Conjecture about some group semiring representations ( and roots of unity ). Let $\Bbb R_+=[0,\infty)$ be a semiring.  $\Bbb R_+[C_n]$ is the group semiring formed by the semiring $\Bbb R_+$ and the cylic group $C_n$.
Let $\Bbb R_+[X_n]$ be the polynomial semiring.
(polynomials of $X_n$ with positive real coef. )
$\Bbb R_+[C_n]$ is isomorphic with $\Bbb R_+[X_n]/(1+X_n+(X_n)^2+...+(X_n)^{n-1})$
For example $\Bbb R_+[C_3]$ is isomorphic to $\Bbb R_+[X_3]/(1+X_3+(X_3)^2)$.
We can give the isomorphism explicit by giving $ X_3 = ( 1^{1/3} )$
$1^{1/3} = \exp\left(\frac{2 \pi i}{3}\right)$
Thus we write : $\Bbb R_+[C_3] => X_3 = (1^{1/3})$
Likewise we get $\Bbb R_+[C_4] => X_4 = ( 1^{1/4} , 1^{2/4} )$
Or in matrix form : $X_4 = \begin{pmatrix} 1^{1/4} & 0  \\ 0 & 1^{2/4}  \end{pmatrix}$
Let $(*, * , ... , * )$ denote the so-called diagonal matrix :
$\begin{pmatrix} * & 0 & \cdots & 0 \\ 0 & * & \cdots& 0\\ \vdots & \vdots & \ddots & 0\\ 0 & 0 & 0 & * \end{pmatrix}$
(by which I mean a square matrix filled with zero's apart from $ * $ on the diagonal.)
The question becomes what the explicit isomorphisms for $X_n$ look like.
$\Bbb R_+[C_n] => X_n = ( "?" , "?" , "?" , ... )$
($"?"$ are roots of unity but which ones ? )
The simplest conjecture is
$M[C_{2n}] => X_{2n} = ( 1^{1/2n} , 1^{2/2n} , 1^{3/2n} , 1^{4/2n} , ... , 1^{n/2n} )$
$ \Bbb R_+[C_{2n+1}]$  => 
$X_{2n+1} = ( 1^{\frac{1}{2n+1}} , 1^{\frac{2}{2n+1}} , 1^{\frac{3}{2n+1}} , 1^{\frac{4}{2n+1}} , ... , 1^{\frac{n}{2n+1}} ) $
But is this conjecture true?
 A: Note that $M[C_n]$ and $M[X]/(1+X+\cdots+X^{n-1})$ are not the same. For one thing, the sum of the first $n$ powers of a generator of $C_n$ (starting with $X^0:=1$) is nonzero by the group semiring construction, whereas the sum of the first $n$ powers of $X$ is zero in $M[X]/(1+X+\cdots+X^{n-1})$; as another thing, the nonzero elements in $M[C_n]$ have no additive inverses, whereas we will see that the quotient polynomial semiring is actually a ring.
In fact, $M[X]/(1+X+\cdots+X^{n-1})\cong{\Bbb R}[X]/(1+X+\cdots+X^{n-1})$. A nice clean isomorphism is induced simply from $M\hookrightarrow {\Bbb R}$; to see that this semiring homomorphism is surjective, try proving that every coset in ${\Bbb R}[X]/(1+X+\cdots+X^{n-1})$ can be represented by a polynomial with positive coefficients (first write a representative with $\deg<n$, then add $1+X+\cdots+X^{n-1}$ to it sufficiently many times). My explanation in your previous question leads to
$${\Bbb R}[X]/(1+X+\cdots+X^{n-1})\cong\begin{cases}{\Bbb C}^{(n-1)/2} & n~{\rm odd} \\ {\Bbb R}\times{\Bbb C}^{n/2-1} & n~{\rm even}\end{cases}$$
as an $\Bbb R$-algebra. If $n=2$ then $M[\zeta_n]=\Bbb R$, else for $n>2$, $M[\zeta_n]=\Bbb C$ (since there will be an $n$th root $\zeta$ with negative real part so $\zeta+\zeta^{n-1}\in M[\zeta_n]$ is a negative real) as subsemirings of $\Bbb C$.
Therefore, $M[X]/(1+X+\cdots+X^{n-1})\cong M[\zeta_n]$ if and only if $n=2$ or $3$. [[Moreover if $A$ is any diagonal complex matrix whose entries are $n$th roots of unity at least one of which is primitive, then $M[A]\cong M[\zeta_n]$ (and this is why working with matrices is superfluous decoration which only serves to obscure what's really going on in this case).]] Correction: if $A$ is a diagonal matrix comprised of at least one of each conjugate pair of $n$th roots of unity (excluding $1$), then the minimal polynomial of $A$ over $\Bbb R$ will be $X^{n-1}+\cdots+X+1$, in which case $$M[A]\cong{\Bbb R}[A]\cong{\Bbb R}[X]/(X^{n-1}+\cdots+X+1)\cong M[X]/(X^{n-1}+\cdots+X+1)$$ The matrix conjecture is therefore true.
