9.4. MORE ON ALGEBRAIC FIELD EXTENSIONS 213

Extending f to g

F f→≃

F̄

p(x) ∈ F [x] f→ p̄(x) ∈ F̄ [x]p(x) = ∑

nk=0 akxk → ∑

nk=0 f (ak)xk = p̄(x)

p(r) = 0 p̄(r̄) = 0F(r) g→

≃F̄(r̄)

rg→ r̄

One such g for each r̄

Definition 9.4.3 Let f : F→ F̄ be an isomorphism. For the sake of convenience, if q(x) is apolynomial bmxm+bn−1xm−1+ · · ·+b1x+b0 in F [x] , then q̄(x) will denote the polynomial

f (bm)xm + f (bn−1)xm−1 + · · ·+ f (b1)x+ f (b0)≡ f (q(x))

Then f defined in this way on F [x] is a homomorphism.

Recall that if p(x) is a monic polynomial of degree at least 1, then F [x]/(p(x)) is acommutative ring. This is from Lemma 3.4.8. I will assume p(x) has degree at least 1because otherwise there isn’t anything new being shown in what follows. Here is a simplelemma.

Lemma 9.4.4 Let F be a field and let G be a commutative ring. Then the multiplicativeidentity and additive identities are unique in both and if there is an isomorphism h : F→G,then G is also a field.

Proof: If 1, 1̄ are multiplicative identities in a commutative ring, then 1 = 11̄ = 1̄ so it isunique. Also the additive identity is unique. This is because if you have 0,0′ both additiveidentities, then 0 = 0+0′ = 0′. h(0)+h(x) = h(0+ x) = h(x) and so h(0) is the additiveidentity in G since a typical thing in G is h(x). Also h(1)h(x) = h(1x) = h(x) so h(1) = 1in G by what was just shown. Now suppose h(x) ̸= 0. Then x ̸= 0 because h(0) = 0 in G.Hence h

(x−1)

h(x) = h(x−1x

)= h(1), the multiplicative identity in G. Hence if something

in G is not 0, then it has a multiplicative inverse and so G is a field. ■If K is a finite field extension of F, this means that [K : F]< ∞. Recall [K : F] is the di-

mension of the vector spaceK having field of scalars F. Then if a basis forK is {k1, ...,kn} ,each of these ki is algebraic and soK= F(k1, · · · ,kn) . The next theorem considers the casewhere you have a simple extension so the basis is of the form

{1,r, ...,rn−1

}.

Theorem 9.4.5 Let f : F→ F̄ be an isomorphism of the two fields. Let r be algebraic overF with minimum polynomial p(x)≡ xn+an−1xn−1+ · · ·+a1x+a0 and suppose there existsr̄ a root of p̄(x)≡ f (p(x)), p̄(r̄) = 0. Then

1. If h : F [x]/(p(x))→ F̄ [x]/(p̄(x)) is defined by

h([q(x)])≡ [q̄(x)]≡ [ f (q(x))] ,

then h is an isomorphism.