Uniform extraction for all distinct-prime tensors

---

Theorem and standing

There is one $\delta_0>0$ such that every finite subset of every tensor $\bigotimes_{p\in J}S_p$, where $J$ is a finite set of distinct odd primes, contains a dissociated subset of relative size at least $\delta_0$.

More generally, all roots of unity, considered in the discrete additive group $\mathbb C_d$, are proportionately dissociated and therefore Sidon. The proof below is author-recorded; two separate reviewers checked it, and it remains subject to external review. No publication-priority claim is made. The roots-of-unity conclusions are also machine-checked in Lean and recorded as the native claims L20 (analytic Sidon) and L21 (one extraction denominator) at tier 2; the Lean proof map lists the modules and the finite Pisier implication they prove in place of the cited theorem.

This is not a counterexample to E0774. The missing theorem is still unbounded full-dissociated coloring. Distinct-prime grids are a more structured candidate than the original large-side schedule; unlike that schedule they include the checked $S_3\otimes S_5\otimes S_7$ three-color obstruction.

1. What the tensor theorem supplies

The proved tensor extension and rounding lemmas apply to any finite list of side sizes at least three, not only the list chosen there. For prime sides set

$$a_p=\frac{p-1+H_{p-1}}p, \qquad Q(P,x)=\pi^2\prod_{P<p\leq x}a_p.$$

Starting at the free singleton, the loss in positive-real Sidon parameter from adding all primes above $P$, through any finite stage, is at most

$$\frac12\sum_{p>P}\frac{Q(P,p^-)}{p^2}. \tag{1}$$

We prove the completely explicit bound

$$P=2^{40}\quad\Longrightarrow\quad \frac12\sum_{p>P}\frac{Q(P,p^-)}{p^2}<2^{-18}. \tag{2}$$

All logarithms on this page are natural.

2. Elementary prime estimates, with constants

Let $\Lambda(n)$ be the von Mangoldt function, $\psi(x)=\sum_{n\leq x}\Lambda(n)$, and $A(x)=\sum_{p\leq x}(\log p)/p$. For every real $x\geq2$,

$$\psi(x)<4x,\qquad |A(x)-\log x|\leq5. \tag{3}$$

For a positive integer $n$, the exponent of a prime in $\binom{2n}{n}$ is a sum of nonnegative terms $\lfloor2n/p^j\rfloor-2\lfloor n/p^j\rfloor$. Each prime power in $(n,2n]$ contributes one. Hence $\psi(2n)-\psi(n)\leq\log\binom{2n}{n}\leq2n\log2$. Sum on dyadic scales and then round $x$ up to the next power of two to obtain $\psi(x)<4x$.

The identity

$$\log(n!)=\sum_{m\leq n}\Lambda(m)\lfloor n/m\rfloor$$

and (3)'s first bound show $\sum_{m\leq n}\Lambda(m)/m\in[\log n-1,\log n+4]$, using $n\log n-n\leq\log(n!)\leq n\log n$. The omitted prime-power terms satisfy

$$\sum_p\sum_{j\geq2}\frac{\log p}{p^j} \leq\sum_{n\geq2}\frac{\log n}{n(n-1)}<3.$$

For the last estimate use $1/[n(n-1)]\leq2/n^2$ and $\sum_{n\geq2}(\log n)/n^2 \leq(\log2)/4+\int_2^\infty(\log t)/t^2\,dt<5/4$. Thus $A(n)\in[\log n-4,\log n+4]$. Passing to $n=\lfloor x\rfloor$ costs less than $\log2<1$, proving (3).

Write $A(x)=\log x+E(x)$, $|E(x)|\leq5$. Partial summation gives

$$B(x):=\sum_{p\leq x}\frac1p =\log\log x+C+R(x),\qquad |R(x)|\leq\frac{10}{\log x}. \tag{4}$$

Explicitly, take $C=1-\log\log2+\int_2^\infty E(t)/(t\log^2t)\,dt$; the remainder is $E(x)/\log x-\int_x^\infty E(t)/(t\log^2t)\,dt$. These estimates are proved here. They also agree with the checked full statements of Theorems 13 and 15 in Tao's author lecture notes.

