Show that every finite field is of prime-power order, that is, it has a prime-power number of elements.
Here are some hints which play a role:
Hint 1: Let E be an extension field of a finite field F, where F has q elements. Let `bb alpha in E` be algebraic of degree n , then `F(alpha)` has `bb q^n` elements.
This hint is crucial for understanding why the number of elements in an extension field becomes a power of the base field's size.
How it connects to the proof: In our proof, we establish that a finite field E contains a prime field `P cong ZZ_p` . We then view F as an extension field of this prime field.
Every element in F is algebraic over P . This means that for any element `alpha in F` there's some polynomial with coefficients in P that has `bb alpha` as a root. This is a fundamental property of finite fields - they are always algebraic extensions of their prime fields.
If we consider F as an extension of P, it means F can be generated by some elements over P. If F is a finite field, it is a finite extension of P. This implies that F can be seen as a vector field over P. The degree n in the hint directly corresponds to the dimension of the vector space F over P.
So our prime field P has p elements , playing the role of "q" in the hint, and the finite field F is an extension of degree n over P. This means that F is an n-dimensional vector space over P , and the total number of elements in F will be `p^n`. This is precisely what the hint states as `q^n` elements.