Conjecture about some group ring representations.

Question

In this link : http://bandtechnology.com/PolySigned/

A set of numbers is described : $P(N)$.

$ P(3),P(4),P(5),... $ are all (algebraicly closed) group rings.

Identify $PN$ with $R[X_N]/(1+X_N+(X_N)^2+...+(X_N)^{N-1})$

Then we get $P3 => R[X_3]/(1+X_3+(X_3)^2) => X_3 = ( 1^{1/3} )$

$P4 => ... => X_4 = ( ( 1^{1/4} , 1^{2/4} )$

Where $(.,.)$ means a couple.

The question becomes

$PN => X_N = ( ? , ? , ? , ... )$

($"?"$ are roots of unity but which ones ?)

The simplest conjecture is

$P(2N) => X_{2N} = ( 1^{1/2N} , 1^{2/2N} , 1^{3/2N} , 1^{4/2N} , ... , 1^{N/2N} )$

$ P(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 ?

Answer 1

Let $M=[0,\infty)$ be a semiring. It is unclear if the polysign numbers $P_n$ described in the link are the "group semiring" construction $M[C_n]$, or if they are the group ring ${\Bbb R}[C_n]$. In the former case, the element $1+(-1)$ (in $M[C_2]$ say) cannot be simplified to $0$, whereas in the latter it can. But in either case, $M[C_n]$ and ${\Bbb R}[C_n]$ will never be a field for $n>1$, since they contain zero divisors and nontrivial idempotents (those besides $0$ and $1$) and $M[C_n]$ is not even a ring.

You are probably asking if/when the group ring ${\Bbb R}[C_n]$ is isomorphic to ${\Bbb R}[\zeta_n]$ where $\zeta_n$ is a primitive root of unity. The answer for $n>1$ is: never. First of all you should familiarize yourself with the basic constructions involved (which I gather you are not, based on our chat):

• A group $G$ is a set with an associative binary operation with an identity element under which all elements are invertible. If we omit the existence of inverses, we have a monoid, and if we further omit the existence of an identity element, we have a semigroup.
• A (unital) ring $R$ is a set with addition, multiplication, $0$, and $1$, satisfying a bunch of properties ($R$ is an abelian group under $+$ with identity $0$, and is a monoid under $\times$ with identity $1$ and "absorption" element $0$, in which $\times$ distributes over $+$, for example is one formulation).
• A polynomial ring $R[x]$ is the ring of all polynomials in a formal variable $x$ with coefficients taken from the ring $R$.
• If $S$ is a ring containing a subring $R$, and $s\in S$, then $R[s]$ is the smallest subring of $S$ containing both $R$ and $s$. Its elements look like polynomials "in $s$" with coefficients from $R$, but it is not the same as a polynomial ring $R[x]$, because if $s$ is algebraic over $R$ then there are polynomials in $s$ equal to zero (if $x$ is a formal variable, no nonzero polynomial in it is $0$.)
• If $G$ is a group and $R$ a ring, $R[G]$ is the set of formal (and finite) "$R$-linear combinations" of the elements from $G$, with coefficients taken from $R$. Integral group rings are $\Bbb Z[G]$, and group algebras are when $R$ is a field. Sometimes the brackets are omitted (e.g. $\Bbb ZG$). The operation of multiplication is determined by imposing distributivity, the usual multiplication between elements of $R$, and the usual group operation between elements of $G$.
• A (left) ideal of a ring $R$ is an additive subgroup which absorbs ambient (left) multiplication; that is a set $I$ such that $I+I$ and $RI$ are subsets of $I$. If $R$ is commutative, then equivalently, an ideal $I\triangleleft R$ is a kernel of some ring homomorphism $R\to S$. The set of all multiples of an element $a\in R$, denoted $Ra$ or $(a)$, is called a principal ideal.
• If $I\triangleleft R$ is an ideal $R$, the quotient ring is the set of "cosets" $r+I=\{r+i:i\in I\}$ with addition $(r_1+I)+(r_2+I):=(r_1+r_2)+I$ and $(r_1+I)(r_2+I):=r_1r_2+I$.
• A principal ideal is the set of multiples of a given element, $(a)=\{ra:r\in R\}$.

This is a lot of information to take in. This takes a lot of algebra background to grok fully and sufficiently, and I recommend you learn some abstract algebra, at least enough to understand the above concepts. For our purposes, let's let our rings and groups be commutative.

One of the most ubiquitous constructions in algebra is the quotient of a polynomial ring. An important part of characterizing these extensions is the (general abstract algebra version) of the Chinese Remainder Theorem: if $I,J\triangleleft R$ with $I+J=R$ ("coprime") then $R/IJ\cong R/I\times R/J$.

In particular, consider two polynomials $p(x),q(x)\in F[x]$ that share no common factor, where $F$ is a field (and hence $F[x]$ is what's called a PID). Then $F[x]/(p(x)q(x))\cong F[x]/(p(x))\times F[x]/(q(x))$.

CRT generalizes to arbitrarily many pairwise coprime polynomials.

Note that for $n>2$, $\Bbb R[\zeta_n]=\Bbb C$. On the other hand, the cyclic group algebras are

$${\Bbb R}[C_n]\cong {\Bbb R}[X]/(X^n-1)\cong{\Bbb R}^a\times{\Bbb C}^b$$

where $a=1$ if $n$ is odd and $a=2$ if $n$ is even, and $b=(n-a)/2$. This can be obtained by considering $X^n-1=\prod(X-\zeta)$ over all $n$th roots of unity $\zeta$, and pairing the complex $\zeta$s according to conjugates in order to obtain the irreducible factors (over $\Bbb R$): $X-1$, $X+1$ if $n$ is even, and $X^2-\cos(2\pi k/n)X+1$ for various $k$ otherwise (via $\zeta_n^k+\bar{\zeta}_n^k=\cos(2\pi k/n)$). If we quotient ${\Bbb R}[x]$ by a linear polynomial (i.e. $x-1$ or $x+1$) we obtain $\Bbb R$, and if we quotient ${\Bbb R}[x]$ by a quadratic irreducible we obtain $\Bbb C$ (this is all up to isomorphism). This tells us that $a=1,2$ depending on if $n$ is odd or even, and to obtain $b$ we simply note that $\Bbb C$ has dimension $2$ as a real vector space and $\Bbb R[C_n]$ has dimension $n$.

Things get more interesting for rationals: $\Bbb Q[C_n]\cong\prod_{d\mid n}\Bbb Q(\zeta_n)$ is a product of cyclotomic fields.