How Does x^{p^n}-x Relate to Monic Irreducible Polynomials in \mathbb{Z}_p[x]?

In summary, the zeros of x^{p^n}-x are also the zeros of x^{p^m}-x. The factorization of x^{p^n}-x over \mathbb{Z}_p consists only of polynomials of degree d dividing n. If \alpha is a zero of x^{p^n}-x, then the minimal polynomial of \alpha has degree d dividing n.
  • #1
ehrenfest
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1

Homework Statement


Show that [itex]x^{p^n}-x[/itex] is the product of all the monic irreducible polynomials in [itex]\mathbb{Z}_p[x][/itex] of a degree d dividing n.

Homework Equations


The Attempt at a Solution


So, I want to prove that the zeros of all such monic polynomials are also zeros of [itex]x^{p^n}-x[/itex] and vice versa. I cannot do either, unfortunately. We know that the elements of GF([itex]p^n[/itex]) are precisely the zeros of [itex]x^{p^n}-x[/itex]. And we know that if f(x) is a monic polynomial of degree m in [itex]\mathbb{Z}_p[x][/itex], then when you adjoin any of its zeros to [itex]\mathbb{Z}_p[/itex], you get a field with p^m elements whose elements are precisely the zeros of [itex]x^{p^m}-x[/itex]. So, I guess if [itex]\alpha[/itex] is a zero of [itex]x^{p^n}-x[/itex], then does the irreducible monic polynomial that [itex]\alpha[/itex] is a zero of need to be a factor of [itex]x^{p^n}-x[/itex]?
 
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  • #2
ehrenfest said:
So, I guess if [itex]\alpha[/itex] is a zero of [itex]x^{p^n}-x[/itex], then does the irreducible monic polynomial that it is a zero of need to be a factor of [itex]x^{p^n}-x[/itex]?
Yes: if m(x) is the minimal polynomial of [itex]\alpha[/itex] over Z_p (what you call the irreducible monic polynomial associated with [itex]\alpha[/itex]), and if f(x) is a another polynomial in Z_p[x] with [itex]\alpha[/itex] as a root, then m(x)|f(x).
 
  • #3
morphism said:
Yes: if m(x) is the minimal polynomial of [itex]\alpha[/itex] over Z_p (what you call the irreducible monic polynomial associated with [itex]\alpha[/itex]), and if f(x) is a another polynomial in Z_p[x] with [itex]\alpha[/itex] as a root, then m(x)|f(x).

If you write out the unique factorization of m(x) and f(x) in [itex]\bar{\mathbb{Z}_p}[x][/itex], then both factorizations must contain [itex]x-\alpha[/itex], 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 [itex]\mathbb{Z}_p[x][/itex]?
 
  • #4
m(x) is the monic polynomial of least degree with [itex]\alpha[/itex] 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([itex]\alpha[/itex])=a([itex]\alpha[/itex])m([itex]\alpha[/itex])+r([itex]\alpha[/itex]), which implies that r([itex]\alpha[/itex])=0. The minimality of m now let's us conclude that r=0.
 
  • #5
So the next step is to show that if [itex]\alpha[/itex] is a zero of [itex]x^{p^n}-x[/itex], then the minimal polynomial of [itex]\alpha[/itex] has degree d dividing n. That would imply that the factorization of [itex]x^{p^n}-x[/itex] over [itex]\mathbb{Z}_p[/itex] consists only of polynomials of degree d dividing n. Let m be the degree of \alpha. Then \alpha is also a solution to [itex]x^{p^m}-x[/itex] but I am not really sure if that even helps.
 
  • #6
anyone?
 
  • #7
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.)
 

Related to How Does x^{p^n}-x Relate to Monic Irreducible Polynomials in \mathbb{Z}_p[x]?

1. What is a monic polynomial?

A monic polynomial is a polynomial where the coefficient of the highest degree term is equal to 1. For example, x^2 + 2x + 3 is a monic polynomial, but 2x^3 + 4x^2 + 1 is not.

2. What does it mean for two polynomials to be a product of monic polynomials?

When two polynomials are a product of monic polynomials, it means that they can be written as a multiplication of two monic polynomials. For example, (x + 2)(x + 3) is a product of monic polynomials, where both x + 2 and x + 3 are monic polynomials.

3. How do you find the product of monic polynomials?

To find the product of monic polynomials, you can use the FOIL method. This stands for First, Outer, Inner, Last and involves multiplying each term in the first polynomial with each term in the second polynomial. Then, you can combine like terms and simplify the resulting polynomial to get the final product.

4. Why are monic polynomials important?

Monic polynomials are important because they are the simplest form of a polynomial. They help to simplify equations and make it easier to solve for unknown variables. Additionally, they have special properties that make them useful in various mathematical applications.

5. Can two non-monic polynomials be a product of monic polynomials?

No, two non-monic polynomials cannot be a product of monic polynomials. This is because the product of two monic polynomials will always result in a monic polynomial, and the coefficients of non-monic polynomials will not be equal to 1.

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