Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $K$ be a field , $a\in K$ , let $d$ be the greatest common divisor, of all the irreducible factors of $x^n-a$ in $K[x]$.

$i)$ Prove that $ d|n$ , and there exist $b\in K$ , such that $a^d =b^n$.

$ii)$ Let's suppose that there exist only one solution $\underline{in K}$ of $x^d=1$. Prove that $x^n-a$ has an irreducible factor of degree $d$.

Well... I have no idea how to attack this problem. If I work over $\mathbb{Q}$ it's easier because I have the cyclotomic polynomials to factorize it. Otherwise I don't know :/ EDITED: Thanks for let me know of the mistake :D

share|improve this question
Did you mean $x^n-a$ instead of $x^n-x$? And did you meant the greatest common divisor of the degrees of the irreducible factors? –  Ted Oct 1 '12 at 4:57
Could be, but at least in the exercise list, is written as $x^n-x$ –  Daniel Oct 1 '12 at 4:59
There's got to be an $a$ somewhere in the polynomial, else the $a$ appears out of nowhere in problem i) and that can't possibly be right. –  Ted Oct 1 '12 at 5:48
Question makes no sense. From first sentence, $d$ is a polynomial; from rest of problem, it's a number. Whoever gave you this exercise list, please go back and ask what the problem is really supposed to be. –  Gerry Myerson Oct 1 '12 at 5:56
It's still a little crazy. Let $K$ be the field of 2 elements, let $a=1$, let $n=2$. Then $d$ is the polynomial $x-1$, and I don't know what to make of $1^{x-1}=b^2$, nor of $x^2-1$ having a factor of degree $x-1$. Even if we're working over the rationals, and the irreducible factors of $x^n-a$ are relatively prime, one can take any nonzero rational for $d$, divisibility is a trivial property in a field, and degree $d$ makes no sense. Where in blazes did you find this problem? –  Gerry Myerson Oct 2 '12 at 0:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.