Show that every finite field is of prime-power order, that is, it has a prime-power number of elements.
Detailed Proof:
Step 2: A finite field is a finite-dimensional vector space over its prime field.
Let F be a finite field and `P cong bb ZZ_p ` be its prime field , as established in Step 1. Since P is a subfield of F , we can consider F as a vector space over P. Because F is a finite field , it has a finite number of elements. Consequently , F must be a finite-dimensional vector space over P. Let the dimension of F over P be n.
Step 3: Determine the order of the finite field.
Since F is an n-dimensional vector space over P , there exists a basis {`alpha_1 , alpha _2 , ... , alpha_n`} for F over P. This means that every element `x in F` can be uniquely expressed as a linear combination of the basis elements
`x = c_1 alpha_1 + c_2 alpha_2 + ... + c_n alpha_n` , where `c_i in P` for all i = 1 , ... , n . Since `P cong ZZ_p` , there are p choices for each coefficient `c_i` . Because there are n such coefficients , and each is independant , the total number of distinct elements in F is `p * p * ... * p` (n times) . Therefore , the order of the finite field is `bb {p^n}`. Since p is a prime number and n is a positive integer , `p^n` is a prime power number.
Conclusion:
We have shown that every finite field F has a prime characteristic p , and it can be viewed as a finite-dimensional vector space of dimension n over its prime field `ZZ_p`. This structure directly implies that the number of elements in F is `p^n` , which is a prime power number. QED