I am trying to solve the following problem: let K be a finite field with 729 elements.
- How many α∈K make K∗=⟨α⟩?
- How many fields E are such that K|E is a field extension? What number of elements have each?
- How many β∈K satisfy K=F3[β]?
- How many irreducible polynomials of degree 2, 3 and 6 are in F3[t]?
And I have argued as follows:
Since K has 729 elements, and 729 is 36, it follows K≅F36, the finite field with 36 elements, so every E such that K|E is field extension needs to satisfy |E|=pk with k|6, so there are, up to isomorphism, 4 field extensions of the form E⊆K, that are F3|K, F32|K, F33|K and F36|K.
Also, since K∗ is cyclic of order pn−1, K∗=⟨u⟩ for some u∈K∗, and each element α∈K∗ can be written α=uk. For α∈K∗ to satisfy K∗=⟨α⟩, it is needed to be gcd(k,pn−1)=1, since
order(uk)=order(u)gcd(k,order(u))
so the number of α's with this property is φ(728)=288.
This should give all the elements such that K=F3[α], since every element β∈K except 0 satisfies β=αk∈F3[α] and |K|=|F3[α]| (is this right?)
For the last, I would only know how to calculate the number of irreducible and monic polynomials, but I don't know how to calculate the whole number of irreducible polynomials with those degrees.
I would appreciate some hints or help. Thanks in advance.
Answer
Most of it looks good, except for the last part. You don't need β to be a generator for F3[β]=F3[α]. What you need is for the minimal polynomial of β over F3 to have degree 6. Equivalently, you need β not to be in any proper subfield of K.
To count the number of irreducible polynomials, use the fact that the irreducible polynomials of these degrees in F3[t] are going to factor completely in K. For example, an irreducible polynomial of degree 2 is going to have two distinct conjugate roots in F32∖F3, so there should be (9−3)/2 of them (that are monic).
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