claimroutine

Polynomial error delta is zero or bounded

Splits into δ=0\delta=0 or ∣δ∣≤CfBD|\delta|\le C_fB^D
polynomial upper bound in the height contradiction
Claim·§2.7 The height contradiction·inside propFinite test success gives integer zero

Statement (paper)

Let δ=f(η(a1),…,η(an))∈∗Z\delta=f(\eta(a_1),\ldots,\eta(a_n))\in{}^*\Z. Either δ=0\delta=0, in which case ff has a zero in Zn\Z^n by elementarity, or δ≠0\delta\ne0. With B=max⁡(1,∣η(a1)∣,…,∣η(an)∣)B=\max(1,|\eta(a_1)|,\ldots,|\eta(a_n)|) and D=max⁡(1,deg⁡f)D=\max(1,\deg f), there is an ordinary constant Cf>0C_f>0 such that ∣δ∣≤CfBD|\delta|\le C_f B^D.

In plain English

Set δ=f(η(a1),…,η(an))\delta=f(\eta(a_1),\ldots,\eta(a_n)) in ∗Z{}^*\Z. Either δ=0\delta=0, which gives an integer zero of ff by elementarity, or δ≠0\delta\neq 0 and ∣δ∣≤CfBD|\delta|\le C_f B^D, where B=max⁡(1,∣η(ai)∣)B=\max(1,|\eta(a_i)|), D=max⁡(1,deg⁡f)D=\max(1,\deg f), and Cf>0C_f>0 is an ordinary constant depending only on ff.

Why it's here

This splits the final argument into two cases. The case δ=0\delta=0 finishes the proposition at once. In the case δ≠0\delta\ne0 the polynomial size bound is the upper half of the height contradiction; the later claims show that δ\delta is divisible by large powers of contact radicals, which forces ∣δ∣|\delta| to be too large to satisfy this bound. The bound follows from the triangle inequality, since each monomial of ff has degree at most DD and so is at most a constant times BDB^D, applied to the tuple obtained from the compactness model of the previous lemma.}
Classified routine:
Direct case split plus the triangle-inequality bound on a polynomial evaluated at a tuple of height at most B (sum of absolute coefficients times B^D), transferred by elementarity to the nonstandard integers from the compactness model of the previous lemma.

Figures

A proof-flow diagram of six boxes connected by arrows. The left column starts with varying $A\in R$ and choosing a representation at precision $K$ with slopes $\widetilde s_i$, then passes to local analysis at each prime $q\mid M_K(\widetilde s_i)$ (pole parity and coarsening), then to the paired elliptic tests giving $v_q(\widetilde a-\eta(a))\ge K$. The right column keeps the root tuple fixed with nonzero $\delta=f(\eta(\mathbf a))$ bounded by $C_fB^D$; an arrow labeled 'polynomial evaluation' and an arrow from the tests feed a box concluding $M_K(\widetilde s_i)^K\mid\delta$. A final wide box applies the five-point height estimate to get $h(\widetilde A)\ll_f B$ uniformly for $A\in R$.
Figure 1diagramp. 16
The use of one nonzero δ\delta for every A∈RA\in R. All slopes shown come from the representation chosen at the fixed ordinary precision KK, and MKM_K uses the fixed exceptional set for KK. The comparison holds for every a∈Ra\in R, so it applies to the fixed root tuple. The implied constant in the final bound is ordinary and independent of AA and a\mathbf a.
What it shows
A proof-flow diagram of six boxes connected by arrows. The left column starts with varying A∈RA\in R and choosing a representation at precision KK with slopes s~i\widetilde s_i, then passes to local analysis at each prime q∣MK(s~i)q\mid M_K(\widetilde s_i) (pole parity and coarsening), then to the paired elliptic tests giving vq(a~−η(a))≥Kv_q(\widetilde a-\eta(a))\ge K. The right column keeps the root tuple fixed with nonzero δ=f(η(a))\delta=f(\eta(\mathbf a)) bounded by CfBDC_fB^D; an arrow labeled 'polynomial evaluation' and an arrow from the tests feed a box concluding MK(s~i)K∣δM_K(\widetilde s_i)^K\mid\delta. A final wide box applies the five-point height estimate to get h(A~)≪fBh(\widetilde A)\ll_f B uniformly for A∈RA\in R.
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Sources

PaperOpen
Section
§2.7 The height contradiction
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sections/02-reduction.tex:762
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