204 CHAPTER 7. VECTOR SPACES AND FIELDS

Example 7.3.24 The polynomial x2 + 1 is irreducible in R [x] , polynomials having realcoefficients. To see this is the case, suppose ψ (x) divides x2 + 1. Then

x2 + 1 = ψ (x) q (x)

If the degree of ψ (x) is less than 2, then it must be either a constant or of the form ax+ b.In the latter case, −b/a must be a zero of the right side, hence of the left but x2 + 1 has noreal zeros. Therefore, the degree of ψ (x) must be two and q (x) must be a constant. Thusthe only polynomial which divides x2 + 1 are constants and multiples of x2 + 1. Therefore,this shows x2 +1 is irreducible. Find the inverse of

[x2 + x+ 1

]in the space of equivalence

classes, R/(x2 + 1

).

You can solve this with partial fractions.

1

(x2 + 1) (x2 + x+ 1)= − x

x2 + 1+

x+ 1

x2 + x+ 1

and so1 = (−x)

(x2 + x+ 1

)+ (x+ 1)

(x2 + 1

)which implies

1 ∼ (−x)(x2 + x+ 1

)and so the inverse is [−x] .

The following proposition is interesting. It was essentially proved above but to emphasizeit, here it is again.

Proposition 7.3.25 Suppose p (x) ∈ F [x] is irreducible and has degree n. Then everyelement of G = F [x] / (p (x)) is of the form [0] or [r (x)] where the degree of r (x) is lessthan n.

Proof: This follows right away from the Euclidean algorithm for polynomials. If k (x)has degree larger than n− 1, then

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

where r (x) is either equal to 0 or has degree less than n. Hence

[k (x)] = [r (x)] . ■

Example 7.3.26 In the situation of the above example, find [ax+ b]−1

assuming a2+ b2 ̸=0. Note this includes all cases of interest thanks to the above proposition.

You can do it with partial fractions as above.

1

(x2 + 1) (ax+ b)=

b− ax

(a2 + b2) (x2 + 1)+

a2

(a2 + b2) (ax+ b)

and so

1 =1

a2 + b2(b− ax) (ax+ b) +

a2

(a2 + b2)

(x2 + 1

)Thus

1

a2 + b2(b− ax) (ax+ b) ∼ 1

and so

[ax+ b]−1

=[(b− ax)]

a2 + b2=b− a [x]

a2 + b2

You might find it interesting to recall that (ai+ b)−1

= b−aia2+b2 .