212 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

If f defined on F [x] is as just described, then is indeed a homomorphism of F [x] andF̄ [x] as claimed. This follows from an elementary computation.

The following is a nice theorem which will be useful.

Theorem 9.4.2 Let F be a field and let r be algebraic over F. Let p(x) be the minimumpolynomial of r. Thus p(r) = 0 and p(x) is monic and no nonzero polynomial havingcoefficients in F of smaller degree has r as a root. In particular, p(x) is irreducible over F.Then define f : F [x]→ F(r) , the polynomials in r by

f

(m

∑i=0

aixi

)≡

m

∑i=0

airi

Then f is a homomorphism. Also, defining g : F [x]/(p(x)) by

g([q(x)])≡ f (q(x))≡ q(r)

it follows that g is an isomorphism from the field F [x]/(p(x)) to F(r) .

Proof: First of all, consider why f is a homomorphism. The preservation of sums isobvious. Consider products.

f

(∑

iaixi

∑j

b jx j

)= f

(∑i, j

aib jxi+ j

)= ∑

i jaib jri+ j

= ∑i

airi∑

jb jr j = f

(∑

iaixi

)f

(∑

jb jx j

)Thus it is clear that f is a homomorphism.

First consider why g is even well defined. If [q(x)] = [q1 (x)] , this means that

q1 (x)−q(x) = p(x) l (x)

for some l (x) ∈ F [x]. Therefore,

f (q1 (x)) = f (q(x))+ f (p(x) l (x))

= f (q(x))+ f (p(x)) f (l (x))

≡ q(r)+ p(r) l (r) = q(r) = f (q(x))

Now from this, it is obvious that g is a homomorphism.

g([q(x)] [q1 (x)]) = g([q(x)q1 (x)]) = f (q(x)q1 (x)) = q(r)q1 (r)

g([q(x)])g([q1 (x)]) ≡ q(r)q1 (r)

Similarly, g preserves sums. Now why is g one to one? It suffices to show that if g([q(x)])=0, then [q(x)] = 0. Suppose then that g([q(x)]) ≡ q(r) = 0 Then q(x) = p(x) l (x)+ρ (x)where the degree of ρ (x) is less than the degree of p(x) or else ρ (x) = 0. If ρ (x) ̸= 0, thenit follows that ρ (r) = 0 and ρ (x) has smaller degree than that of p(x) which contradictsthe definition of p(x) as the minimum polynomial of r. Hence, [q(x)] = 0 and g is one toone. Since p(x) is irreducible, F [x]/(p(x)) is a field. It is clear that g is onto. Therefore,F(r) is a field also. (This was shown earlier by different reasoning.) ■

Here is a diagram of what the following theorem says.