Given an algebraic equation of degree $n$ of form: $$x^{n} - a\_{n-1}x^{n-1} - a\_{n-2}x^{n-2} - \dots - a\_{0} = 0$$ where each $a\_{i} \in \Bbb Q^{+}$ and atleast one positive root, how does one decide efficiently if the largest magnitude positive root lies in a cyclotomic extension? 



Deciding if the largest absolute value real root lies in a cyclotomic extension
(1) Factoring polynomials over $\mathbb{Q}$ is easy both in theory and practice so you might as well assume that you know the minimal polynomial of the root.


(2) Let $r$ be the largest real root. The assumption that your polynomial is of the minimal form implies that all the other roots have magnitude $\leq r$ (as I assume you realize). Conversely, if $r$ is a positive real algebraic number all of whose Galois conjugtes have norm $<r$, then it obeys a polynomial of this form. Crossed out claim is false; $2+\sqrt{3}$ is a counterexample. Draw a picture of the points $((2+\sqrt{3})^n, (2-\sqrt{3})^n)$ to see why.


(3) Suppose that $f(x)$ is irreducible with integer coefficients. Let $p$ be a prime not dividing the discriminant of $f$ and let $f(x) = \prod g\_i(x)$ be the factorization of $f$ in $\mathbb{F}\_p(x)$. If the roots of $f$ lie in a cyclotomic field, then all the $g\_i$ have the same degree. (This follows from the using relation between factoring polynomials modulo primes and the Galois group; see for example [here](http://sbseminar.wordpress.com/2007/08/24/factoring-polynomials-and-the-frobenius/), or Chapter 2 in Villegas' *Experimental Number Theory*, as suggested by Felipe Voloch.) This gives a rapid way to prove that the roots of a polynomial are NOT cyclotomic. 


Unfortunately, the converse doesn't hold: The above criterion will be true whenever the splitting field of $f$ is generated by a single root of $f$. **I suspect there are clever ways to use reduction modulo primes to rapidly reject more $f$'s.**


(4) From basic class field theory, the roots of $f$ are in $\mathbb{Q}(e^{2 \pi i/N})$ if and only if the factorization of $f$ modulo $p$ is determined by the value of $p$ modulo $N$. Also, all the primes dividing such an $N$ will occur in the discriminant of $f$. I suspect that this is a good way to guess $N$ in practice. 


(5) See [this question](https://math.stackexchange.com/questions/36536) for the question of how to find $N$ if we are promised that some such $N$ exists. In particular, the methods there will give you an $N$ which will work if any $N$ works so, by [Trager's method](http://en.wikipedia.org/wiki/Factorization_of_polynomials#Factoring_over_algebraic_extensions_.28Trager.27s_method.29), the problem is computable. **I do not know what the best method is in practice.**


(6) See theorem 1.0.5 of [Calegari-Morrison-Snyder](http://arxiv.org/abs/1004.0665) for a complete classification when the $a\_i$ are integers and the largest root is less than $76/33 \approx 2.303$. I believe that Morrison and Snyder (both of whom are very active on MO) have since published stronger results.


