214 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

2. p̄(x) is the minimum polynomial for r̄

3. There exists a unique isomorphism g : F(r)→ F(r̄) which agrees with f on F andg(r) = g(r̄).

Proof: Define h : F [x]/(p(x))→ F̄ [x]/(p̄(x)) by h([q(x)]) ≡ [q̄(x)] ≡ [ f (q(x))] . Iclaim this is an isomorphism.

First, why is h well defined? If [q(x)] = [q1 (x)] is [q̄(x)] = [q̄1 (x)]? This is equivalentto verifying that if [q(x)] = 0, then [q̄(x)] = 0. Does this happen? If q(x) = p(x) l (x) , isq̄(x) = p̄(x) l̄ (x)? This is true since f is a homomorphism. Thus this h is well defined.

Also, h is a homomorphism because

h([q(x)] [q1 (x)]) = h([q(x)q1 (x)])≡ [ f (q(x)q1 (x))] = [ f (q(x))] [ f (q1 (x))]

h is clearly onto because f is onto F̄.Is h one to one? If h([q(x)])≡ [q̄(x)]≡ [ f (q(x))] = 0, does it follow that [q(x)] = 0?

To say that [q̄(x)] = 0 is to say that f (g(x)) = q̄(x) = l̄ (x) p̄(x) = f (l (x) p(x)) . Thusf (l (x) p(x)) = f (q(x)) . However, since f is one to one on F, this requires l (x) p(x) =q(x) and so [q(x)] = 0 showing that h is one to one. Hence h is an isomorphism.

Thus F̄ [x]/(p̄(x)) is a field by Lemma 9.4.4 because F [x]/(p(x)) is a field due to p(x)being the minimum polynomial for r which forces p(x) to be irreducible. Indeed, if p(x) =k (x) l (x) where each of these two have smaller degree than p(x) , then 0 = p(r) = k (r) l (r)and since k (r) , l (r) are in the field F(r) , it follows that one of these is 0 which contradictsp(x) being the minimum polynomial for r. It follows from Lemma 3.4.8 that p̄(x) mustalso be irreducible. Hence p̄(x) is the minimum polynomial for r̄.

Then from Theorem 9.4.2, the following diagram holds

F(r) α−1→ F [x]/(p(x)) h→ F̄ [x]/(p̄(x)) ᾱ→ F̄(r̄)

where ᾱ is the isomorphism described by ᾱ ([q̄(x)]) ≡ q̄(r̄) . Thus all mappings are iso-morphisms and so you can let g = ᾱ ◦h◦α−1 and this shows the existence of an extensionof f as an isomorphism from F(r) to F̄(r̄) which satisfies

g(q(r))≡ ᾱ ◦h◦α−1 (q(r))≡ ᾱ ◦h([q(x)])≡ ᾱ ([q̄(x)])≡ q̄(r̄) .

In particular, if q(r) = r, then since f (1) is the multiplicative identity in F̄,

g(r) = g(q(r))≡ ᾱ ◦h([q(x)])≡ ᾱ ◦h([x])≡ ᾱ ([ f (1)x]) = ᾱ ([x]) = r̄

If q(r) ∈ F, then q(r) = q0 ∈ F and so g(q(r)) = g(q0) =

ᾱ ◦h◦α−1 (q0)≡ ᾱ ◦h([q0])≡ ᾱ ([q̄0])≡ q̄0 ≡ f (q0) .

so g agrees with f on F.Now suppose g : F(r)→ F̄(r̄) is an isomorphism which agrees with f on F where

p̄(r̄) = 0 and g(r) = r̄. How many can there be? There is one by the above. I need to showit must have the above form. To do this, note that

g◦α = ᾱ ◦h on F [x]/(p(x))