202 CHAPTER 7. VECTOR SPACES AND FIELDS

Remark 7.3.19 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 7.3.20 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 degree 1. Henceyou would have x2 − 2 = (x+ a) (x+ b) which requires that a + b = 0 so this factorizationis 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 exists

n (x) ,m (x) such that1 = n (x)

(x2 − 2

)+m (x)

(x2 + x

)Thus [m (x)] =

[x2 + x

]−1. How could you find these polynomials? First of all, it suffices

to consider only n (x) and m (x) having degree less than 2.

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 =1

2, b = −1

2, c = −1

2, d = 1

Then the desired inverse is[− 1

2x+ 1]. To check,(

−1

2x+ 1

)(x2 + x

)− 1 = −1

2(x− 1)

(x2 − 2

)Thus

[− 1

2x+ 1] [x2 + x

]− [1] = [0].

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

Definition 7.3.21 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.

Proposition 7.3.22 Let F ⊆ K ⊆ L be fields. Then [L : F ] = [L : K] [K : F ].

Proof: Let {li}ni=1 be a basis for L over K and let {kj}mj=1 be a basis of K over F . Thenif l ∈ L, there exist unique scalars xi in K such that

l =

n∑i=1

xili