# Thread: reducible polynomial x^p+a in Zp

1. ## reducible polynomial x^p+a in Zp

Let p be a prime and a be an element of Zp. Show that x^p+a is reducible in Zp[x]

im not sure if this is the best way to approach the question but
as x^p+a is reducible there must be a factor
therefore x^p+a=0
and x^p=-a for some x

would it then suffice to say that as Zp contains a primitive root,
hence there exists an element of Zp congruent to -a, and therefore there is a root in Zp? This doesn't seem like a very good solid solution

2. Originally Posted by i_never_noticed
Let p be a prime and a be an element of Zp. Show that x^p+a is reducible in Zp[x]

im not sure if this is the best way to approach the question but
as x^p+a is reducible there must be a factor
therefore x^p+a=0
and x^p=-a for some x

would it then suffice to say that as Zp contains a primitive root,
hence there exists an element of Zp congruent to -a, and therefore there is a root in Zp? This doesn't seem like a very good solid solution

It really isn't. Try to show the following:

=== The map $a\to a^p$ in $\mathbb{Z}/p\mathbb{Z}$ is a bijection (in particular it is onto) ===