Determine all of the monic irreducible polynomials in $\mathbb Z_3 [x]$ of degree $4.$

Question

Determine all of the monic irreducible polynomials in $\mathbb Z_3 [x]$ of degree $4$. Prove that you have found them all and that the ones you found are irreducible.

I am looking for some sort of way to figure this out without having to list everything.

I can figure this out for degree 3 in z3 just fine, but I'm having difficulties with degree 4. Can someone please show me a step by step to possibly get an answer to this?

Thank you!

We have a test coming up, and I have a feeling this is going to be on it, and I'm trying to understand this completely.

Answer 1

I don't think there's any way to avoid doing a fairly large amount of listing (given that the final answer you are looking for is a list of $18$ degree $4$ polynomials...). However, you at least can avoid going through all $81$ monic polynomials one by one and trying to factor them all. The following is one strategy you might follow.

If a polynomial $p(x)$ of degree $4$ is reducible, then either it has a linear factor or it is a product of two irreducible quadratic polynomials. If $p(x)$ has a linear factor, then we must have $p(0)=0$, $p(1)=0$, or $p(-1)=0$; each of these cases corresponds to a simple linear relationship between the coefficients of $p$. It should not be too hard to list out all of the monic polynomials such that none of these relations hold (there are $24$ of them).

Now you just have to eliminate the products of two irreducible quadratic polynomials. To do this, first write down all the irreducible monic quadratic polynomials by the same method as above (by ruling out any linear factor). You should get that there are only $3$ of them. Now just compute all of the products of pairs of these (there are $6$ such products, $3$ where the two factors are the same and $3$ where they are different). Crossing these off your earlier list of polynomials with no linear factor, you now have a list of all the irreducible polynomials (there should be $18$ of them).

Answer 2

I don’t see how to do it without any listing at all. I would do it this way, but I’d be using a fairly primitive symbolic-computation package to help me. First, I’d work over $k=\Bbb F_9$, and find an element $z\in S=\Bbb F_{81}\setminus\Bbb F_9$. Then I would do a listing of the elements of $S$, namely all $a+bz$ with $a,b\in k$ but $a\ne0$. Then I’d find all $4$-tuples $A_y=\{y,y^3,y^9,y^{27}\}$, taking care not to repeat any. Since I started with $72=81-9$ things in $S$, there’d be $72/4=18$ disjoint sets $A_y$. Then I’d multiply out $(X-y)(X-y^3)(X-y^9)(X-y^{27})\in\Bbb F_3[X]$ to get the irreducible quartic corresponding to each $A_y$.