How many irreducible polynomials of degree exactly 6 are there over the finite field with 3 elements?
Solution
Work inside \(\FF_{3^6}\).
1. Exactly\[ 3^6-3^3-3^2+3=696 \]elements of \(\FF_{3^6}\) have degree exactly \(6\) over \(\FF_3\).
For an element \(\alpha\in\FF_{3^6}\), the degree \([\FF_3(\alpha):\FF_3]\) divides \(6\). Hence an element whose degree is less than \(6\) has degree \(1\), \(2\), or \(3\), and therefore lies in \(\FF_{3^2}\) or \(\FF_{3^3}\). Conversely every element of either of these proper subfields has degree less than \(6\).
The two subfields satisfy \[ \FF_{3^2}\cap\FF_{3^3}=\FF_3, \] because the degree of their intersection over \(\FF_3\) divides both \(2\) and \(3\). Thus the number of elements lying in a proper subfield is \[ 3^2+3^3-3=9+27-3=33. \] Subtracting from \(|\FF_{3^6}|=3^6=729\) gives \[ 729-33=696. \]
2. There are exactly\[ \frac{696}{6}=116 \]monic irreducible polynomials of degree \(6\) over \(\FF_3\).
Every element \(\alpha\) counted in step 1 has a monic irreducible minimal polynomial of degree \(6\) over \(\FF_3\). Such a polynomial has six distinct roots in \(\FF_{3^6}\), \[ \alpha,\alpha^3,\alpha^{3^2},\ldots,\alpha^{3^5}, \] and all six have the same minimal polynomial. Conversely every root of a monic irreducible sextic has degree \(6\) over \(\FF_3\). Thus the \(696\) degree-six elements are partitioned into sets of six roots, one set for each monic irreducible sextic. Hence there are \(696/6=116\) such monic polynomials.
3. The number of irreducible polynomials of degree exactly \(6\) over\(\FF_3\) is\[ \boxed{232}. \]
Every irreducible polynomial is a unique nonzero scalar multiple of a monic irreducible polynomial. Since \(\FF_3^\times\) has two elements, each of the \(116\) monic irreducible sextics from step 2 has exactly two nonzero scalar multiples. Therefore the total number is \[ 2\cdot116=232. \]
4. Q.E.D.
Step 3 gives the requested number.