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TauCeti.Algebra.Homology.EulerCharacteristic.ExtEuler.DualNumbers

The dual numbers: Ext-finite but not Ext-bounded #

Let k be a field, let A = k[ε] be the dual numbers k[ε]/(ε²), and let S = A/(ε) be the residue field of A, viewed as an A-module. Every Ext group of the pair (S, S) is a one-dimensional k-vector space (TauCeti.extDualNumberResidueEquiv), so TauCeti.IsExtFinite holds while none of the groups vanishes and TauCeti.IsExtBounded fails: the alternating sum ∑ n, (-1)ⁿ dim_k Extⁿ(S, S) is not a finite sum. This is the example that separates the two halves of TauCeti.IsEulerAdmissible. Since TauCeti.extEuler takes a proof of TauCeti.IsEulerAdmissible as an argument, it cannot be instantiated for this pair and exposes no totalised fallback value.

Main results #

References #

No degree is a vanishing bound for the Ext groups of k[ε]/(ε) against itself, because none of them vanishes.

The dual-numbers rejection. Extⁿ(S, S) never vanishes, so the pair (S, S) is not Ext-bounded.

Every Ext group of k[ε]/(ε) against itself is a finite-dimensional k-vector space.

The dual-numbers rejection. The residue field S of k[ε] is not Euler-admissible against itself: its Ext groups are all one-dimensional, so Ext-finiteness holds (TauCeti.isExtFinite_dualNumberResidue), but none of them vanishes, so the alternating sum ∑ n, (-1)ⁿ dim_k Extⁿ(S, S) is not a finite sum. Because TauCeti.extEuler consumes a proof of TauCeti.IsEulerAdmissible, it cannot be instantiated for this pair and exposes no totalised fallback value.