0
$\begingroup$

This is an exercise from the book "Einführung in die algebraische Zahlentheorie" by Alexander Schmidt. Let $p$ be a prime number. And let $f \in \mathbb{Z}[X]$ be a polynomial with integer coefficients such that it is not the zero polynomial modulo $p$ ($f \not\equiv 0 \pmod p$). But such that $f(a)\equiv 0 \pmod p$ for all $a\in \mathbb{Z}$.

$\endgroup$
4
  • 1
    $\begingroup$ Multiply that by $x$? $\endgroup$ Jun 19, 2019 at 6:07
  • $\begingroup$ Good point. Thanks for the hint. I will remove the $f(x) = x^{p-1} - 1$ from the question to allow for a more non-trivial answer. $\endgroup$
    – ahartel
    Jun 19, 2019 at 6:16
  • 5
    $\begingroup$ Try the product $\prod_{i=0}^{p-1} (x-i)$. $\endgroup$ Jun 19, 2019 at 6:30
  • 1
    $\begingroup$ Indeed, the set of such polynomials is exactly the set of multiples of $x^p-p = \prod_{i=0}^{p-1} (x-i)$. $\endgroup$ Jul 3, 2019 at 4:59

1 Answer 1

1
$\begingroup$

Let's assume that $f\in F_p[X]$ is a non-zero polynomial with all $a\in F_p$ as roots. When $a\in F_p$ is a root, $(x-a)$ is a factor, thus $f(x)=g(x)\prod_{i=0}^{p-1} (x-i)$ for some non-zero $g\in F_p[X]$. We also see that every polynomial on the form $g(x)\prod_{i=0}^{p-1} (x-i)$ has every $a\in F_p$ as a root. We conclude that the polynomials are precisely the ones on the form: $$g(x)\prod_{i=0}^{p-1} (x-i),$$ where $g\not\equiv 0$.

$\endgroup$
1
  • $\begingroup$ Welcome, Frederick. In effect, the theory of finite fields says that every finite extension of the prime field $\mathbb F_p$ has $p^n$ elements and that in the algebraic closure of $\mathbb F_p$ there exists a unique subfield with as many elements as the considered extension of $\mathbb F_p$ and that is constituted by the roots of the polynomial $f(X)=X^{p^n}-X$. $\endgroup$
    – Piquito
    Jul 3, 2019 at 21:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .