Prove Polynomial is Irreducible

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Homework Help Overview

The discussion revolves around proving the irreducibility of the polynomial f(x) = x^p + x^{p-1} + ... + x - 1 over the field Z_p, where p is a prime number. Participants are exploring various approaches to establish this property.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster notes that f(x) has no roots in Z_p but expresses uncertainty about further steps. Another participant attempts to analyze the polynomial by considering its reciprocal and exploring transformations, yet also encounters difficulties. A third participant reflects on the necessity of the reciprocal approach, suggesting an alternative derivation of f(x+1).

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and attempts without reaching a consensus. Some guidance is offered through the exploration of the polynomial's properties, but no definitive solution has been established.

Contextual Notes

Participants are working under the constraints of proving irreducibility in the context of polynomial equations over a finite field, specifically Z_p, and are navigating through various mathematical transformations and assumptions related to the problem.

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Homework Statement


Show the polynomial
<br /> f(x) = x^p + x^{p-1} + ... + x - 1<br />
is irreducible over Z_p where p is a prime.

Homework Equations


The Attempt at a Solution


I know f(x) has no roots in Zp, but other than that, I'm stuck. Thanks for the help.
 
Last edited:
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I tried:
Let g(x) = x^{p}f(1/x) be the reciprocal of f(x). g(x) is irreducible iff f(x) is irreducible. We have
g(x) = -x^p + x^{p-1} + ... + x + 1
-g(x) = x^p - x^{p-1} - ... - x - 1
-(x - 1)g(x) = x^{p+1} - 2x^p + 1
-xg(x+1) = (x+1)^{p+1} - 2(x+1)^p + 1 = x^{p+1} - x^p + x
-g(x+1) = x^p - x^{p-1} + 1

So, showing this polynomial is irreducible over Z_p is equivalent to the original problem. But again, I'm stuck here :).
 
BTW, no real need for the reciprocal as I just realized that you can derive f(x+1) = x^p + x^{p-1} - 1 similarly.
 
In case anyone is interested in the solution:

The reciprocal of f(x+1) is g(x) = 1 + x - x^p. But -g(x) is an irreducible trinomial. It follows f(x) is irreducible.
 

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