Kaplansky’s zero-divisor conjecture

Can two nonzero weighted sums of reversible moves multiply to zero when repeating any nonidentity move never gives the do-nothing move? A group records how moves compose and undo each other; its identity is the do-nothing move. To multiply two finite weighted sums, compose every pair of moves and add the weights of equal results; negative weights can cancel positive ones. If repeating a nonidentity move three times does give the identity, complete cancellation is possible. The conjecture asks whether forbidding all such finite-order moves prevents it, for every field: a coefficient system with the usual arithmetic laws, including division by nonzero coefficients.[1]

Can two nonzero finite weighted sums of reversible moves multiply to zero when repeating any nonidentity move never gives the identity operation? This is Kaplansky’s zero-divisor question. A group records how moves compose and undo each other. A field supplies coefficient arithmetic with division by nonzero coefficients. In the group algebra $K[G]$, multiply basis elements using the group law and distribute over their coefficients. For a nonidentity element $g$ of order three, $(1-g)(1+g+g^2)=1-g^3=0$: the symbols $1,g,g^2$ are distinct, so neither factor is zero, in any characteristic. By contrast, an infinite-order generator $t$ gives $(1-t)(1+t+t^2)=1-t^3\ne0$. The torsion-free hypothesis excludes the first mechanism; the conjecture concerns all possible finite supports and coefficients, not only powers of one generator.[1]

Many families of groups have been shown to avoid this cancellation, but the statement for every torsion-free group and every field remains open. Finding repeated ways to compose the same move is not enough: the weights must cancel at every resulting group element. Counterexamples to the different unit conjecture concern unexpected multiplicative inverses, rather than a product of two nonzero sums equaling zero.[1] [2]

Division-ring embeddings prove stronger conclusions in covered classes. Non-unique-product examples alone are not zero divisors.[1] [2]

An individual move can always be undone, but a weighted sum of moves may lose that cancellation property. If two sums become equal after multiplication by the same nonzero sum, can we recover that they were equal to begin with? Zero divisors are exactly the obstruction. The problem connects reversible group operations with the reliability of algebraic cancellation after addition is introduced.[1]

The connection to cancellation is exact. If $ab=ac$, distributivity gives $a(b-c)=0$. Absence of zero divisors would therefore allow cancellation of any nonzero left factor; the corresponding argument works for right factors. This transfers a basic feature of the group law to finite weighted sums, where different products can coincide. Division-ring methods establish it in covered group classes, but many coinciding products alone do not decide whether coefficients can cancel everywhere.[1] [2]