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I am trying to show the following statement:

Let F be a finite field. Show that there exists a non-constant polynomial $f \in F[x]$ that has no roots in F.

My idea was to simply construct such a polynomial since this is only an existence proof. But my solution seems to be too banal. Below is what I came up with:

Let $g$ be a polynomial in F with $g(x) = 0 \forall x\in F$. We can now simply add a constant factor from F to g s.t

$g_{1}(x) = g(x) + a$ where $a$ is constant $a\in F$. It is now simple to see that $g_{1}(x) \neq 0 \forall x\in F$.

Is this right? It seems to me that $g_{1}(x)$ is just a constant in this case and not a polynomial, hence the proof is wrong... some help and tipps will be much appreciated.

thank you

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    $\begingroup$ Well, you have to prove that $g(x)$ exists (though you could just take $x^{p^n}-x$). The polynomial you construct thereby is not a constant, though it only takes one value on the given field. It takes different values in extensions of the field. Side note: I wouldn't use $g'(x)$ as a notation here...that notation is for derivatives. $\endgroup$ – lulu Jul 6 '17 at 12:11
  • $\begingroup$ what is p? just some constant or a prime? $\endgroup$ – DariusTheGreat Jul 6 '17 at 12:15
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    $\begingroup$ $p$ is the characteristic of your field, and $p^n$ the order of the field. Every finite field has order $p^n$ for some prime $p$ and natural number $n$. $\endgroup$ – lulu Jul 6 '17 at 12:16
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    $\begingroup$ There is nothing wrong with your proof. (It is a polynomial, $x^{p^n}-x+a$, it need not be a constant value a priori for all values in the field, but the axioms of the field necessitate that it be a constant non-zero value). Also notice that any irreducible polynomial of degree not dividing the degree of the field has no root in the field. (If it had then it would be divisible by the minimal polynomial of that root). $\endgroup$ – Siddharth Joshi Jul 6 '17 at 12:42
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You can proceed like this:

Let $p(x)=\prod_{a\in F}(x-a)$. As $F$ is a finite field, it contains $F_p$ for some $p\in \mathbb P$. Pick $\overline 1\in F_p\subset F$. The polynomial $$q(x)=p(x)+\overline 1$$ can't have any roots in $F$.

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    $\begingroup$ Doesn't $p(x)+a$ do the job for any nonzero $a\in F$, as OP suggests? $\endgroup$ – MPW Jul 6 '17 at 12:16
  • $\begingroup$ furthermore is $p$ a prime? Why do we have to take a prime? $\endgroup$ – DariusTheGreat Jul 6 '17 at 12:17
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    $\begingroup$ You have to take a prime, otherwise $F_p$ won't be a field. But it's true that it also works for every $a\neq 0$. $\endgroup$ – A. Salguero-Alarcón Jul 6 '17 at 12:20

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