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i'm working on a problem and i need some help.

Problem:

Let $p$ be a prime number, show that in the polynomial ring $(\mathbb{Z} / p \mathbb {Z})[x]$ for every $d \in \mathbb N,d \ge 2$ exists a polynomial $f$ with $\deg(f)=d$ with no root in $ \mathbb Z/p \mathbb Z $.

My Idea:

I already shown it for $p = 3$ so there is no need to talk about that.

for $p \ge 5$ I know that i have to differ the even and odd $\deg(f)$ and i know that i have to use $ \prod_{i = 1} ^d (x-i)+1  $

I can show it for general $d$ but i do not know, how to show it for odd AND even $d$.

I would appreciate your help.

user26857
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Tinkerbell_ye
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    Doesn't $\prod_{i = 0} ^{d-1} (x-i)+1$ always work for $d\ge p$? – lhf Nov 26 '21 at 21:34
  • For every $d\ge 1$ there is a field $\mathbb F=\mathbb F_{p^d}$ with $p^d$ elements. The characteristic of that field is $p$, and the degree is $d$. The multiplicative group of $\mathbb F$ is finite cyclic, so $\mathbb F$ is generated by one element $a$. The minimal polynomial of $f(x)$ of $a$ over $\mathbb F_p$ has degree $d$ and is irreducible over the prime field. Hence, in particular, $p(x)$ has no roots in $\mathbb F_p$. – markvs Nov 26 '21 at 21:34
  • @markvs, sure, but the OP does not need $f$ to be irreducible. – lhf Nov 26 '21 at 21:35
  • He gets irreducibility as a free bonus. – markvs Nov 26 '21 at 21:35
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    Does this answer your question? Polynomial with no roots over Field $ \mathbb{F}_p $. – plop Nov 26 '21 at 21:44
  • @markvs Thanks a lot for your answer but it does not cover my question about the odd and even deg. Any ideas about that? – Tinkerbell_ye Nov 26 '21 at 21:47
  • My answer (comment) works for even and odd degrees equally well. – markvs Nov 26 '21 at 21:49
  • I am confused. These two of your statements seem to be in conflict: "I can show it for general $d$" vs. "I do not know, how to show it for odd AND even $d$" – Hagen von Eitzen Nov 26 '21 at 21:50
  • @Hagen von Eitzen I am told that the answer should differ for odd and even deg. I already show it without odd and even differentiation but I do not know how to include odd and even deg in my proof – Tinkerbell_ye Nov 26 '21 at 21:55
  • @Tinkerbell_ye I do not see any reason for that. For example, you can always show that there exists $a$ such that $x^d-x-a$ does the trick, no matter what $d$ is (as long as it is $>1$). Incidentally, $p$ need not even be prime. – Hagen von Eitzen Nov 26 '21 at 21:58
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    @HagenvonEitzen's answer is clearly the most elegant and elementary solution. – reuns Nov 27 '21 at 01:28

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