Let be the ring homomorphism defined by Since the characteristic of is the kernel of must be and the image of must be a subfield of isomorphic to We will denote this subfield by Since is a finite field, it must be a finite extension of and, therefore, an algebraic extension of Suppose that is the dimension of where is a vector space. There must exist elements such that any element in can be written uniquely in the form
where the 's are in Since there are elements in there are possible linear combinations of the 's. Therefore, the order of must be
We will prove this lemma using mathematical induction on We can use the binomial formula (see Chapter 2, Example 2.4) to verify the case for that is,
If then
must be divisible by since cannot divide Note that is an integral domain of characteristic so all but the first and last terms in the sum must be zero. Therefore,
Now suppose that the result holds for all where By the induction hypothesis,
Therefore, the lemma is true for and the proof is complete.
Let be separable. Then factors over some extension field of as where for Taking the derivative of we see that
Hence, and can have no common factors.
To prove the converse, we will show that the contrapositive of the statement is true. Suppose that where Differentiating, we have
Therefore, and have a common factor.
Let and let be the splitting field of Then by Lemma 22.5, has distinct zeros in since is relatively prime to We claim that the roots of form a subfield of Certainly 0 and 1 are zeros of If and are zeros of then and are also zeros of since and We also need to show that the additive inverse and the multiplicative inverse of each root of are roots of For any zero of we know that is also a zero of since
provided is odd. If then
If then Since the zeros of form a subfield of and splits in this subfield, the subfield must be all of
Let be any other field of order To show that is isomorphic to we must show that every element in is a root of Certainly 0 is a root of Let be a nonzero element of The order of the multiplicative group of nonzero elements of is hence, or Since contains elements, must be a splitting field of however, by Corollary 21.36, the splitting field of any polynomial is unique up to isomorphism.
Theorem 22.7.
Every subfield of the Galois field has elements, where divides Conversely, if for then there exists a unique subfield of isomorphic to
Let be a subfield of Then must be a field extension of that contains elements, where is isomorphic to Then since
To prove the converse, suppose that for some Then divides Consequently, divides Therefore, must divide and every zero of is also a zero of Thus, contains, as a subfield, a splitting field of which must be isomorphic to
Let be a finite subgroup of of order By the Fundamental Theorem of Finite Abelian Groups (Theorem 13.4),
where and the are (not necessarily distinct) primes. Let be the least common multiple of Then contains an element of order Since every in satisfies for some dividing must also be a root of Since has at most roots in On the other hand, we know that therefore, Thus, contains an element of order and must be cyclic.
Let be a generator for the cyclic group of nonzero elements of Then
Example 22.13.
The finite field is isomorphic to the field Therefore, the elements of can be taken to be
Remembering that we add and multiply elements of exactly as we add and multiply polynomials. The multiplicative group of is isomorphic to with generator