specific example of a field We consider set of sums:
$$a_0z_0+\ldots +a_{n-1}z_{n-1},$$
where $a_i\in\mathbb{Q}$ and $z_i$ are n-th complex roots of the unity. I have to chceck if this set with standard operations $+$ and $\cdot$ is a field or not.
The answer is yes, which confused me. For all coefficients $a_i=1$ we have the sum 
$$\sum_{k=0}^{n-1}z_{k}$$ which is equal to $0$ for any $n\in\mathbb{N}$. Does it mean that the above sum is a neutral element of an addiction? But then we get $0$ for each $n$ from another combination (roots looks different). What will be the unity of multiplication and how can we show that any element has an inverse?
 A: So you have the set $C$ of all sums
$$a_0 + a_1 \omega + \cdots + a_{n-1} \omega^{n-1}$$ where $a_0$, $\ldots$, $a_{n-1} \in \mathbb{Q}$ ( $\omega= e^{\frac{2\pi i}{n}}$). It is easy to see the set is closed under addition and multiplication ( use $\omega^n = 1$). We need to show that every $\ne 0$ element in this set has an inverse in this set. It is enough to show that every element $x$ in this set satisfies an equation of form 
$$x^n + b_{n-1} x^{n-1} + \cdots + b_0= 0$$ with $b_0$, $\ldots$, $b_{n-1} \in \mathbb{Q}$.
Assume that we showed that. Let $x \in C$, $x \ne 0$. In the equality above, consider the smallest $k$ so that $b_k\ne 0$. So we have
$$x^n + b_{n-1} x^{n-1} + \cdots + b_k x^k = 0$$ or
$$x^k( x^{n-k} + b_{n-1} x^{n-k-1} + \cdots + b_k) = 0$$
Since $x\ne 0$, the expression in the bracket must be $0$.
We get $x^{n-k} + b_{n-1} x^{n-k-1} + \cdots + b_k = 0$, and so 
$$x ( x^{n-k-1} + b_{n-1} x^{n-k-2} + \cdots + b_{k+1} )= - b_k\ne 0$$
We see that $x$ has an inverse in $C$.
In order to find an equation for an $x= a_0 + a_1 \omega + \cdots + a_{n-1} \omega^{n-1}$, we'll find an $n\times n$ matrix with rational elements and eigenvalue $x$. Indeed, consider the matrix $M=c_{ij} = a_{j-i}$, where the indexes are considered in $\mathbb{Z}/n=\{0,1, \ldots, n-1\}$. One now checks that the vector $(1, \omega,\ldots,\omega^{n-1})$ is an eigenvector of $M$ for the eigenvalue $\sum_{k=0}^{n-1} a_k \omega^k$. 
Indeed, we have
$$\sum_{j=0}^{n-1} c_{ij} \omega^j = \sum_{j=0}^{n-1} a_{j-i}\omega^j = \sum_{k=0}^{n-1} a_k \omega^{k+i} = (\sum_{k=0}^{n-1} a_k \omega_k) \cdot \omega^i$$
Therefore, $x$ is a root for the characteristic polynomial of $M$.
$\bf{Added:}$ We also get a "fun" way to produce the inverse of some elements in $C$. Let $0 \ne x = \sum a_k \omega^k$ and assume moreover that the circular matrix $A=(a_{i-j})$ is invertible ( this is equivalent to : all the numbers $x_m = \sum a_k \omega^{m k}$ are $\ne 0$, for $m=0,1,\ldots m-1$). The inverse of $A$ will be again a circulant matrix $A'=(a'_{j-i})$ and $x'=\sum a'_k \omega^k$ is the inverse of $x$. 
