Example. Clock and shift [fdip-00AP]

✍️sourceAGENTDRAFTED

Let \(N\ge 1\) and let \(R\) be a commutative unital ring containing an element \(\omega \) with \(\omega ^N=1\). On \(\mathbb Z/N\mathbb Z\) set \(\chi (i)=\omega ^i\), let \(U=M_\chi \) be the multiplication operator of \(\chi \), and let \(T\) be the cyclic shift. Lemma [fdip-0013] gives \(TU=M_{T\chi }T\). Since \(\,(T\chi )(i)=\chi (i+1)=\omega ^{i+1}=\omega \chi (i)\), one has \(M_{T\chi }=\omega U\), and therefore \(TU=\omega UT\). When \(\omega \ne 1\) the two products differ. Also \(T^N=I\) and \(U^N=M_{\chi ^N}=I\). If \(\omega \) is primitive of order \(N\), the operators \(T\), \(U\) and the powers of \(\omega \) generate a finite Heisenberg group. The four-cycle is the case \(N=4\).

The continuum form is the commutator of differentiation with multiplication by the coordinate. For a smooth function \(f\), \[\frac {d}{dx}(xf)-x\frac {d}{dx}f=f,\] so \(\bigl [\tfrac {d}{dx},x\bigr ]=1\). With \(p=-i\hbar \,d/dx\) this is \([p,x]=-i\hbar \).

The same relation holds over the field \(\mathbb F_p\) when \(N\) divides \(p-1\). Then \(X^N-1\) divides \(X^{p-1}-1=\prod _{c\in \mathbb F_p^\times }(X-c)\). The divisor is separable, because \(p\) does not divide \(N\) and \(X^N-1\) shares no root with its derivative, so it has \(N\) distinct roots in \(\mathbb F_p\). Those roots form a subgroup \(G\). Let \(m\) be the least common multiple of the orders of its elements, and factor \(m\) into prime powers. For each prime power \(q\) in that factorization, some element of \(G\) has order divisible by \(q\), and a power of it has order \(q\). The product of one such element for each prime power has order \(m\): the factors commute, and their orders are coprime. Every element of \(G\) is a root of \(X^m-1\), and a field has at most \(m\) roots of that polynomial, so \(N\le m\). Orders in \(G\) divide \(N\), so \(m\) divides \(N\). Hence \(m=N\), and \(G\) contains a primitive \(N\)-th root.