If a monic rational polynomial of degree $p-1$ has $p$-th root of unity as a root, is it the cyclotomic polynomial? If a monic rational polynomial of degree $p-1$ has a $p$-th root of unity as a root, where $p$ is prime, does that make it the cyclotomic polynomial $x^{p-1}+...+1$?
I think this is the same as asking whether a rational polynomial could have a real common factor with the cyclotomic polynomial, such as $x^2+2Re(\zeta)x+1$ where $\zeta$ is a $p$-th root of unity. So we can ask what rational vectors is contained in 
$span_\mathbb{R}\{(1,2Re(\zeta),1,0,...),(0,1,2Re(\zeta),1,0,...),...,(0,...,0,1,2Re(\zeta),1)\}$, but I'm not sure where to go from here.
 A: If you also require the polynomial to be monic, then yes. This is because the cyclotomic polynomial of degree $p-1$ is the minimal polynomial over $\mathbb Q$ for all (nontrivial) $p$-th roots of unity.
A: Your statement is true.
Suppose that there is a monic rational polynomial $f(x)$ of degree less than $p$ such that $f(x)$ has a $p$-th root of unity as a root. Then, consider greatest common divisor of polynomials $f(x)$ and $g(x)=x^{p-1}+x^{p-2}+\ldots+x+1$. Note that $d(x)=\gcd(f(x), g(x))$ is a rational polynomial of degree at least $1$ (since $f$ and $g$ has a common root). Also, $d(x)$ divides $g(x)$. However, the polynomial $g(x)=x^{p-1}+x^{p-2}+\ldots+x+1$ is irreducible (over $\mathbb{Q}$). Hence, $d(x)=cg(x)$ for some constant $c\in\mathbb{Q}$.
Therefore, $f(x)$ is divisible by $g(x)$ because $d(x)|f(x)$. Now, note that $\deg g=p-1$ and $\deg f<p$, so $f(x)=ag(x)$. Since $f(x)$ is monic we obtain that $f(x)=g(x)$, as desired.
To prove that $g(x)$ is irreducible over $\mathbb{Q}$ one may use the Eisenstein's criterion. Consider 
$$
h(x)=g(x+1)=\frac{(x+1)^p-1}{(x+1)-1}=\frac{1}{x}\sum\limits_{k=1}^{p}\binom{p}{k}x^k=\sum\limits_{k=1}^{p}\binom{p}{k}x^{k-1}.
$$
Now, it's easy to see that all coefficients of $h(x)$, except the leading coefficient, are divisible by $p$, but the last coefficint isn't divisible by $p^2$. Thus, $h(x)$ is irreducible over $\mathbb{Q}$, so $g(x)$ is also irreducible over $\mathbb{Q}$.
