Irreducible polynomial in $Z[X]$

Asked Mar 04 '17 at 19:50

Active Mar 04 '17 at 19:50

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Prove that $f(x)=x^p-x+k$ is irreducible in $Z[X]$,where $p$ is a prime and $k$ is an integer not divisible by $p$.

I think I solved it, but it seems very easy,so the proof could be incorrect. Proof: Let $f(x)=g(x)h(x)$ and let $\alpha$ be root of $g$,now reduce $f$ modulo $p$ and we get $f^{*}=k$ and $g^{*}$ divides $f^{*}$,therefore $g=pq(x)+k_1$,where $k_1$ is not divisible by $p$ and $q$ is with integer coefficients,hence because $\alpha$ is an algebraic integer $p$ must divide $k_1$,which is a contradiction.

asked Mar 04 '17 at 19:50

Kristiyan Vasilev

Do the same as here, which is for $k=-1$. – Dietrich Burde Mar 04 '17 at 19:53
My proof is the same as the one there. – Kristiyan Vasilev Mar 04 '17 at 19:56
Then everything is fine. – Dietrich Burde Mar 04 '17 at 21:16
@KristiyanVasilev: It is not true that when you reduce $X^p - X$ modulo $p$ you get $0$; you still get $X^p - X$. What you get, though, is that the polynomial function $x^p-x : \Bbb F_p \to \Bbb F_p$ is $0$, indeed, but do not mistake a polynomial for its associated polynomial function! – Alex M. Mar 05 '17 at 16:00
@DietrichBurde: That proof does not explain why $x^p-x-1$ is irreducible in $\Bbb F_p$, and I don't find this statement obvious. – Alex M. Mar 05 '17 at 16:04
@KristiyanVasilev: Who says that $f$, $g$ or $h$ must have a root in $\Bbb Z$ or in $\Bbb F_p$? – Alex M. Mar 05 '17 at 16:11
1

@AlexM. Indeed, that is not obvious, but an even more frequent duplicate. – Dietrich Burde Mar 05 '17 at 17:21
@AlexM: Can you give a solution to the problem? – Kristiyan Vasilev Mar 05 '17 at 17:22
@KristiyanVasilev: By intelligently adapting the answers by Elaqqad and cat in the links given by Dietrich Burde, you should be able to solve the problem by yourself. If you don't make it, signal it to me in a comment. – Alex M. Mar 05 '17 at 17:38

Irreducible polynomial in $Z[X]$

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