How to show that $x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$ is irreducible over $\mathbb{Z}$?

Asked Jan 24 '20 at 21:57

Active Jan 24 '20 at 21:57

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How can I show that $f(x) = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$ is irreducible over $\mathbb{Z}$ ?

It's easy to see that $f$ has no roots in $\mathbb{Z}$ -- so the factorization of $f$ over $\mathbb{Z}[x]$ will involve no linear factors.

But, it gets very tedious to try and show that $f(x)$ does not factor as a quadratic times a sixth degree polynomial, a cubic times a fifth degree polynomial, or the product of two fourth degree polynomials. It's also not promising to try and Eisenstein's Criterion -- even if one makes the substitutions $x \longmapsto x+1$, $x \longmapsto x-1$, or $x \longmapsto x+2$ for example, Eisenstein's Criterion still can't be applied.

Is there a relatively easy way to see why $f$ is irreducible over $\mathbb{Z}$ ?

Thanks!

asked Jan 24 '20 at 21:57

michiganbiker898

1,499

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Note that $f(z)$ is the $30$-th cyclotomic polynomial, and they are irreducible (https://www.lehigh.edu/~shw2/c-poly/several_proofs.pdf) (I don't know if there is an easy fact to prove, I just think of it) – Alejandro Tolcachier Jan 24 '20 at 22:03
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this question contains a discussion of he irreducibility of cyclotomic polynomials. – lulu Jan 24 '20 at 22:10
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Looks rather more to be the fifteenth cyclotomic polynomial, @AleTolcachier . – Lubin Jan 24 '20 at 23:07
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@Lubin you're right, I didn't see the signs, Heh. – Alejandro Tolcachier Jan 24 '20 at 23:12
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@AleTolcachier Thanks. I didn't even think to recognize the polynomial as cyclotomic. That's a great catch. I appreciate your time. (= – michiganbiker898 Jan 25 '20 at 01:57

How to show that $x^8 - x^7 + x^5 - x^4 + x^3 - x + 1$ is irreducible over $\mathbb{Z}$?

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