Example. The half-turn in characteristic two [fdip-000N]
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  • October 17, 2026

The coefficient assumption in the two-sector decomposition is essential. Let \(k\) now be a nonzero commutative ring of characteristic two. Use the same \(C_2\), group-basis element \(g\), coordinate modules \(V_k,W_k\), and maps \(P,J\). The faithful algebra actions still exist, but the algebra is \[k[C_2]\cong k[z]/(z^2),\] and its idempotents are exactly the scalar idempotents of \(k\). In particular, if \(k\) is a field, it has no idempotents other than \(0,1\), so it cannot be isomorphic to \(k\times k\).

The Group Algebra still has its group basis, as in [sengupta2010representations, Sections 3.1-3.2, pp. 39-41]. The character projectors in [sengupta2010representations, Section 4.5, p. 63] require the group order to be invertible in the field. The following calculation identifies what replaces that product when the order is zero in \(k\).

Proof. 

  • October 17, 2026

Put \(\eta =g-1\). Its coefficient at \(g\) is \(1\), so \(\eta \ne 0\); nevertheless \[\eta ^2=g^2-2g+1=0.\] Since \(g=1+\eta \), the pair \(1,\eta \) is a basis: \[c+d\eta =(c-d)+dg,\] so coefficient comparison proves independence as well as spanning. Evaluation at \(\eta \) therefore induces a unital \(k\)-algebra map \[\begin {aligned} \Psi :k[z]/(z^2)&\longrightarrow k[C_2],\\ [f(z)]&\longmapsto f(\eta ). \end {aligned}\] It is well defined because \(\eta ^2=0\). Every polynomial has a unique remainder \(c+dz\) modulo the monic polynomial \(z^2\): write \(f=c+dz+z^2q\), and observe that a multiple of \(z^2\) has zero constant and linear coefficients. Hence the quotient has basis \([1],[z]\), even when \(k\) has zero divisors. The map sends it to the basis \(1,\eta \), so it is bijective, with inverse \(a+bg\mapsto [(a+b)+bz]\). Evaluation and its inverse preserve the unit and scalars. The algebra \(k[z]/(z^2)\) is the algebra of dual numbers over \(k\); its nonzero element \([z]\) has square zero.

If \(c+d\eta \) is idempotent, commutativity and characteristic two give \[(c+d\eta )^2=c^2.\] Comparison in the basis \(1,\eta \) forces \(d=0\) and \(c^2=c\); conversely every such scalar is idempotent. If \(k\) is a field, or more generally has no zero divisors, \(c(c-1)=0\) implies \(c=0\) or \(c=1\). Both are idempotent. In contrast, \((1,0)\) is a nonzero idempotent different from the unit in \(k\times k\). An algebra isomorphism preserves these properties, which rules out the product. The field hypothesis matters: a general characteristic-two ring can already have nontrivial scalar idempotents. For example \(k=\mathbb F_2\times \mathbb F_2\) has 4 scalar idempotents, including \((1,0)\) and \((0,1)\); all 4 remain idempotent in \(k[z]/(z^2)\). Thus the assertion of exactly two idempotents does not extend to every characteristic-two ring, nor does it rule out all product decompositions in that generality.

On the vertex module, the same half-turn still satisfies \(P^2=I\), while \(-1=1\). Its two proposed eigenspaces now coincide: \[\begin {aligned} \ker (P-I)&=\ker (P+I)\\ &=\{(x,y,x,y):x,y\in k\}. \end {aligned}\] Its coordinate basis is \(\delta _0+\delta _2,\delta _1+\delta _3\): coordinates zero and one recover the coefficients. It is free of rank two (dimension two over a field); adding it to itself does not recover the full module, because \(\delta _0\) is not in it. On the retained module, \[\begin {aligned} \ker (J-I)&=\ker (J+I)\\ &=\{(x,x):x\in k\}, \end {aligned}\] whose basis is \(\delta _0^R+\delta _2^R\), with its coefficient recovered at zero. It has free rank one, and dimension one over a field. These equalities follow directly from the coordinate formulas for \(P,J\).

The nonzero nilpotent is visible in the faithful action as well. For \(N=P-I\), \[\begin {aligned} N^2&=P^2-2P+I=0,\\ N\delta _0&=\delta _2-\delta _0\ne 0. \end {aligned}\] The last vector has coefficient \(1\) at coordinate \(2\). Similarly, \(J-I\) is nonzero on \(\delta _0^R\) and has square zero: \((J-I)\delta _0^R=\delta _2^R-\delta _0^R\ne 0\). Both coordinate injectivity proofs for the algebra actions in Definition [fdip-000L] still apply. Thus the nilpotent \(\eta \) has a faithful nonzero action on both modules. The expressions \((I\pm P)/2\) and \((I\pm J)/2\) are undefined because \(2=0\). The action and projection intertwiner survive, but the complementary projectors and eigenspace direct sum do not. This is a change of coefficient ring for the algebra and modules; the real Gaussian law is not being extended to characteristic two.

TauCeti's MonoidAlgebra.isIdempotentElem_cyclicTwo_iff proves the idempotent obstruction more generally for a commutative nonzero ring without zero divisors in which \(2\) is not a unit. A characteristic-two field meets these hypotheses. The generator is identified with \(\operatorname {Multiplicative.ofAdd} (1:\mathbb Z/2\mathbb Z)\) as in Theorem [fdip-003O]. Indeed, if \(a+bg\) is idempotent, coefficient comparison gives \(a^2+b^2=a\) and \(2ab=b\). Hence \(b(2a-1)=0\). If \(b\ne 0\), absence of zero divisors gives \(2a=1\), contradicting the nonunit hypothesis. Therefore \(b=0\) and \(a\in \{0,1\}\). This argument does not require characteristic two, but does require absence of zero divisors. The dual-number presentation and the coordinate nilpotent above are separate characteristic-two calculations.