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Let $q$ the power of a prime number $p$ count the number of polynomials $f(X)\in F_q[X]$ such that $F(X)$ vanishes in $F_q$

Attempt

I know that given $f$ then $f$ determines a unique splitting field, but I´m not sure if given a splitting if I have two polynomials $f$ and $g$ wich split in it then $f$ and $g$ must have some characteristic.

I try think in $F_q[x]$ for $q$ the power of a prime number $p$. Which I know is the splitting field of $f(X)=X^q-X$ and suppose that I have a polynomial $g(X)\in F_q[x]$ such that $V(g)=F_q$ I try show that $g(X)= uF(X)$ for $u$ unit of $F_q$. but I dont conclude any.

Some advice?

Alan Jr
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  • Aren't there trivially infinitely many polynomials $f(x)\in\Bbb{F}_q[x]$ such that $f(a)=0$ for all $a\in\Bbb{F}_q$. For example all the polynomials of the form $x^{q+j}-x^{j+1}$, $j$ an arbitrary natural number. Or all the polynomials of the form $g(x)(x^q-x)$. – Jyrki Lahtonen Sep 16 '21 at 08:35
  • And if we restrict to have degree exactly $q$? – Alan Jr Sep 16 '21 at 13:44
  • for example $x^{q+j}-x^{j+q}$ have degree $q+j$ and hence we don´t should consider it – Alan Jr Sep 16 '21 at 13:48
  • and $g(x)(x^q-x)$ have degree $\geq q$ so $g$ in this case must be a constant $c\in F_q$ – Alan Jr Sep 16 '21 at 13:50
  • Oh, I don't see the degree constraint in the question body. Anyway, yes for a polynomial to vanish everywhere in $\Bbb{F}q$ it has to be a multiple of $x^q-x$. This is because $x^q-x=\prod{a\in\Bbb{F}_q}(x-a)$. So then you only have constant multiples of $x^q-x$. – Jyrki Lahtonen Sep 16 '21 at 14:33
  • I´cant understand why $x^q-x=\prod_{a\in F_q}(x-a)$ can you explain why? I have some intuition but I don´t get the correct idea – Alan Jr Sep 16 '21 at 15:47
  • I suppose a polynomial $f$ of degree $q$ and i know that since $0$ is in $F_p$ then $f(x)=xg(x)$ where $g$ have degree $q-1$ but I don´t get how should be $g$ – Alan Jr Sep 16 '21 at 15:49
  • I think that since $x^q-x=x(x^{q-1}-1)$ and $x^{q-1}-1=(1+x+x^2+\cdots x^{q-2})(x-1) $ it is $x^q-x=x(x-1)(1+x+x^2+\cdots+x^{q-2})$ but I can´t see where is the product of all the roots $a$ in $F_q$ – Alan Jr Sep 16 '21 at 15:52
  • Every non-zero element $a$ from $\Bbb{F}_q$ satisfies $a^{q-1}=1$ by Lagrange's theorem (from group theory). Therefore every element of $\Bbb{F}_q$ is a zero of $x(x^{q-1}-1)=x^q-x$. Therefore every polynomial of the form $x-a$ is a factor of $x^q-x$. – Jyrki Lahtonen Sep 16 '21 at 17:10
  • The last piece you may be missing (if any) is that a degree $q$ polynomial cannot have more than $q$ zeros. We found $q$ zeros, so the product $\prod_a(x-a)$ must be a factor of $x^q-x$. The degrees match, as do the leading coefficients, so they are equal. – Jyrki Lahtonen Sep 18 '21 at 03:12

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