3. The logarithmic saving makes the full prime tail summable

Put $\beta=\log(3/2)$, so $1/3<\beta<1/2$. Convexity of $1/x$ on each interval $[j-1/2,j+1/2]$ gives

$$H_{p-1}-1=\sum_{j=2}^{p-1}\frac1j \leq\int_{3/2}^{p-1/2}\frac{dx}{x} <\log p-\beta.$$

Consequently $\log a_p\leq(\log p-\beta)/p$. From (3)–(4), for $x\geq P\geq e^{20}$,

$$\begin{aligned} \log\frac{Q(P,x)}{\pi^2} &\leq A(x)-A(P)-\beta(B(x)-B(P))\\ &\leq\log(x/P)-\beta\log\frac{\log x}{\log P} +10+\frac{20\beta}{\log P}\\ &<\log(x/P)-\beta\log\frac{\log x}{\log P}+11. \end{aligned}$$

Thus

$$Q(P,x)\leq\pi^2e^{11}\frac xP \left(\frac{\log P}{\log x}\right)^\beta. \tag{5}$$

One more partial summation, now integrating $(\log t)^{-1-\beta}$ against $dA(t)$, gives

$$\sum_{p>P}\frac1{p(\log p)^\beta} \leq(\log P)^{-\beta} \left(\frac1\beta+\frac{10}{\log P}\right) <4(\log P)^{-\beta}. \tag{6}$$

Indeed the main term is the integral against $d\log t$; the boundary error and error integral are each at most $5(\log P)^{-1-\beta}$. This also proves convergence, without the prime number theorem.

Since $a_p\geq1$, (5) may bound $Q(P,p^-)$ by $Q(P,p)$. Combining (5)–(6) bounds (1) by $2\pi^2e^{11}/P<2^{22}/P$. The last numerical inequality follows from $\pi^2<10$, $e<3$, and $20\cdot3^{11}<2^{22}$. For $P=2^{40}$, $\log P>20$, proving (2).

4. One extraction constant for all finite prime sets

Nest the tensors over the increasing primes above $P$ using the primitive slice embeddings of the original tensor theorem. By (2), their union has positive-real Sidon parameter at least $1-2^{-18}$. Pisier's Theorem 2 applies once to that union, giving a single $\delta>0$ for every finite subset. Any finite prime-tail tensor is a slice of a sufficiently large initial tensor and inherits this same constant.

Let $M=\prod_{p\leq P}p$, a fixed finite integer. In a tensor containing small primes, partition a given subset by its small-prime coordinates. There are at most $M$ slices. One contains at least $1/M$ of the subset; it is a scalar-tensor copy of a prime-tail configuration. Extract in that slice. Thus $\delta_0=\delta/M$ works for all distinct-prime tensors. The constant is small and unquantified, but crucially it is independent of the finite prime set. No finite-stage sequence of Pisier constants is used.

5. Cyclotomic interpretation and prime powers

For squarefree $r$, multiplication identifies $\bigotimes_{p\mid r}\mathbb Q(\zeta_p)$ with $\mathbb Q(\zeta_r)$: the map is surjective and both dimensions are $\prod_{p\mid r}(p-1)=\varphi(r)$. The frequency configuration becomes exactly the $r$-th roots of unity $\mu_r$. For $p=2$, the two antipodal columns are dealt with by the small-coordinate slice selection; the analytic extension never uses $S_2$.

For arbitrary $n$, put $r=\operatorname{rad}(n)$. The $n/r$ distinct multiplicative $\mu_r$-cosets span subspaces $\zeta\mathbb Q(\zeta_r)$. They sum to $\mathbb Q(\zeta_n)$, and the sum of their dimensions is $(n/r)\varphi(r)=\varphi(n)$. Hence their sum is direct: every rational, and therefore every signed, relation separates by these cosets.

The cyclotomic degree fact is Proposition 1.2 of the checked publisher copy of Ramsey–Graham, pp. 330 and 338 of the held copy; the same coset separation is their Theorem 2.12, whose proof was also checked.

Apply $\delta_0$-extraction in each coset and unite the extracted sets. Coset separation makes the union dissociated. Every finite set of roots of unity lies in some $\mu_n$, proving the global assertion. This is ordinary Sidonicity, not positive-real Sidonicity of the full root set: antipodal pairs make its positive-real parameter zero.

6. Exact E0774 obligation remaining

Let $p_1=3,p_2=5,\ldots$ be all odd primes, and let $E_k=\bigotimes_{i\leq k}S_{p_i}$. The proved extraction constant $\delta_0$ is uniform. These grids have no zero or antipodal pair, so the positive-integer realization and separated-block assembly apply exactly. It now suffices to prove

$$\sup_k\chi_{\rm diss}(E_k)=\infty. \tag{7}$$

Equation (7) is unproved. The divergence of $\sum_p1/p$ only establishes unbounded rational-independent covering, not (7). The existence of all regular-polygon relations likewise does not by itself establish a coloring obstruction. This package does not claim that all roots of unity, or the associated positive-integer block union, fail a finite dissociated partition.