A Counterexample to Kaplansky's Direct-Finiteness Conjecture in Odd Characteristic

1. Fix the coefficient field and low-rank graph factorization

If $p\leq m$, then $p$ divides both $m!$ and $(m!)^2+1$, a contradiction; hence $p>m$. Also $(m!)^2=-1$ in $\mathbb F_p$. Choose the least nonsquare $\eta\in\mathbb F_p^*$ and set

The norm $N_{K_0/\mathbb F_p}(\xi)=-\eta$ is nonsquare, so both extensions are quadratic and $|K|=p^4$. Fix $M=p^4-1$ and $\zeta\in K^*$ of order $M$.

The vertices of $\Gamma$ are the 600-subsets $I_i$ of $[1200]$, with adjacency defined by $|I_i\cap I_j|<200$. Three pairwise adjacent vertices would have a union of at least 1203 elements, so the graph is triangle-free. Multilinearizing the degree-400 polynomial

gives $F_{ij}=\sum_{T\in\mathcal T}\alpha_i(T)\beta_j(T)$, where $\mathcal T=\{T\subset[1200]:|T|\leq400\}$. Then $F_{ii}=1$, $F_{ij}=0$ for distinct nonneighbors, and $t=|\mathcal T|$ satisfies $t/m<1/100$. The denominators are invertible because $p>m$.

Lemma 2.1; Proposition 2.2 · pp. 3–4

2. Glue the directed annihilations while retaining the finite groups

Put $P=(\mathbb F_p)^3$, $A_0=(\mathbb Z/M\mathbb Z)^P$, and $H=A_0\rtimes P$. Let $c$ be the identity-coordinate function, $u(s)=(0,s)$, and $v(s)=cu(s)c^{-1}$. Then $U=u(P)$ and $V=cUc^{-1}$ have $U\cap V=1$. For $\ell=(p-1)/2$, the local idempotent satisfies

where $w=|H|$ and $d=\dim_K eK[H]$. Along each edge $i<j$, identify $v(1,\bar j,\bar j^2)$ in $H_i$ with $u(1,\bar i,\bar i^2)$ in $H_j$ to form $G_0$. Calling the shared element $y$, the right annihilator in $K[\langle y\rangle]=K[X]/(X^p)$ gives $e_i=qX^{p-\ell}$; since $p-\ell=\ell+1$, this implies $e_ie_j=0$. No reverse ordered product is assumed.

Proposition 4.2 establishes $H_i\hookrightarrow G_0$. Nonvanishing of the required local words of length at most three follows from Vandermonde independence, using $p>m$, and $U\cap V=1$. In Lemma 4.1, a hypothetical kernel yields a minimal spherical map with exactly one nonidentity face. Neutral edges, bridges, equal-type faces, and consecutive equal labels are reduced. Triangle-freeness and the short-word condition then force ordinary vertex degrees and face lengths to be at least four. The Euler sum is eight, whereas the exceptional face contributes at most three in the loop-free case; the remaining single-loop case contributes at most seven in total. Ordinary graph-of-groups normal form does not replace the injectivity proof for this first quotient.

Corollary 3.3 · p. 8

Lemma 4.1; Proposition 4.2 · pp. 9–13

3. Common-ring identities yield a certified matrix defect

Proposition 5.1 supplies $G_0\hookrightarrow G_1$ and a nonzero central idempotent $f$. In $S=fK[G_1]$, whose identity is $f$, put $P_i=fe_i$. The identities $Y_iJ_i=fI_d$, $J_iY_i=P_iI_w$ hold simultaneously at every vertex.

For $C_0=(\beta_j(T)P_j)_{T,j}$ and $D_0=(\alpha_i(T)P_i)_{i,T}$, the product is $\mathcal P+N$, where $\mathcal P=\operatorname{diag}(P_i)$ and $N$ is strictly lower triangular. Thus $N^m=0$ and $\mathcal PN=N=N\mathcal P$. A finite geometric sum corrects $D_0C_0$ to $\mathcal P$; taking $w$ copies and composing with $J_i,Y_i$ gives

The inequalities here are ordinary integer inequalities. Let $\iota:S^r\to S^{N_0}$ and $\pi:S^{N_0}\to S^r$ include and project onto the first $r$ coordinates. Set

Then $A_RB_R=I_r$. If $u_*$ has last coordinate $f$ and all others zero, $v_*=C^\sharp u_*\ne0$, $D^\sharp v_*=u_*$, and $A_Rv_*=0$. Therefore $(B_RA_R-I_r)v_*=-v_*\ne0$ certifies the reverse defect without deciding equality of group words.

Lemmas 6.1–6.2 · pp. 18–20

4. Preserve the defect under scalarization and finish the finite prescription

Choose the least $q\geq0$ with $n=M^q\geq r$, and pad $A_R,B_R$ to size $n$ by identity matrices. The finite Heisenberg group $\Lambda=(\mathbb Z/M\mathbb Z)^q\times(\mathbb Z/M\mathbb Z)^q\times\mathbb Z/M\mathbb Z$ has multiplication

The matrix units $\Omega_{uv}$ of Proposition 7.1 form a basis of a nonzero central sector $f'K[\Lambda]\simeq M_n(K)$. A shift/modulation representation sends them to standard matrix units, proving linear independence and hence injectivity of $\Phi:M_n(K[G_1])\hookrightarrow K[G_1\times\Lambda]$. Multiplication formulas alone would not establish injectivity.

