Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

Question

Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

This exercise I received on a test, and I didn't resolve it. I would be curious in any demonstration with explanations. Thanks!

Please make your post self-contained; I know it seems redundant, but it's best to include the actual question inside the body of the post, even if it's already in the title. — Zev Chonoles, May 29 '13 at 06:41
Ok, sorry! I'm newbie on this forum, and I'm learning every day on how to have a decent behavior in all aspects. — Florin M., May 29 '13 at 06:42
No worries, I know there are a lot of things to get used to on a new forum :) — Zev Chonoles, May 29 '13 at 06:44
https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomials) might be useful. — Lucenaposition, Jul 15 '24 at 00:01

Zev Chonoles · Answer 1 · 2013-05-29T07:12:41.713

9

Hint: Make a Newton polygon for the prime $p=3$. Use the corollary at the top of page 2 in these notes by Paul Garrett (alternatively, here are screenshots: page 1, page 2).

edited May 29 '13 at 07:12

answered May 29 '13 at 06:25

Zev Chonoles

132,937

This is the approach I would have taken, though perhaps involving more advanced techniques than OP had at the fingertips during the test mentioned. – Lubin May 30 '13 at 03:07

score 7 · Accepted Answer · edited Jul 14 '24 at 18:13

7

For binomials there is a classical irreducibility test (below). It implies that $\,x^{105}-c\,$ is irreducible over a field $\,F\,$ if $\,c\,$ is not a third, fifth, or seventh power in $\,F,\,$ since $\,105 = 3\cdot 5\cdot 7.$

Theorem $\ $ Suppose $\:c\in F\:$ a field, and $\:0 < n\in\mathbb Z.$

$\quad x^n - c\ $ is irreducible over $\:F \!\iff\! c \not\in F^p\,$ for all primes $\,p\mid n,\,$ and $\,c\not\in -4F^4$ when $\, 4\:\!|\:\!n$

Proofs can be found in many Field Theory textbooks, e.g. see Lang's Algebra, or see Karpilovsky, Topics in Field Theory, Theorem 8.1.6, excerpted in this answer.

edited Jul 14 '24 at 18:13

Bill Dubuque

282,220

answered May 29 '13 at 19:07

Key Ideas

4,352

@Zev Typos now fixed, thanks. – Key Ideas May 30 '13 at 02:26
So, $x^{105}-9$ is irreducbile over $Z$ ? – Florin M. May 30 '13 at 11:23
Yes, since $,9,$ is neither a 3rd, 5th nor 7th power of a rational, the theorem implies that the polynomial $\ X^{\large 3^i 5^j 7^k}!!-9\ $ is irreducible over $,\Bbb Q.\ \ $ – Key Ideas May 30 '13 at 13:18

Is the polynomial $x^{105} - 9$ reducible over $\mathbb{Z}$?

2 Answers2

Linked