Do polynomials of a root of unity constitute a field? For some (strictly) positive integer $n$, let $\omega_n = e^{2\pi i/n}$ be the principal $n$th root of unity, and let $F$ be a subfield of $\mathbb C$ (e.g. $\mathbb R$ or $\mathbb Q$ or $\mathbb C$ itself). Let $P_n(F)$ be the set of polynomials in $\omega_n$ with coefficients in $F$; that is, the elements of $P_n(F)$ are
$$
 \sum_{k=0}^{n-1} a_k\omega_n^k,\qquad a_k\in F.
$$
Furthermore, let $P_n(F)$ be equipped with the usual addition and multiplication operations of complex numbers.
This clearly results in a commutative ring, but my question is: Is P_n(F) actually a field?
(Clearly, most field properties are automatically satisfied. The only non-obvious feature is invertibility.)
Note: this question has some similarities to this one.
 A: This looks like an application of isomorphism theorem as you have already observed that $P_n(F)$ is a commutative ring. Consider the map $\alpha:F[x] \to P_n(F)$ defined by $\alpha(f)= f(\omega_n)$. It is easy to see that $\alpha $ is a ring homomorphism and it's kernel $I$ is the ideal generated by minimal polynomial of $\omega_n$. Further observe that $ \alpha$ is onto and $I $ is Maximal and it follows that $F[x]/I \cong P_n(F)$ is a field as $I$ is Maximal.
Update:
Say $g$ is a least degree monic polynomial satisfied by $\omega_n$ with coefficients in $F$ (termed as minimal polynomial of $\omega_n$), then the ideal $I=(g(x))$ in $F[x]$ is maximal. To see this prove this you may use the following

*

*Every ideal in $F[x]$ is generated by a single element (PID).

*Division algorithm.

Here you can see a proof.
For the last part,there is a theorem which says that if $R$ is a commutative ring with unity, $I$ is maximal ideal iff $\frac{R}{I}$ is a field. Proofs can be found here.
A: OK, I realized the answer to this question as I was typing it up, but I decided to post it anyway so as to not waste the effort, and in case anyone is interested.
First note that the proof is trivial for $n=1$, since $P_1(F) = F$.
With $n>1$, and given $a_k,k=1,\ldots,n-1$ such that $\sum_{k=0}^{n-1}a_k\omega_n^k\neq 0$, our goal is then to find $b_k,k=1,\ldots,n-1$ such that
$$
    \left(\sum_{k=0}^{n-1}a_k\omega_n^k\right)\left(\sum_{k=0}^{n-1}b_k\omega_n^k\right)=1.
$$
Expanding out this product and grouping by powers of $\omega_n$, we find
$$
    \left(\sum_{k=0}^{n-1}a_kb_{n-k}\right)\omega_n^0 + \left(\sum_{k=0}^{n-1}a_kb_{n-1-k}\right)\omega_n^1 + \ldots + \left(\sum_{k=0}^{n-1} a_kb_{n-m-k}\right)\omega_n^m + \ldots + \left(\sum_{k=0}^{n-1} a_kb_{1-k}\right)\omega_n^{n-1}.
$$
(Throughout this proof, index addition is performed modulo $n$.)
We require that the $m=0$ sum be 1, and all the others 0.[1]
This can be put as the vector equation
$$
   \mathbf A\boldsymbol b = \boldsymbol e,
$$
where $\boldsymbol b=(b_{n-1},\ldots,b_1,b_0)$, $\boldsymbol e=(1,0,\ldots,0)$, and $\mathbf A_{ij} = a_{i+j}$, i.e. $\mathbf A$ is the matrix whose row (and column) vectors are cyclic permutations of $(a_0,a_1,\ldots,a_{n-1})$, with the identity permutation in the top row. I claim that either this has a solution $\boldsymbol b$, or $\sum_{k=0}^{n-1}a_k\omega_n^k=0$.
I will do this by showing that if $\sum_{k=0}^{n-1}a_k\omega_n^k\neq 0$, then the top row of $\mathbf A$ is linearly independent of the other rows.
If that is the case, then there exists a vector $\hat{\boldsymbol b}$ that is orthogonal to all row vectors of $\mathbf A$ except the top one.
Scaled appropriately, this is our solution $\boldsymbol b$.[2]
Let us then assume that two rows in $\mathbf A$ are identical.
By the construction of $\mathbf A$, this implies that $n$ has a divisor $m$, with $1\leq m<n$, such that $a_{k+m}=a_k$ for all $k$.
However, then
$$
    \sum_{k=0}^{n-1}a_k\omega_n^k = \left(\sum_{\ell=0}^{n/m-1}\omega_n^{\ell m}\right)\sum_{k=0}^{m-1}a_k\omega_n^k = 0,
$$
since the parenthesis contains the sum of all $(n/m)$th roots of unity, which is zero.
Thus, we conclude that no two rows of $\mathbf A$ are identical.
Since clearly no row is a multiple of another, this gives $\mathrm{rank}(\mathbf A)\geq 2$ and implies the existence of a row that is linearly independent of all other rows.
This proves that $P_n(F)$ is a field, since all its nonzero elements have inverses.
It also provides a straightforward construction of those inverses.
This proof makes decisive use of the way roots of unity sum to zero.
This raises questions about possible generalizations: can we similarly build a field out of some group other than that of $n$th roots of unity, given that we can equip the group elements with a sensible behaviour under the addition operation of $F$?

[1] If $n$ is even, we can also let the $m$th sum cancel the $(m+\frac n2)$th sum, but guided by the knowledge that the inverse, if it exists, is unique, we quickly find that doing this does absolutely nothing to help us find the inverse. Thus, we ignore it, and tackle even $n$ the same way as we do odd ones.
[2] $\hat{\boldsymbol b}$ can be found in practice by first setting it equal to the top row, and then applying Gram-Schmidt or similar to make it orthogonal to all the other rows. If $\mathbf A$ is invertible, this reduces to solving the linear system $\mathbf A\boldsymbol b = \boldsymbol e$, but our method works also if $\mathbf A$ is singular.
