0
$\begingroup$

I want to find a valid polynomial for a finite field $\mathbb{Z}_p[x]_{f(x)}$ with $d=deg(f(x))$. For this definition to hold, it can be deduced that $p$ must be prime and the polynomial $f(x)$ irreducible, which is equivalent to the following condition:

$$f(k)\neq0\enspace\forall k\in\mathbb{Z}_p[x]_{f(x)}$$

Then, we can use the form of $f(x)=a_0\cdot x^{d}+a_1\cdot x^{d-1}+\cdots+a_{d}$ to establish a set of conditions like:

$$f(0)=a_0\cdot 0+a_1\cdot 0+\cdots+a_{d}\implies a_{d}\not\equiv0\pmod{p}$$ $$f(1)=a_0\cdot 1+a_1\cdot 1+\cdots+a_{d}\implies \sum_{i=0}^{d} a_{i}\not\equiv0\pmod{p}$$ $$\vdots$$ $$f(p-1)=a_0\cdot (p-1)^{d}+a_1\cdot (p-1)^{d-1}+\cdots+a_{d}\implies \sum_{i=0}^{d} a_{i}\cdot(p-1)^{d-i}\not\equiv0\pmod{p}$$

So, is there any way to obtain the coefficients $a_{i}$ of the polynomial trough a closed form that spans every valid solution with an approach similar to the above conditions?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ Your notation is somewhat confusing, but if I'm interpreting it correctly: to check that a polynomial is irreducible it is not sufficient to check that it has no zero. A reducible polynomial can also be a product of several irreducible factors of degree $>1$. $\endgroup$
    – Achim Krause
    Commented Sep 19 at 17:01
  • $\begingroup$ You need to define your notation. Normally a number theorist writes $\mathbb Z_p$ for the ring of $p$-adic integers and $\mathbb F_p$ for the finite field with $p$ elements. Further, I have no idea what you mean by $\mathbb Z_p[x]_{f(x)}$. The most likely meaning using "standard" notation is some sort of localization of the polynomial ring. But it seems more likely that you mean the quotient ring $\mathbb Z_p[x]/f(x)\mathbb Z_p[x]$. Finally, you seem to indicate that a polynomial is irreducible if it has no roots. That is not true in general. $\endgroup$
    – Joe Silverman
    Commented Sep 19 at 17:02
  • $\begingroup$ Like $x^4+1$ mod $5$. $\endgroup$
    – Achim Krause
    Commented Sep 19 at 17:03
  • 2
    $\begingroup$ And, if @LSp has the correct interpretation, then I think the question is asking for too much. There are algorithms for finding such polynomials, but, so far as I know, no all-encompassing explicit formula. $\endgroup$
    – Gerry Myerson
    Commented Sep 19 at 21:58
  • 2
    $\begingroup$ Re, wait, given $p$ and $d$, you want every polynomial such that $\mathbb F_p[x]/(f(x))$ is isomorphic to $\mathbb F_{p^d}$? This is just asking for all of the irreducible polynomials, filtered by degree, and that's definitely too much to hope for. $\endgroup$
    – LSpice
    Commented Sep 20 at 19:34

0

Reset to default

You must log in to answer this question.