Kenneth Kuttler

390 CHAPTER 16. VECTOR SPACES

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−1i=0 cixi which is impossible unless each ci = 0 be-

cause p(x) has degree m. ■The last assertion in the proof follows from the definition of addition and multiplication

in F [x]/(p(x)) and math induction. If each ai ∈ F,[anxn +an−1xn−1 + · · ·+a1x+a0

]= [an] [x]

n +[an−1] [x]n−1 + · · · [a1] [x]+ [a0] (16.7)

Remark 16.4.19 The polynomials consisting of all polynomial multiples of p(x) , denotedby (p(x)) is called an ideal. An ideal I is a subset of the commutative ring (Here the ringis F [x] .) with unity consisting of all polynomials which is itself a ring and which has theproperty that whenever f (x) ∈ F [x] , and g(x) ∈ I, f (x)g(x) ∈ I. In this case, you couldargue that (p(x)) is an ideal and that the only ideal containing it is itself or the entire ringF [x]. This is called a maximal ideal.

Example 16.4.20 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.

Thus the above is an illustration of something general described in the following defi-nition.

Definition 16.4.21 Let F ⊆ K be two fields. Then clearly K is also a vector space overF. Then also, K is called a finite field extension of F if the dimension of this vector space,denoted by [K : F ] is finite.

There are some easy things to observe about this.

390 CHAPTER 16. VECTOR SPACESThus [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 inF [x] /(p (x)). Then from the division algorithm,F(x) = p(x) a(x) +r)where r(x) is either 0 or has degree less than the degree of p(x). Thusbut clearly [r (x)] € span (1 wo s["") . Thusspan ((1],---[x]""') = Fla] /(p(x)).Then {1 yo yr} is a basis if these vectors are linearly independent. Suppose thenthatm—1 . m1 :Y? cilx]’ =| ¥ cix'| =0i=0 i=0Then you would need to have p(x) /2"5' c:x' which is impossible unless each c; = 0 be-cause p(x) has degree m.The last assertion in the proof follows from the definition of addition and multiplicationin F [x] / (p(x)) and math induction. If each a; € F,1[anx" +dy—1x" | +++» +.ayx+ao]} = [an] [x] + [ani] x)" | +++ [ai] x] + [ao] (16.7)Remark 16.4.19 The polynomials consisting of all polynomial multiples of p(x) , denotedby (p(x)) is called an ideal. An ideal I is a subset of the commutative ring (Here the ringis F [x] .) with unity consisting of all polynomials which is itself a ring and which has theproperty that whenever f (x) € F [x], and g(x) €1, f (x)g(x) EL. In this case, you couldargue that (p(x)) is an ideal and that the only ideal containing it is itself or the entire ringF |x]. This is called a maximal ideal.Example 16.4.20 The polynomial x? —2 is irreducible in Q|x]. This is because if xr2 = 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. NowQa] / (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[a] / (x —2) is of the form a+ by2.Thus the above is an illustration of something general described in the following defi-nition.Definition 16.4.21 Let F C K be two fields. Then clearly K is also a vector space overF. Then also, K is called a finite field extension of F if the dimension of this vector space,denoted by [K : F] is finite.There are some easy things to observe about this.