Put $G=G_1\times\Lambda$, $g=\Phi(I_n)$, $a=1-g+\Phi(\widetilde A_R)$, and $b=1-g+\Phi(\widetilde B_R)$. Cross terms with the complement vanish, giving $ab=1$; injectivity gives $ba-1=\Phi(\widetilde B_R\widetilde A_R-I_n)\ne0$. The coefficient field remains the same $K$ throughout. Finitely many finite groups and the specified identifications generate $G$, while the embedded $H_i$ retain torsion. Every choice of field element, character, basis, or finite inverse occurs in an ordered finite set.

Proposition 7.1; Lemma 7.2; Corollary 7.3 · pp. 20–22

Proof of Theorem 1.1; finite prescriptions · pp. 22–23

1. Augmentation degree annihilates columns and rows

For translations $L_s$ on $K^P$, set $X_j=L_{\epsilon_j}-I$. Then $X_j^p=0$ and the vectors $X_1^{n_1}X_2^{n_2}X_3^{n_3}e_0$, $0\leq n_j<p$, form a basis. Put $D=3(p-1)$ and let $B$ consist of the basis columns of total degree at least $D-\ell+1$. Then

The first equality raises degree to at least $D+1$; the second follows by applying the same argument to $-s$ and transposing. Transpose involves no scalar conjugation.

Proposition 3.1; §3.1 · pp. 5–6

2. Vary the weights to obtain enough nondegenerate pairings

For a character $\lambda$ of $A_0$, put $D_\lambda=\operatorname{diag}((x\lambda)(c))$ and $N_\lambda=B^{\mathsf T}D_\lambda^{-1}B$. By Cauchy–Binet, $\det(B^{\mathsf T}\operatorname{diag}(\tau_x)B)$ has squares of maximal minors as coefficients of distinct monomials. Full column rank makes it nonzero. The degree-$h$ grid bound gives singular pairings a fraction at most $h/M<1/4$. Characters with nontrivial stabilizer also occupy a fraction below $1/4$, so more than half have free orbits and invertible $N_\lambda$. Choosing one representative from each orbit containing such a character gives $|\mathcal R|>|A_0|/(2p^3)$. Admissibility need not be constant on an orbit.

Lemma 3.2; §3.2 · pp. 6–7

3. Convert representation rank to regular rank

Fourier matrix units identify a free-character-orbit block with $M_{p^3}(K)$. Place

in each selected block and zero in the complement to define $e$. Since $T_\lambda B=B$, it is idempotent of rank $h$. Its left-multiplication image consists of matrices whose $p^3$ columns lie in $\operatorname{im}B$, so its regular rank is $hp^3$. Hence $d=hp^3|\mathcal R|$ and $d/w>h/(2p^3)>1/96$. The weighted row factor annihilates the conjugate translations $v(s)$ on the prescribed side. Only the prime-to-$p$ subgroup $A_0$ is diagonalized; semisimplicity of the entire modular group algebra is not assumed.

§3.3; Equations (3.6)–(3.7) · pp. 7–8

1. Align character sectors and control their stabilizers

Attach $L_i=(C_M)^{H_i}\rtimes H_i$ to $\langle z\rangle\times G_0$, identifying $z$ with the constant function $\delta_i$. Matching free actions orbit by orbit over $\langle z\rangle\times H_i$ proves that each factor embeds. Thus $z$ retains its order and the identity coefficient $M^{-1}$ of $f$ is nonzero.

The stabilizer $T_\theta$ of a character with $\chi(z)=\zeta$ meets $A_0$ trivially. Otherwise an element of order $q>1$ would force the exponent sum one to lie in $q(\mathbb Z/M\mathbb Z)$, a contradiction. Hence $T_\theta\hookrightarrow(\mathbb F_p)^3$ is elementary abelian, possibly nontrivial and nonnormal. Explicit matrix units and dimension comparison give

§§5.1–5.2; Lemma 5.2 · pp. 14–16

2. Conjugate to a coordinate projection over a nilpotent ideal

In one block put $D_\theta=K[T_\theta]$ and $E=\rho_\theta(P_i)$. Its augmentation ideal satisfies $\mathfrak a^{3(p-1)+1}=0$. Choose bases of the image and complementary image after augmentation, and form $Q$ from the corresponding columns of $E$ and $I-E$. Write $Q=Q_0(I+N)$ with entries of $N$ in $\mathfrak a$; a finite geometric sum gives its inverse. Thus $Q^{-1}EQ=\operatorname{diag}(I_{d_0},0)$.

Comparison with the original regular representation gives ordinary integer equalities $d=d_0|T_\theta|$, $w=s_\theta|T_\theta|$, and therefore $wd_0=s_\theta d$. Pairing these equally numerous coordinates constructs the rectangular matrices blockwise, and pulling back all blocks produces $J_i,Y_i$. The proof does not jump from numerical $K$-linear ranks to a free-module isomorphism over $S$.

§5.3; Equations (5.5)–(5.7) · pp. 17–18

1. Composition order and injectivity

Expanding two finite sums gives $T_cT_d=T_{cd}$. Thus $T_aT_b=T_{ab}=\operatorname{id}$, proving injectivity of $T_b$. The finite support of $b$ supplies a memory set, so it is a cellular automaton.

2. Characterize the image and exclude point indicators

If $y=T_bx$, then $T_{ba}y=T_bT_aT_bx=y$. Conversely, $T_{ba}y=y$ implies $y=T_b(T_ay)$, so $y$ lies in the image. Finally $(T_{ba}\delta_h-\delta_h)(1)=(ba-1)_h\ne0$ excludes $\delta_h$. These computations hold on the full function space, so allowing infinitely supported preimages does not change the conclusion.

Proposition 8.1; Equation (8.1) · p. 23