Abstract Algebra: Theory and Applications · Problem 21.5.27

Finite Fields — Thomas Judson

Mathematics — Abstract Algebra · transcendental-elements-rational-functions

Problem

Let

E

be an extension field of

F

and

\alpha \in E

be transcendental over

F

. Prove that every element in

F(\alpha)

that is not in

F

is also transcendental over

F

. Hint. Let

\beta \in F(\alpha)

not in

F

. Then

\beta = p(\alpha)/q(\alpha)

, where

p

and

q

are polynomials in

\alpha

with

q(\alpha) \neq 0

and coefficients in

F

. If

\beta

is algebraic over

F

, then there exists a polynomial

f(x) \in F[x]

such that

f(\beta) = 0

. Let

f(x) = a_0 + a_1 x + \cdots + a_n x^n

. Then

0 = f(\beta) = f\!\left(\frac{p(\alpha)}{q(\alpha)}\right) = a_0 + a_1\left(\frac{p(\alpha)}{q(\alpha)}\right) + \cdots + a_n\left(\frac{p(\alpha)}{q(\alpha)}\right)^n.

Now multiply both sides by

q(\alpha)^n

to show that there is a polynomial in

F[x]

that has

\alpha

as a zero.