Factoring Polynomials in Z_p: Finding Degree 'd

[barbiemathgurl]

help me its so hard

working in the finite field Z_p show that the all the factors of polynomial x^{p^n}-x have degree "d" such that d|n.

thanx

 

[mathwonk]

use the binomial theorem and the fact that mod p, (x-a)^p = x^p-a^p.

 

[Hurkyl]

Let a be a root of that polynomial. What does that tell you about Z_p[a]?
(or, did I mean to ask that the other way around?)

Oh, even better: do you know the splitting field of that polynomial?

 

[Kummer]

help me its so hard

working in the finite field Z_p show that the all the factors of polynomial x^{p^n}-x have degree "d" such that d|n.

thanx

I think you mean all irreducible factors.

This can be avoided without splitting fields. But it is a little more lengthy.

We can find $$\bar{\bold{Z}_p}$$ this is the algebraic closure of $$\bold{Z}$$. Given $$p(x)$$ an irreducible factor we can choose $$\alpha \in \bar{\bold{Z}_p}$$ that is a zero. Now if we adjoin $$\alpha$$ to $$p(x)$$ we get a field $$\bold{Z}_p(\alpha)$$ with $$[\bold{Z}_p(\alpha):\bold{Z}_p]=d$$ because that is the degree of $$p(x)$$. Let $$F$$ be the set of all zeros in $$\bar{\bold{Z}_p}$$ to the polynomial $$x^{p^n}-x$$, this makes $$F$$ a field with $$p^n$$ elements. Since $$\bold{Z}_p \subset F$$ we have that $$[F:\bold{Z}_p]=n$$. Finally notice that $$\alpha \in F$$ since $$p(\alpha)=0$$ and $$p(x)|\left( x^{p^n}-x \right)$$. This immediately implies that $$\bold{Z}_p (\alpha) \subseteq F$$. Now we have succesfully established the fact that $$\bold{Z}_p\subseteq \bold{Z}_p(\alpha) \subseteq F$$. Now using theorem for degrees of finite field extensions we have $$[F:\bold{Z}_p(\alpha)][\bold{Z}_p(\alpha):\bold{Z}_p]=[F:\bold{Z}_p]$$. So $$[F:\bold{Z}_p(\alpha)]\cdot d = n$$ which shows that $$d|n$$.