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Hironaka.Resolution.Algebraic.MarkedOrderReduction.NonmonomialInvariant

Two marked ideals with the same nonmonomial part keep it along a common sequence #

In Step 2 of the proof of [Kol07, Theorem 107] ([Kol07, 111, Step 2]) a round of order reduction for N(I)^m + I^s is at once a smooth blow-up sequence of order ≥ s for (N(I), s) and of order ≥ m for (I, m); after it, "we stop when cosupp(I_r, m) ∩ cosupp(N(I_r), s) = ∅", which compares the nonmonomial part N(I_r) of the new marked ideal with the marked transform N_r := Π^{-1}_*(N(I), s) of the old one. The comparison is the invariant proved here, not in the sources as such: for two marked ideals (I, m) and (J, s) on the same triple with N(I) = N(J), along a smooth blow-up sequence of order ≥ m for the first and ≥ s for the second, the nonmonomial parts of the two induced marked ideals agree at the end. One blow-up is the identity N(π^{-1}_*(I, m)) = N(π^* N(I)) of Hironaka/Resolution/Algebraic/MarkedOrderReduction/Transform.lean (nonmonomialPart_markedTransform), applied to both marked triples: N(I_{i+1}) = N(π^* N(I_i)) = N(π^* N(J_i)) = N(J_{i+1}). The variant with the second sequence unmarked at the exact order, used for the rounds of Step 1, is nonmonomialPart_markedTransformSeq_last_eq in Hironaka/Resolution/Algebraic/MarkedOrderReduction/Step1NonmonomialPart.lean.

For two marked ideals (I, m) and (J, s) on the same triple with N(I) = N(J), along a smooth blow-up sequence of order ≥ m for (I, m) and of order ≥ s for (J, s), the nonmonomial parts of the induced marked ideals agree at the end, with respect to the induced boundary: nonmonomialPart_markedTransform applied to both marked triples at every stage (the identity "N((Π_1)^{-1}_*(I, m)) = (Π_1)^{-1}_* N(I)" of [Kol07, 111, Step 1], read modulo N). Step 2 of the proof of Theorem 107 uses it at J = N(I) and s the separation order: N(I_r) = N(Π^{-1}_*(N(I), s)). Not in the sources as such.