# Is $x^{p}+1 \in Z_{p}[x]$ irreducible or reducible? [closed]

If $p$ is a prime number and $x^{p}+1 \in \mathbb Z_{p}[x]$. Show that $x^{p}+1$ isn't irreducible in $\mathbb Z_{p}[x]$. I need know how can show that, is for a homework that I send tomorrow to the University. Only I know that I need use the little theorem of Fermat.

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## closed as off-topic by TMM, Matthew Conroy, Michael Albanese, YACP, azimutJan 1 at 11:02

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Hint: it is $\,f(x) - f(-1)\,$ so has root $\,x - ?\,$ by the Factor Theorem. –  Bill Dubuque Jan 1 at 1:41
$(x+1)^p\equiv x^p+1\pmod{p}$ by Binomial-theorem. –  Siddharth Prasad Jan 1 at 1:41
Thanks, but I dont know how I can show that. Plz help –  Vader4891 Jan 1 at 1:59

$x^p+1 = (x+1)^p$ in characteristic $p$ :)

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Thanks, but I dont know how I can show that. Plz help –  Vader4891 Jan 1 at 1:47
Show $p\mid \binom{p}{k}$ for $1\leq k \leq p-1$. –  Pedro Tamaroff Jan 1 at 2:05
why I need show that? –  Vader4891 Jan 1 at 2:21
You don't really need that actually, it is enough to just show that $-1$ is a root, which will show that the polynomial is reducible (but not give a complete factorisation). –  fkraiem Jan 1 at 2:24
$-1$ is a root because $(x+1)^{p}=0$ if $x=-1$. Thanks to you –  Vader4891 Jan 1 at 2:39

$f=X^p+1$ $$f(-1)=(-1)^p+1=(-1)^{p-1}(-1)+1=1\cdot(-1)+1=-1+1=0$$

$=>(X+1)$ divide $X^p+1$

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The suggestions thus far are very good. A simply way of solving this without use of the binomial theorem is we know $x^p+1$ is reducible if it has a root in $\mathbb{Z}_p$. For any prime $p$, $-1=1 \in \mathbb{Z}_p$

(Technically this is not so. However, this simplifies the matter and it is simple to find the correct corresponding positive element in $\mathbb{Z}_p$ for $-1$ as $\mathbb{Z}_p \cong \mathbb{Z}/\langle p\rangle$. I simply find it cleaner to use $-1$.).

In any case, if $p$ is an odd prime, then notice when $x=-1$, we have $$x^p+1=(-1)^p+1=-1+1=0$$ so that $-1$ is a root. If $p$ is an even prime (that is $p=2$), we are looking a the ring $\mathbb{Z}_2$, then notice $x=1$ yields $$x^p+1=(-1)^2+1=1+1=0$$ in $\mathbb{Z}_2$ so that again $x^p+1$ has a root and is reducible.

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Thanks, nice answer –  Vader4891 Jan 1 at 3:26