Let $p$ be an odd prime and $\zeta \neq 1$ be a root of the equation $x^p-1$. Prove that $ \operatorname{irr}(\zeta, \mathbb{Q})=1+x+x^2+x^3+\cdots+x^{p-1}$ and then conclude that that the extension $ \ E=\mathbb{Q}(\zeta) \ $ is an algebraic extension of $\mathbb{Q}$ of degree $p-1$.
Answer:
Let $f(x)=1+x+x^2+x^3+\cdots+x^{p-2}+x^{p-1}$
We will use Einstein's criterion to show that this is irreducible.
Look at ,
$$f(x+1)=1+(x+1)+(x+1)^2+\cdots+(x+1)^{p-2}+(x+1)^{p-1} \ \\ \Rightarrow f(x+1)=\sum_{k=0}^{p-1} \sum_{j=0}^k \binom{k}{j} y^{k-j} $$
Now foreach $0 \leq n \leq p-1$, the coefficient on $y^n$ is
$$ \sum_{k=n}^{p-1} \binom{k}{k-n} =\sum_{k=n}^{p-1} \binom{k}{n} =\binom{p}{n+1}$$
Thus we see that
$ f(x+1)=x^{p-1}+px^{p-2}+\cdots+ \frac{p(p-1)}{2} x+p \ $ , which is a monic polynomial.
Clearly this polynomial staisfies Einstein's criterion of irreducibility .
Thus $f(x+1)$ is irreducible over $\mathbb{Q}$
This implies $f(x)$ is irreducible over $\mathbb{Q}$
So $\operatorname{irr}(\zeta , \mathbb{Q} )= 1+x+x^2+x^3+\cdots+x^{p-1}$
But I do not use the condition $\zeta \neq 1$.
Am I right ?
What about the next part?
$1$ is not a root of $1+x+..+x^{p-1}$ but so is $\zeta$.