# Why is $\alpha$, a primitive pth root of unity, a root of the polynomial $1+t+\ldots +t^{p-1}$?

I'm currently reading a proof that the minimal polynomial of $$\alpha$$, a primitive pth root of unity, over $$\mathbb{Q}$$ is $$1+t+\ldots +t^{p-1},$$ yet the author simply states "Clearly $$\alpha$$ is a zero." in the proof, and I can't seem to prove that bit myself.

• Oh, now I see that if I multiply the polynomial by $(t-1)$ I get $t^p-1$, which zeroes are clearly $1, \alpha , \ldots , \alpha ^{p-1}$. – Leo Apr 29 at 0:27

Note $$(1+t+\cdots + t^{p-1} ) (1-t)= t^p -1$$, and that $$\alpha$$ is a root of $$t^p -1$$, since it is a pth root of unity, but not of $$1-t$$, since it is primitive. Then it must be a root of $$1+\cdots+t^{p-1}$$, as the author says.