this vector space,denoted by [K : F ] is finite.

3.4. POLYNOMIALS AND FIELDS 67Then you would need to have p(x) /"5' c:x’ which is impossible unless each c; = 0 be-cause p(x) has degree m.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.Definition 3.4.10 A commutative ring is just a field in which the assumption that multi-plicative inverses for nonzero elements may not exist. An ideal I in a commutative ringR is a subset of R closed with respect to addition and additive inverses such that rI CImeaning that something in I multiplied by r € R will yield something in I. Then F |x] is acommutative ring and (p(x)) is an example of an ideal. A maximal ideal is an ideal forwhich the only ideal containing it is itself or the entire ring.Example 3.4.11 The polynomial x* — 2 is irreducible in Q(x). This is because if x* —2 = p(x)q(x) where p(x) ,q(x) both have degree less than 2, then they both have degree1. Hence you would have x* —2 = (x+a)(x+b) which requires that a+b =0 so thisfactorization is of the form (x—a)(x-+a) and now you need to have a= V2 ¢ Q. NowQ(x) / (x? —2) is of the form a+b |x] where a,b € Q and [x]’ —2 =0. Thus one can regard[x] as V2. Q(x) / (x* —2) is of the form a+ by2.In the above example, [x? +x] is not zero because it is not a multiple of x2 —2. Whatis [x? +x] ~'? You know that the two polynomials are relatively prime and so there existsn(x) ,m(x) such thatl=n(x) (x —2)+m/(x) (x +x)Thus [m (x)] = [x? +.] ~' How could you find these polynomials? First of all, it sufficesto consider only n(x) and m (x) having degree less than 2. Otherwise, reiterating the above,m(x) = p(x)l(x)+r(x) where r(x) has degree smaller than the degree of p(x) and youcould simply use r(x) in place of m (x).1 = (ax+b) (x? —2)+(cx+d) (x? +x)1 = ax —2b+bx* +x? +0x° +.dx* —2ax+dxNow you solve the resulting system of equations.1 1 1b=—=,c=-~,d=1a= 2 2’2Then the desired inverse is [- 5x+ 1] . To check,(-3x+1) (x? +x) -l= —F (1) (2-2)Thus [—5x+1] [x? +x] — [1] = [0].The above is an example of something general described in the following definition.Definition 3.4.12 Let F C K be two fields. Then clearly K is also a vector space over F.Then also, K is called a finite field extension of F if the dimension of this vector space,denoted by [K : F] is finite.