Let's denote a polynomial of degree $k$: $f(x) = (a_1,\ldots ,a_k) := a_1x^k+\cdots +a_k$. Find all irreducible monic polynomials over $\mathbb{Z}_3$ of degree at most $4$.

For $\deg f = 1$, we have $f = (1,0),(1,1),(1,2)$. This serves as a generator for all Reducible monics of degree $2$. We have total of 6 ways to uniquely multiply them to obtain a monic of degree $2$ and since there are $9$ total, this leaves us $3$, thus:
For $\deg f=2$, we have $f = (1,0,1),(1,1,2),(1,2,1)$.

For $\deg f=3$, we have total of $27$ monics and the generator of reducibles of $\deg 3$ is the union of the irreducibles of $\deg 1$ and $\deg 2$ and so we have multiply them in "all possible ways" that yield something of degree $3$.
The counting gets messy here:
For degree $1$ irreducibles, we can cube each of them, $3$ ways. We can Square one and multiply by another, $6$ ways and we can multiply them all for a total of $10$ ways.
Since we want the product to be of degree $3$ , we need to multiply $\deg f=1$ irreducibles with $\deg f=2$ irreducibles, that gives us another $9$ possibilities.
Are there $19$ reducibles of degree $3$? (i.e 8 irreducibles?)

Concern: Are all these products unique, that is, do we count everything exactly once?

For $\deg f=4$, we can (hopefully) count analogously.

I can write tables for small degrees and test everything out by hand and find irreducibles in some amount of effort. What to do when we find all irreducibles of , say, at most degree $8$ over $\mathbb{Z}_{17}$?

Is there a quicker method of determining irreducibility or weeding out the irreducibles of a given degree? The end of goal is still explicitly finding all of them.

  • 2
    $\begingroup$ See this entry on wikipedia. $\endgroup$ – WimC Oct 29 '16 at 9:12
  • $\begingroup$ @WimC ah, thanks, that confirms my observations. Do you have any tips on how to explicitly write them all out, though? For the fourth degree, $N(3,4) = 18$. It would be bothersome to write all the $81$ polynomials out and see what fits $\endgroup$ – Alvin Lepik Oct 29 '16 at 9:35

You can do the following: The Galois Group of $\mathbb{F}_{81}$ is the cyclic group $C_4$. There are $18$ orbits of size $4$, the others are of order $2$ or $1$. These orbit are of the form $\{z^k,z^{3k}, z^{9k}, z^{27k} \}$, for $z$ a primitive element and the exponents taken $\operatorname{mod} 81$. An irreducible polynomial is then given by $(z-z^k)(z-z^{3k})(z-z^{9k})(z-z^{27k})$.

EDIT after a comment of Jyrki Lahtonen

The elements of $\mathbb{F}_{81}$ are, by Galois theory, invariant for a subgroup $H$ of the Galois group $G$, there is only such (non trivial) subgroup namely the sqyare of the generator of $G$, the Frobenius exponential map of order $3$. This means that we are to look out for values $k$ for which $z^k$ is invariant under the Frobenius map of order $9$, i.e. the $k$ for which $$ z^{9k} = z$$, or, equivalently $z^{8k} = 1$. Since the order of $z$ is $80$ this gives rise to the values $k \in \{10, 20, 30, 40, 50, 60, 70, 80\}$ .

  • $\begingroup$ A good start, but there is the (minor) catch that if $k$ is a multiple of ten, then $z^k$ is in the subfield $\Bbb{F}_9$. Those have quadratic minimal polynomials. Or, if $10\mid k$, then $z^{9k}=z^k$. $\endgroup$ – Jyrki Lahtonen Oct 29 '16 at 13:56
  • $\begingroup$ @Jyrki Indeed, $10$ does indeed give an orbit of size $2$. As there are $81$ points in total not all the orbits can have a size of $4$. I only thought that it was the wish of the OP not to wade through a total of $81$ polynomials. I'll edit my answer to reflect the values for $k$ that give rise to an orbit of size $<4$. $\endgroup$ – Marc Bogaerts Oct 29 '16 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.