gpt 6 astra ultra attempt on Math-289

Kaplansky's Idempotent Conjecture

Definitions and mathematical statement

For a field $K$ and group $G$, the ordinary group ring consists of finite sums $\sum_g a_gg$, with multiplication extended bilinearly from $G$. An idempotent is an element $e$ satisfying $e^2=e$. The assertion is \[ G\text{ torsion-free},\quad e\in K[G],\quad e^2=e \quad\Longrightarrow\quad e=0\text{ or }e=1. \] Torsion-free means no nonidentity element has finite order. Fields of every characteristic and groups without finite-generation hypotheses are included. Matrix rings, operator-algebra completions, and infinite formal sums are not in scope. In particular, an idempotent in one of those larger objects is not a counterexample to this scalar finite-support statement.

Formulation and concept references: [S1].

Short English statement

Are zero and one the only idempotents in the ordinary group ring of every torsion-free group over every field?

Research attempt: leading terms, coefficient reduction, and traces

GPT-6 Astra Ultra, 27 September 2026. Status: UNRESOLVED.

The group ring in the question consists of scalar elements with finite support. I retain that restriction throughout the algebraic arguments. A leading-term proof settles an ordered-group subcase over every field; a specialization argument reduces the possible coefficient fields; and a trace calculation explains a useful sufficient condition in characteristic zero. None of these arguments proves that torsion-freeness alone supplies the missing condition.

A complete proof for bi-orderable groups.

Suppose $G$ has a total order invariant under multiplication on both the left and the right. For nonzero finite sums \[ a=\sum_{g\in A}a_g g,\qquad b=\sum_{h\in B}b_h h, \] let $g_0,h_0$ be the respective largest support elements. If $g\le g_0$ and $h\le h_0$, right and left invariance give \[ gh\le g_0h\le g_0h_0. \] If either input inequality is strict, one of these inequalities is strict. Thus $g_0h_0$ is obtained uniquely from the pair $(g_0,h_0)$. Its coefficient in $ab$ is $a_{g_0}b_{h_0}\ne0$ because $K$ is a field. Hence $K[G]$ has no zero divisors.

An idempotent satisfies $e(e-1)=0$. The domain conclusion therefore gives $e=0$ or $1$. This proves the conjecture for bi-orderable groups in every characteristic. In particular, lexicographic order on $\mathbb Z^d$ gives the usual Laurent-polynomial-ring instance.

The same noncancellation argument works whenever every two nonempty finite subsets $A,B$ have at least one uniquely represented element of $AB$; this is the unique-product property. One does not need an order in that argument. But torsion-freeness is not the unique-product property, and it does not justify choosing a multiplicatively invariant order. The broader graded-ring results in [S1] explicitly distinguish these hypotheses. In a general torsion-free group, several support pairs can give the same product, so coefficient cancellation is precisely what the leading-term proof fails to control.

Any counterexample is already finitely generated.

If $e$ has finite support $A$, let $H=\langle A\rangle$. Then $e\in K[H]$, and the equation $e^2=e$ is unchanged when read in that subring. If $G$ is torsion-free, so is $H$. Thus no infinite-generation phenomenon is needed for a counterexample: it would occur in a finitely generated torsion-free subgroup.

There is a parallel coefficient reduction. Let $e$ be nontrivial. Some coefficient $a_g$ with $g\ne1$ is nonzero, since a scalar idempotent over a field is $0$ or $1$. Form the algebra over the prime field generated by the finitely many coefficients of $e$ and by $a_g^{-1}$. The idempotency equations are finite polynomial identities in this algebra.

In characteristic zero, the fraction field of that finitely generated algebra embeds in $\mathbb C$: choose images of a finite transcendence basis algebraically independent over $\mathbb Q$, then extend across the finite algebraic extension. Coefficientwise embedding carries $e$ to a nontrivial idempotent in $\mathbb C[H]$.

In characteristic $p$, take a maximal ideal of the finitely generated $\mathbb F_p$-algebra. By the field form of the finite-generation lemma, its residue field is a finite extension of $\mathbb F_p$, hence a finite field. Reduction preserves $e^2=e$, while $a_g$ remains nonzero because its inverse was included. The reduced idempotent is still nontrivial.

Consequently it would suffice to prove the conjecture for finitely generated torsion-free groups over $\mathbb C$ and over all finite fields. This is a genuine reduction of quantifiers, not a proof of either remaining case. In particular, a characteristic-zero theorem alone does not automatically settle positive characteristic: the specialization argument runs from a positive-characteristic counterexample to a finite field of the same characteristic.

A trace criterion in characteristic zero.

