The polynomial is irreducible iff the condition is satisfied

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Discussion Overview

The discussion revolves around the irreducibility of the polynomial $f(x)=x^n+1$ within the context of algebra, specifically focusing on conditions under which this polynomial is irreducible over the rational numbers $\mathbb{Q}[x]$. Participants explore hints and approaches to demonstrate that irreducibility occurs if and only if $n=2^k$ for some integer $k \geq 0$.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant requests hints for showing the irreducibility of the polynomial $f(x)=x^n+1$.
  • Another participant suggests demonstrating a factorization of $x^{2n+1} + 1$ as a hint towards the solution.
  • A different participant notes that $n$ can be expressed as $2^k \ell$, where $k \ge 0$ and $\ell$ is odd, leading to a discussion on the implications of $\ell$ being greater than 1 or equal to 1 for the irreducibility of $x^n + 1$.
  • One participant proposes using Eisenstein's criterion to show irreducibility when $\ell = 1$, while another participant argues that simpler polynomial factorizations and Gauss's lemma could suffice without needing Eisenstein's criterion.

Areas of Agreement / Disagreement

Participants express differing views on the methods to demonstrate irreducibility, with some favoring Eisenstein's criterion and others advocating for simpler approaches. There is no consensus on the best method to use.

Contextual Notes

The discussion includes various assumptions about the form of $n$ and the applicability of different mathematical criteria, which remain unresolved. The effectiveness of the proposed methods is also debated without reaching a definitive conclusion.

mathmari
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Hey! :o

I need some help at the following exercise:

Show that the polynomial $f(x)=x^n+1 \in \mathbb{Q}[x]$ is irreducible if and only if $n=2^k$ for some integer $k \geq 0$.

Could you give me some hints what I could do?? (Wondering)
 
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This is a fairly straightforward algebra exercise. Hint : show that

$$x^{2n+1} + 1 = (x + 1)\left ( x^{2n} - x^{2n-1} + \cdots - x + 1 \right)$$
 
mathmari said:
Hey! :o

I need some help at the following exercise:

Show that the polynomial $f(x)=x^n+1 \in \mathbb{Q}[x]$ is irreducible if and only if $n=2^k$ for some integer $k \geq 0$.

Could you give me some hints what I could do?? (Wondering)

Hi mathmari,

Note that $n$ is of the form $2^k \ell$, where $k \ge 0$ and $\ell$ is odd. Thus $x^n + 1 = x^{2^k \ell} + 1 = (x^{2^k})^{\ell} + 1$. Hence, if $\ell > 1$, $x^{2^k} + 1$ is a proper factor of $x^n + 1$. In case $\ell = 1$, show that $(x + 1)^n + 1$ is irreducible by application of Eisenstein's criterion with $p = 2$. Show that this implies $x^n + 1$ is irreducible.
 
Eh, Eisenstein's criterion is a bit too much. Some easy polynomial factorizations, Gauss lemma and a pack of popcorns does the trick.
 

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