Thursday, January 18, 2018

Properties of the finite field with 729 elements



I am trying to solve the following problem: let K be a finite field with 729 elements.




  • How many αK make K=α?

  • How many fields E are such that K|E is a field extension? What number of elements have each?


  • How many βK 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 KF36, 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 EK, that are F3|K, F32|K, F33|K and F36|K.



Also, since K is cyclic of order pn1, K=u for some uK, and each element αK can be written α=uk. For αK to satisfy K=α, it is needed to be gcd(k,pn1)=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 βK except 0 satisfies β=αkF3[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 F32F3, so there should be (93)/2 of them (that are monic).


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