For this paragraph take $K=\mathbb C$. The left regular action embeds $\mathbb C[G]$ into its group von Neumann algebra $\mathcal N(G)$, with faithful tracial state \[ \tau\Big(\sum_g a_g g\Big)=a_1. \] An algebraic idempotent need not be self-adjoint, so positivity of its trace should not be asserted without an argument. Put \[ Q=e^*e+(1-e^*)(1-e). \] For every vector $v$ in the regular Hilbert space, \[ \langle Qv,v\rangle=\|ev\|^2+\|(1-e)v\|^2 \ge\tfrac12\|v\|^2. \] Thus $Q$ is positive and boundedly invertible. A direct calculation using $e^2=e$ gives \[ Qe=e^*e=e^*Q. \] With $S=Q^{1/2}$, the element $P=SeS^{-1}$ is a self-adjoint idempotent, hence an orthogonal projection. Functional calculus keeps $S$ and its inverse in $\mathcal N(G)$. Traciality gives \[ \tau(e)=\tau(P)\in[0,1]. \tag{1} \] Faithfulness of the trace shows that $\tau(e)=0$ forces $e=0$, and $\tau(e)=1$ forces $e=1$, by applying the same reasoning to $1-P$. Therefore a nontrivial idempotent would satisfy $0<\tau(e)<1$.

It follows that the additional assertion \[ \tau(e)\in\mathbb Z\quad\text{for every idempotent }e\in\mathbb C[G] \tag{2} \] would prove the characteristic-zero conjecture. More generally, any theorem forcing the range dimension of the associated projection to be an integer supplies $2$. This is the form in which suitable integral $L^2$-dimension results can be relevant; no such theorem for arbitrary torsion-free groups is assumed here.

There is an unconditional boundary case: for $e\in\mathbb Z[G]$, its identity coefficient is already an integer. Equation $1$ then proves that $\mathbb Z[G]$ has only trivial idempotents, for every group $G$. This does not extend to rational coefficients by merely clearing denominators. If $ne\in\mathbb Z[G]$, its equation is $(ne)^2=n(ne)$ rather than $(ne)^2=ne$; the trace of $e$ may still be a nonintegral rational number.

Torsion and finite quotients are useful stress tests.

If a group contains an element $g$ of finite order $m>1$ and $\operatorname{char}K\nmid m$, then \[ e=\frac1m\sum_{j=0}^{m-1}g^j \] is a nontrivial idempotent. Every power of $g$ occurs exactly $m$ times in the square, so $e^2=e$. Over $\mathbb C$, its trace is $1/m$, confirming that $1$ alone cannot force triviality. The torsion-free hypothesis must enter a further argument, such as the leading-term property or $2$.

Finite quotients can have such idempotents even when the original group is torsion-free. For example, $\mathbb Z\to\mathbb Z/2\mathbb Z$ sends $t$ to an involution. The quotient contains $(1+\overline t)/2$. Its natural lift in $\mathbb Q[t,t^{-1}]$ satisfies \[ \left(\frac{1+t}{2}\right)^2-\frac{1+t}{2} =\frac{t^2-1}{4}\ne0. \] Thus a finite quotient idempotent does not automatically lift. Conversely, finding projections in an operator-algebra completion does not show that their Fourier expansion has finite support. The completion was used above only to study an idempotent already present in the ordinary group ring.

Direct finiteness is also insufficient by itself: the matrix algebra $M_2(\mathbb C)$ is directly finite, but contains $\operatorname{diag}(1,0)$. This is not a group-ring counterexample; it shows that a general ring-theoretic implication from direct finiteness to absence of nontrivial idempotents is false. Likewise a nontrivial unit is not an idempotent. Gardam's counterexample to the unit conjecture [E1] must not be promoted to a counterexample to the separate statement here.

Verification and outcome.

The exact checks multiply finite-support Laurent polynomials, verify the finite-cyclic averaging idempotents and the failed lift, and verify $Qe=e^*Q$ for a nonsymmetric rational idempotent matrix. The matrix test checks the trace argument's algebra; it is not put forward as an element of $K[G]$.

The proved output is the ordered/unique-product subcase, the finite-generation and coefficient-field reduction, and the trace criterion with its integral-group-ring consequence. The first missing step is controlling cancellation for arbitrary torsion-free groups or, in characteristic zero, deriving the necessary trace integrality. No argument for that universal step, and no counterexample in the stated scalar finite-support setting, is supplied. The conjecture remains unresolved.

Sources

• S1 Johan Öinert, Units, Zero-divisors and Idempotents in Rings Graded by Torsion-Free Groups (arXiv:1904.04847v3), Problem 1(c); Gardam $2021$, introduction. https://arxiv.org/pdf/1904.04847. Formulation source.
• E1 G. Gardam, A counterexample to the unit conjecture for group rings. This concerns units and is cited only to distinguish that result from the idempotent problem. https://arxiv.org/abs/2102.11818.
• E2 J. Öinert and S. Wagner, Complex group rings and group $C^*$-algebras of group extensions. Distinguishes ordinary group-ring statements from their completion analogues. https://arxiv.org/abs/1811.06424.