A valuation on a power of a spanned ideal #
If v ≤ δ on a generating set s and v ≤ 1 on the ideal Ideal.span s, then
v ≤ δ ^ n on Ideal.span s ^ (n + 1).
s is an arbitrary Set R: the bound needs no finiteness. Finiteness enters only at the
intended call site, Wedhorn's Theorem 7.10, where δ is chosen as the maximum of v over a
finite generating set of an ideal of definition — a maximum that needs the set to be finite,
whereas this bound does not.
Why the two hypotheses, and why the shift #
Both are needed and neither can be dropped. An element of Ideal.span s ^ (n + 1) is a sum of
terms c * t₀ * ⋯ * tₙ with tᵢ ∈ s and the coefficient c arbitrary, so a bound on the
generators alone says nothing: v c can be as large as it likes. The trick is to attach the
coefficient to one generator — c * t₀ lies in the ideal, so v (c * t₀) ≤ 1 — and to bound the
remaining n bare generators by δ each. That is exactly where the shift comes from: n + 1
factors give δ ^ n, because one of them is spent absorbing the coefficient.
It is also why the proof does not simply induct along I ^ (n + 1) = I ^ n * I: splitting off an
element of I rather than a generator only ever yields the bound 1, so the exponent never
grows. Instead the power is rewritten as the span of the pointwise power s ^ n
(Submodule.span_pow), which keeps the generators visible.
Main results #
References #
- T. Wedhorn, Adic Spaces, arXiv:1910.05934v1, Theorem 7.10.
A valuation on a power of a span. If v ≤ δ on a generating set s and v ≤ 1 on
Ideal.span s, then v ≤ δ ^ n on Ideal.span s ^ (n + 1).
One of the n + 1 factors is spent absorbing the arbitrary coefficient, which is why the exponent
on the right is n and not n + 1; see the module docstring.