Let F be an extension of a field with q elements and let E b

Let F be an extension of a field with q elements and let E be an extension of F. Suppose that E is algebraic over F. Prove that |F()| = q^n for some positive integer n.

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Solution

It should be no surprise that
pi is also transcendental over Q. If it were algebraic over Q then there would be a polynomial
f(x) 2 Q[x] such that f(p)=0
. Note that all the odd degree terms would then have a common
factor of p, Move these terms to the other side of the equation, factor out the p

and then square both sides. Then you have a polynomial that has as a zero which we know is impossible.

so for F(alpha) = q^2   , if   F (apha) for n

F(alpha) = q^n    proved