# Product of monic polynomials

1. Mar 2, 2008

### ehrenfest

1. The problem statement, all variables and given/known data
Show that $x^{p^n}-x$ is the product of all the monic irreducible polynomials in $\mathbb{Z}_p[x]$ of a degree d dividing n.

2. Relevant equations

3. The attempt at a solution
So, I want to prove that the zeros of all such monic polynomials are also zeros of $x^{p^n}-x$ and vice versa. I cannot do either, unfortunately. We know that the elements of GF($p^n$) are precisely the zeros of $x^{p^n}-x$. And we know that if f(x) is a monic polynomial of degree m in $\mathbb{Z}_p[x]$, then when you adjoin any of its zeros to $\mathbb{Z}_p$, you get a field with p^m elements whose elements are precisely the zeros of $x^{p^m}-x$. So, I guess if $\alpha$ is a zero of $x^{p^n}-x$, then does the irreducible monic polynomial that $\alpha$ is a zero of need to be a factor of $x^{p^n}-x$?

Last edited: Mar 2, 2008
2. Mar 2, 2008

### morphism

Yes: if m(x) is the minimal polynomial of $\alpha$ over Z_p (what you call the irreducible monic polynomial associated with $\alpha$), and if f(x) is a another polynomial in Z_p[x] with $\alpha$ as a root, then m(x)|f(x).

3. Mar 2, 2008

### ehrenfest

If you write out the unique factorization of m(x) and f(x) in $\bar{\mathbb{Z}_p}[x]$, then both factorizations must contain $x-\alpha$, but why does m(x) need to contain factors with all of the other zeros of f(x)? Does it have something to with the fact that <m(x)> is a maximal ideal in $\mathbb{Z}_p[x]$?

4. Mar 2, 2008

### morphism

m(x) is the monic polynomial of least degree with $\alpha$ as a root (prove this if you don't already know it). Because Z_p is a field, Z_p[x] has a division algorithm. So let's write f(x)=a(x)m(x)+r(x), where r is either zero or its degree is less than that of m. Now note that f($\alpha$)=a($\alpha$)m($\alpha$)+r($\alpha$), which implies that r($\alpha$)=0. The minimality of m now lets us conclude that r=0.

5. Mar 3, 2008

### ehrenfest

So the next step is to show that if $\alpha$ is a zero of $x^{p^n}-x$, then the minimal polynomial of $\alpha$ has degree d dividing n. That would imply that the factorization of $x^{p^n}-x$ over $\mathbb{Z}_p$ consists only of polynomials of degree d dividing n. Let m be the degree of \alpha. Then \alpha is also a solution to $x^{p^m}-x$ but I am not really sure if that even helps.

6. Mar 4, 2008

anyone?

7. Mar 4, 2008

### morphism

If you adjoin \alpha to Z_p, you'll get a subfield of GF(p^n). But now d=[Z_p(alpha) : Z_p] is a divisor of n=[GF(p^n) : Z_p].

(If the minimal polynomial of \alpha over F has degree d, then alpha generates an extension of degree d of F.)