There's this exercise that really has kept me stuck for a day by now, will you please help me figure out:

let's consider polynomials in $\mathbb Z_3$:

  • characterize degree 2 not irreducible monic polynomials. How many are they?
  • characterize degree 2 irreducible monic polynomials. How many are they?
  • how many degree 4 monic polynomials have no roots and are not irreducible?

I have no idea how to characterize those polynomials, I just figured that all polynomials with $\Delta = 0$ or $\Delta = 1$ are not irreducible, if $\Delta = 2$ then the polynomial is actually irreducible. I couldn't help but enumerate all monic degree 2 polynomials and find one by one for myself if they met the condition I've set. Is there a better way to count them without being forced to find them all? What about the last question, that really left me with no idea.

  • $\begingroup$ Hint: A reducible degree 2 monic polynomial has the form $p(x) = (x-a)(x-b)$, as there are ... linear polynomials, there are ... of them :) ... Regarding the last question: The same, but we use degree 2 irreducible polynomials as factors ... HTH $\endgroup$ – martini Jul 18 '12 at 9:48
  • $\begingroup$ And ... BTW: What is $\Delta$? $\endgroup$ – martini Jul 18 '12 at 9:58
  • $\begingroup$ I don't know why the more stupid is the question, the more complicated explainations I try to find out. When solving a degree 2 equation like $ax^2+bx+c=0$, you get $\Delta = b^2-4ac$ to find the roots using the formula $\frac{-b \pm \sqrt{\Delta}}{2a}$ so I have noticed that if $\Delta = 2$ than there's no $x^2 \equiv_3 2$ therefore (from my hopefully not point of view) the polynomial is irreducible. $\endgroup$ – haunted85 Jul 18 '12 at 10:05
  • $\begingroup$ Ok, but you should've written that in your question ... $\endgroup$ – martini Jul 18 '12 at 10:17
  • $\begingroup$ I didn't because I thought it was proof of nothing. My bad. $\endgroup$ – haunted85 Jul 18 '12 at 10:24

As I wrote in my comment, it is more easy to count the reducible degree two polynomials. Such a polynomial is a product of the monic dregree one polynomials. There are three such ($x$, $x-1$ and $x-2$). There are six different products of two of them, hence there are six recucible degree two polynomials. As there are nine monic degree two polynomials alltogether, there are three irreducible ones.

To count the reducible monic degree four polynomials without zero, we argue along the same lines: Such a polynomial is a product $p(x)q(x)$ where $p$, $q$ are monic irreducible degree two polynomials. As we found above, there are three such, we get six reducible, monic, degree four polynomials.


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