How many irreductible polynoms of degree $n=3$ are there over $\mathbb Z_{3}=\{\overline{0}, \overline{1}, \overline{2}\}$? How many irreductible polynoms of degree $n=3$ are there over $\mathbb Z_{3}=\{\overline{0}, \overline{1}, \overline{2}\}$?
Please, check my solution:
$f(x)$ is a polynom of degree $n=3 (\delta f=3) \Rightarrow f(x)=a_0 + a_1 x +a_2 x^2 + a_3 x^3$. We have three choices for each coeficient, so we can build $3^4 = 81$ polynoms over $\mathbb Z_{3}$. The reductible polynoms are in the shape $p(x)=(x- \alpha)(b_0 + b_1 x + b_2 x^2)$, so we can build $3.(2^3)=24$ reductible polynoms over $\mathbb Z_{3}$. Conclusion: there are 57 irreductible polynoms.
Thank you!
 A: First I count the number of irreducible monic degree-2 polynomials over $\def\zt{\Bbb Z_3}\zt$.  It suffices to count the monic polynomials and then multiply by 2.  A reducible monic polynomial of degree 2 has the form $(x-a)(x-b)$ for $\{a,b\}\subset\{\bar0,\bar1,\bar2\}$. There are 6 possible choices for $\{a,b\}$. There are 9 monic polynomials ($x^2+px+q$ for $\langle p,q\rangle\in\{\bar0,\bar1,\bar2\}^2$), so 3 are irreducible. 
Now I count the number of irreducible degree-3 polynomials over $\zt$.  There are $27$ monic polynomials.  A reducible monic polynomial factors as either $(x-a)P_2(x)$ where $P_2$ is irreducible or as $(x-a)(x-b)(x-c)$.  There are $9$ of the former ($3$ choices of $a$ times 3 choices of $P_2$) and $10$ of the latter, totaling $19$.  This leaves $27-19=8$ monic irreducible polynomials, so $16$ irreducible polynomials in all.
A: I assume you mean degree 3, not degree at most 3. We can assume they are monic. If you want the non-monic ones just multiply all these by 2. So I get these irreducible monic polynomials of degree 3 over $F_3$:
$x^3+2x+1$
$x^3+2x+2$
$x^3+x^2+2$
$x^3+x^2+x+2$
$x^3+x^2+2x+1$
$x^3+2x^2+1$
$x^3+2x^2+x+1$
$x^3+2x^2+2x+2$
Being lazy, I just asked Mathematica to print them out, but I seem to have less than you, so maybe I asked in a silly way.
