Question

Let E/F be a field extension, and let α be an element of E that is algebraic over F.

Let p(x) = irr(α, F) and n = deg p(x).

(a) For f(x) ∈ F[x], let r(x) (∈ F[x]) be the remainder of f(x) when divided by p(x).

Prove that f(x) +p(x)= r(x)+p(x)in F[x]/p(x).

(b) Prove that if |F| < ∞, then | F[x]/p(x)| = |F|n. (For a set A, we denote by |A| the number of elements in A.)

Answer 1

a.

By the division algorithm, if a(x)a(x) and b(x)b(x) are any polynomials, and a(x)≠0a(x)≠0, then there exist unique q(x)q(x) and r(x)r(x) such that

b(x)=q(x)a(x)+r(x),r(x)=0 or deg(r)<deg(a).b(x)=q(x)a(x)+r(x),r(x)=0 or deg⁡(r)<deg⁡(a).

Let b(x)=p(x)b(x)=p(x), and a(x)=x−ca(x)=x−c. Then r(x)r(x) must be constant (since it is either zero or of degree strictly smaller than one), so

b(x)=q(x)(x−c)+r.b(x)=q(x)(x−c)+r.

Now evaluate at x=cx=c.

b.

The field F is naturally embedded in F[x]/hp(x)i by
a 7→ a + hp(x)i for a ∈ F. Thus, we may consider
F[x]/hp(x)i as an extension field of F.
and deg p (x)=n thus F (x)/p (x )= Fn