This is a claim in the book "Abstract algebra" in one of the examples, can someone explain this please ?
I know that the derivative of this polynomial is identically $0$ so there is a multiple root, but why all the roots are multiple ?
This is a claim in the book "Abstract algebra" in one of the examples, can someone explain this please ?
I know that the derivative of this polynomial is identically $0$ so there is a multiple root, but why all the roots are multiple ?
If $n=kp$ then $x^n -1= (x^k -1)^p$, because $x\mapsto x^p$ is an automorphism. So, the roots of $x^n -1$ are exactly the roots of $x^k -1$ and appear with multiplicity at least $p$.
Because $x^{dp}-1 = (x^d-1)^p$