Irreducibility of $x^{2^n}+x+1$ over $\mathbb{Z}_2$

Question

I'm trying to solve this problem from Hungerford V.5.9.

I have to show $x^{2^n}+x+1$ is irreducible over $\mathbb{Z}_2$ if n>2.

I would appreciate some hint cause I don't know how to start with it.Thanks!

EDITED: This question is completely different to On irreducible factors of $x^{2^n}+x+1$ in $\mathbb Z_2[x]$ The reason is obvious, i m asking if this polynomial is irreducible. In the other question, we assume it is reducible without showing it.

It doesn't seem true: $(x^6+x^5+x^3+1)(x^2+x+1)=x^{2^3}+x+1$. — zozoens, Feb 04 '14 at 15:16
For any odd $n>1$ its wrong: $x^{2^n}+x+1=\sum_{k=0}^{n-1} (x^2+x+1)^{2^k}=(x^2+x+1)\cdot(\cdots)$ (the first equality is a telescoping sum by freshman's dream) — J.R., Feb 04 '14 at 15:18
@zozoens, MAPLE tells me that you re not right, this multiplication of polynomials gives $x^8+x^4+x^3+x^2+x+1$ — Lotus, Feb 04 '14 at 15:21
@John sorry, I forgot the $x^2$ term, so we have $(x^6+x^5+x^3+x^2+1)(x^2+x+1)$ — zozoens, Feb 04 '14 at 15:22
I'm a little confused cause it seems obvious that @zozoens is right but this implies Hungerford exercise is wrong... — Lotus, Feb 04 '14 at 15:25
@zozoens: No the same trick works for any $n$ that has an odd factor. So it can only be true for $n=2^m$. — J.R., Feb 04 '14 at 15:33
I don't understand the edit. In the other question it is shown that the irreducible factors of this polynomial are of degree $\ge 2n$. So if $2^n>2n$, i.e. if $n>2$, then this polynomial is reducible, and any exercise asking you to show that it is irreducible is just wrong. — Jyrki Lahtonen, Feb 06 '14 at 22:10

J.R. · Accepted Answer · 2014-02-05T13:18:35.917

4

The claim is not true. Assume $n>1$ is odd. Then

$$ \begin{align} x^{2^n}+x+1 &= (x^{2^n}+x^{2^{n-1}}+1)+(x^{2^{n-1}}+x^{2^{n-2}}+1)+\cdots+(x^2+x+1)\\ &=\sum_{k=0}^{n-1}(x^{2^{k+1}}+x^{2^k}+1) = \sum_{k=0}^{n-1} (x^2+x+1)^{2^k}\\ &= (x^2+x+1)\left(\sum_{k=0}^{n-1} (x^2+x+1)^{2^k-1}\right) \end{align} $$

We used that $n$ is odd in the first equality, where there are $n$ ones on the right side, thus summing to $1$.

Note: The same argument can be used to show that $x^{2^n}+x+1$ is reducible whenever $n$ has an odd factor greater than $1$. Therefore it can only be irreducible if $n=2^m$.

edited Feb 05 '14 at 13:18

answered Feb 04 '14 at 15:25

J.R.

17,904

This works for odd $n$, as the third roots of unity are then zeros of this polynomial, and your calculation verifies that the third cyclotomic polynomial is a factor. However, the polynomial is never irreducible, when $n>2$. – Jyrki Lahtonen Feb 06 '14 at 22:09
@JyrkiLahtonen why is the polynomial never irreducible, when n>2? – No One Mar 31 '15 at 03:45
@TiWen: See this answer to the question this is a duplicate of. Tip about the site: If a question is marked as a duplicate of another, older question, the older version is often (but not always!) a good source for extra information. – Jyrki Lahtonen Mar 31 '15 at 03:58

Irreducibility of $x^{2^n}+x+1$ over $\mathbb{Z}_2$

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