Extension of Finite Fields: Proving the Number of Elements in F(\alpha)

[Elzair]

Homework Statement

Let E be an extension of a finite field F, where F has q elements. Let \alpha \epsilon E be algebraic over F of degree n. Prove F \left( \alpha \right) has q^{n} elements.

Homework Equations

An element \alpha of an extension field E of a field F is algebraic over F if f \left( \alpha \right) = 0 for some nonzero f\left(x\right) \epsilon F[x].

The Attempt at a Solution

I do not know how to begin. Is F \left( \alpha \right) a simple extension field?

 

[MathematicalPhysicist]

The answer is rather simple, F(\alpha)=\{a_0+a_1\alpha+...+a_{n-1} \alpha ^{n-1} : a_0,...,a_{n-1} \in F\}
Now count the number of options to choose each a's and multiply them, to get your answer.

 

[Elzair]

Thanks! I just have one question, though. Why is n-1 the highest exponent? Doesn't F \left( \alpha \right) have degree n?

 

[MathematicalPhysicist]

That becasue when you show that F(\alpha) is spanned by \{ 1,\alpha,..,\alpha ^{n-1} \} you use the fact that alpha is algebraic with minimal polynomial of degree n when you show that every polyonimal with degree higher than n-1 we can write in terms of a polynomial of degree n-1 at most. And from the minimality of the minimal polynomial we show that this set is independent.

From there we conclude what I wrote in my first post.