How many distinct monic irreducible polynomials of degree \(2\) are there over \(\mathbb F_5\)?
How many distinct monic irreducible polynomials of degree \(3\) are there over \(\mathbb F_5\)?
Solution
1. A polynomial of degree \(2\) or \(3\) over a field is reducible if and only if it has a root in that field.
If such a polynomial has a root \(a\), then it is divisible by \(x-a\) and is reducible. Conversely, any nontrivial factorization of a polynomial of degree \(2\) or \(3\) has a factor of degree \(1\), and a degree-\(1\) factor supplies a root in the field.
2. There are exactly \(15\) reducible monic quadratic polynomials over \(\FF_5\).
For each \(a\in\FF_5\), let \(A_a\) be the set of monic quadratic polynomials having \(a\) as a root. Every element of \(A_a\) has the form \[ (x-a)(x-b), \qquad b\in\FF_5, \] so \[ \abs{A_a}=5. \] If \(a\neq b\), then \[ A_a\cap A_b = \{(x-a)(x-b)\}, \] so every pairwise intersection has size \(1\). No quadratic has three distinct roots. Hence inclusion–exclusion gives \[ \abs{\bigcup_{a\in\FF_5}A_a} = 5\cdot5-\binom52 = 25-10 = 15. \] By step 1, this union is exactly the set of reducible monic quadratics.
3. There are exactly\[ 25-15=10 \]monic irreducible quadratic polynomials over \(\FF_5\).
A monic quadratic is determined by its two nonleading coefficients, so there are \[ 5^2=25 \] monic quadratics in total. Subtract the \(15\) reducible ones counted in step 2.
4. There are exactly \(85\) reducible monic cubic polynomials over \(\FF_5\).
For each \(a\in\FF_5\), let \(B_a\) be the set of monic cubic polynomials having \(a\) as a root. Such a polynomial is \[ (x-a)q(x), \] where \(q\) is an arbitrary monic quadratic, so \[ \abs{B_a}=5^2=25. \]
If \(a\neq b\), then a polynomial in \(B_a\cap B_b\) has the form \[ (x-a)(x-b)(x-c), \qquad c\in\FF_5, \] and therefore \[ \abs{B_a\cap B_b}=5. \] If \(a,b,c\) are three distinct field elements, then \[ B_a\cap B_b\cap B_c = \{(x-a)(x-b)(x-c)\}, \] so every triple intersection has size \(1\). No cubic has four distinct roots. Inclusion–exclusion gives \[ \begin{aligned} \abs{\bigcup_{a\in\FF_5}B_a} &= 5\cdot25 -\binom52\cdot5 +\binom53\\ &= 125-50+10\\ &= 85. \end{aligned} \] By step 1, these are precisely the reducible monic cubics.
5. There are exactly\[ 125-85=40 \]monic irreducible cubic polynomials over \(\FF_5\).
A monic cubic is determined by its three nonleading coefficients, so there are \[ 5^3=125 \] monic cubics in total. Subtract the \(85\) reducible ones counted in step 4.
6. Q.E.D.
Steps 3 and 5 give the answers \(10\) and \(40\), respectively.