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Let $\mathbb{F}:= \{a_1, \ldots , a_n \}$ be a finite field. Consider the following polynomial $$f = \prod_{i = 1}^n (X-a_i) + 1 $$ Is this polynomial irreducible? If not, is it irreducible in $\mathbb{F}_p$, a finite field with $p$ Elements for a prime $p$?

Constantin K
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    @DietrichBurde: I think you are mistaken. That question is about the existence of some irreducible polynomial of degree$~n$, which is definitely true, but this is about a particular one. That one happens to have no roots in the finite field, but this does not suffice to show it is irreducible. As far as I can see nobody at that question said whether this one is always irreducible or not (I would be the latter). – Marc van Leeuwen Mar 29 '17 at 09:27
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    @DietrichBurde: could you be more precise about where the argument for irreducibility is given? I might be sleepy this morning, but I really could not find any after looking twice. – Marc van Leeuwen Mar 29 '17 at 09:32
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    We have $f=(x^{p^n}-x)+1$, which is irreducible over $\mathbb{F}_p$ only when $n=1$, or $n=p=2$. This is answered here. – Dietrich Burde Mar 29 '17 at 09:52
  • The question isn't answered, or is it? – Constantin K Mar 29 '17 at 12:11
  • For $n\ge 2$, or $n\ge 3$ and $p\ge 3$ it is reducible over $\mathbb{F}_p$, hence also over $\mathbb{F}$. So the question is answered, I think. – Dietrich Burde Mar 29 '17 at 13:00
  • Sorry, I don't understand why? Why is the polynomial of @DietrichBurde the same as mine? – Constantin K Mar 29 '17 at 13:10
  • The product of all linear factors $(x-a_i)$ in a finite field equals $x^{p^n}-x$, see here. So $f-1=(x -a_1)\cdots (x -a_{p^n})=x^{p^n}-x$. – Dietrich Burde Mar 29 '17 at 13:15
  • If $\mathbb{F}$ is a finite field then it has $p^n$ elements and it is the splitting field of $x^{p^n}-x = \prod_{a \in \mathbb{F}} (x-a)$ whose coefficients are in $\mathbb{F}_p$ so yes it is answered – reuns Mar 29 '17 at 14:18

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