Linear Algebra, Theory And Applications, 2012a

460 FIELDS AND FIELD EXTENSIONS
¯p (x) will be the polynomial in ¯F [x] defined as
¯p (x) ≡
n∑
f (a k ) x k .
Also consider f as a homomorphism of F [x] and ¯F [x] in the obvious way.
k=0
f (p (x)) = ¯p (x)
The following is a nice theorem which will be useful.
Theorem F.4.2 Let F beafieldandletr be algebraic over F. Let p (x) be the minimal
polynomial of r. Thus p (r) = 0 and p (x) is monic and no nonzero polynomial having
coefficients 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
( m
)
∑
m∑
f a i x i ≡ a i r i
Then f is a homomorphism. Also, defining g : F [x] / (p (x)) by
i=0
i=0
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 is
obvious. Consider products.
⎛
⎞ ⎛
⎞
f ⎝ ∑ a i x ∑ i b j x j ⎠ = f ⎝ ∑ a i b j x i+j ⎠ = ∑ a i b j r i+j
i j
i,j
ij
= ∑ i
a i r i ∑ j
b j r j = f
( ∑
i
⎛
)
a i x i f ⎝ ∑ j
b j x j ⎞
⎠
Thus it is clear that f is a homomorphism.
First consider why g is even well defined. If [q (x)] = [q 1 (x)] , this means that
for some l (x) ∈ F [x]. Therefore,
q 1 (x) − q (x) =p (x) l (x)
f (q 1 (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)] [q 1 (x)]) = g ([q (x) q 1 (x)]) = f (q (x) q 1 (x)) = q (r) q 1 (r)
g ([q (x)]) g ([q 1 (x)]) ≡ q (r) q 1 (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