Show that $\displaystyle x^{10}+x^9+x^8+...+x+1$ is irreducible over $\displaystyle Q$ - set of all rational numbers.

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- Feb 8th 2010, 11:36 AMArczi1984Irreducible polynomial
Show that $\displaystyle x^{10}+x^9+x^8+...+x+1$ is irreducible over $\displaystyle Q$ - set of all rational numbers.

- Feb 8th 2010, 11:43 AMTheEmptySet
- Feb 8th 2010, 11:50 AMArczi1984
Thank You for help:) I think that I'm too tired because of this I've problems with so easy task:(

- Feb 8th 2010, 11:58 AMBruno J.
Another way to see it is that your polynomial is just $\displaystyle \frac{x^{11}-1}{x-1}$; and the roots of $\displaystyle x^{11}-1$ are the 11-th roots of unity, none of which lie on the real axis except for $\displaystyle x=1$.

- Feb 9th 2010, 12:02 AMOpalg
Being irreducible is a much stronger property than having no roots (in the given field). Having no roots means that there are no linear factors. But for a polynomial to be irreducible it must have no nontrivial factors at all.

As Bruno J. points out, the roots of the given polynomial in the complex field are the complex 11th roots of unity. These can be grouped in complex conjugate pairs to form quadratic factors $\displaystyle x^2 - 2\cos(2k\pi/11)*x + 1$ (for k = 1,2,3,4,5) over the real field. To show that the polynomial is irreducible over the rationals, you would have to show that none of these can be grouped together to form a rational polynomial. I don't know how to do that, but presumably it must involve showing that those cosines are seriously irrational. - Feb 9th 2010, 12:08 AMynj