A finite height bound for rank-one short presentations

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Let $P$ be a primitive positive integer set, meaning $\gcd(P)=1$, and put $d=d(P)\ge2$. Let $R$ be the rational span of its zero ternary relations with support at most $d+1$ in $\mathbb Q^P$. If

$$\dim_{\mathbb Q}(\mathbb Q^P/R)=1,$$

then

$$\boxed{\max P\le d(d+1)^{d-1}(d-1)^{(d-1)/2}.} \tag{1}$$

Choose a maximum dissociated set $D=\{b_1,\ldots,b_d\}$. For each $x\in P$, choose a ternary representation $v_x\in\{-1,0,1\}^d$ on $D$, taking $v_{b_i}=e_i$. Every defining relation $e_x-\sum_i(v_x)_i e_{b_i}$ has support at most $d+1$, so belongs to $R$.

Eliminating all generators outside $D$ therefore identifies the short presentation with the quotient of $\mathbb Q^d$ by the rows

$$u=\sum_x\varepsilon_x v_x$$

coming from short relations. Each entry of each row has absolute value at most $d+1$. The rows span a $(d-1)$-dimensional space, by the rank-one hypothesis.

Choose $d-1$ independent rows as a matrix $U$. Its kernel is the line spanned by $b=(b_1,\ldots,b_d)$. Moreover $\gcd(b_1,\ldots,b_d)=1$: $D\subset P$ and the ternary representations show that $D$ and $P$ generate the same subgroup of $\mathbb Z$.

The nonzero vector of signed maximal minors of $U$ is consequently an integer multiple of $b$. Every minor has $d-1$ rows of Euclidean norm at most $(d+1)\sqrt{d-1}$. Hadamard's determinant inequality gives

$$\max_i b_i\le(d+1)^{d-1}(d-1)^{(d-1)/2}.$$

Finally, each positive ternary combination of $D$ is at most $\sum_i b_i\le d\max_i b_i$, proving (1).

By the short-relation reduction, a primitive least-dimensional failure of the strong odd bound satisfies the hypothesis. For example, a least failure in dimension five would have all its weights at most $103680$.

This makes each fixed least-failure dimension a finite problem. It does not bound that dimension, provide an exclusion certificate, or resolve the logarithmic lower bound. No search of the universe in (1) was undertaken.