Euler class group presentation

PDF p. 3: Expected Euler class comparison

Let $k$ be an algebraically closed field and let $A$ be a reduced, finitely generated commutative $k$-algebra of Krull dimension $d\ge 2$. The height of an ideal is the least height of a prime containing it; the height of a prime is the supremum of the lengths of chains of primes ending there. An ideal $J$ is $\mathfrak{m}$-primary if $\sqrt{J}=\mathfrak{m}$ and $ab\in J$, $a\mathrm{\notin }J$ imply $b\in \mathfrak{m}$.

Let $F(A)$ be the free abelian group on the set of pairs $(J,{\omega }_J)$ in which $J$ is an $\mathfrak{m}$-primary ideal of height $d$ for a maximal ideal $\mathfrak{m}$, and ${\omega }_J:(A\mathrm{/}J{)}^d\twoheadrightarrow J\mathrm{/}{J}^2$ is an $A\mathrm{/}J$-linear surjection. The map is called an orientation here; distinct surjections are distinct generators.

For an ideal $I$ of height $d$ and an orientation ${\omega }_I:(A\mathrm{/}I{)}^d\twoheadrightarrow I\mathrm{/}{I}^2$, write its unique irredundant primary decomposition as $I={\bigcap }_i{J}_i$, where $J_i$ is ${\mathfrak{m}}_i$-primary for distinct maximal ideals. The Chinese remainder identifications induce orientations ${\omega }_{J_i}:(A\mathrm{/}{J}_i{)}^d\twoheadrightarrow J_i\mathrm{/}{J}_i^2$. Put $[(I,{\omega }_I)]={\sum }_i[(J_i,{\omega }_{J_i})]\in F(A)\mathrm{.}$ An orientation is globally lifted if there is an $A$-linear surjection $\tilde{\omega }:A^d\twoheadrightarrow I$ whose induced map modulo $I$ is ${\omega }_I$. Let $K(A)$ be the subgroup generated by all $[(I,{\omega }_I)]$ with globally lifted orientation. The Euler class group in this presentation is the quotient $E(A)=F(A)\mathrm{/}K(A)$. This definition asserts no comparison with a motivic group.

Given: $A$, $d$, $I$, and ${\omega }_I$ as above.

Goal: Show that the component sum and quotient are well defined.

1.1 The ring $A$ is Noetherian because it is a finitely generated algebra over a field. Every prime containing $I$ has height at least $d$ and at most $\dim A=d$, so it is maximal. Thus $B=A\mathrm{/}I$ is a Noetherian ring of dimension zero. [assume-hyp]

1.2 The spectrum of a Noetherian ring is a Noetherian topological space: ascending chains of ideals stabilize. Such a space is a finite union of irreducible closed subsets; otherwise a maximal closed counterexample could be split into two smaller closed subsets. Every nonempty irreducible closed subset of $\mathrm{Spec}B$ has a generic prime, and that prime is maximal. Hence $B$ has finitely many maximal ideals ${\mathfrak{n}}_1,\dots ,{\mathfrak{n}}_r$. The empty-spectrum case gives the zero ring and the empty sum. [step 1.1]

1.3 The nilradical $N$ of $B$ is the intersection of these maximal ideals. It is finitely generated, and each of its generators is nilpotent, so $N^s=0$ for some $s$. Distinct maximal ideals are comaximal, so their powers ${\mathfrak{n}}_i^s$ are pairwise comaximal and their product is $N^s=0$. The Chinese remainder map therefore gives $B\cong {\prod }_{i}B\mathrm{/}{\mathfrak{n}}_i^s$. Each factor is local with nilpotent maximal ideal, hence its zero ideal is primary: an element outside the maximal ideal is a unit. Pulling the kernels of the factor projections back to $A$ gives an irredundant decomposition $I={\bigcap }_i{J}_i$ with $J_i$ primary and radical the inverse image ${\mathfrak{m}}_i$ of ${\mathfrak{n}}_i$. [step 1.2, construct]

1.4 This decomposition is unique. Indeed, in any irredundant primary decomposition of $I$, the radicals are precisely the maximal ideals containing $I$: their intersection is $\sqrt{I}$, and primality and maximality force the two finite sets of radicals to coincide. Localizing at ${\mathfrak{m}}_i$ makes every other component the unit ideal. An ${\mathfrak{m}}_i$-primary ideal is the contraction of its localization at ${\mathfrak{m}}_i$, since multiplication by an element outside ${\mathfrak{m}}_i$ cannot carry an element outside the ideal into it. Thus its component is necessarily $I{A}_{{\mathfrak{m}}_i}\cap A$. [step 1.3]

1.5 Write $L_i={\bigcap }_{j\mathrm{\ne }i}{J}_j$. The ideals $J_i$ and $L_i$ are comaximal, so $I=J_i{L}_i$, $I+J_i^2=J_i$, and $I\cap J_i^2=I{J}_i$. Consequently inclusion induces an isomorphism $(I\mathrm{/}{I}^2){\otimes }_{A\mathrm{/}I}A\mathrm{/}{J}_i\cong I\mathrm{/}I{J}_i\cong J_i\mathrm{/}{J}_i^2\mathrm{.}$ Tensoring ${\omega }_I$ with $A\mathrm{/}{J}_i$ and using this isomorphism yields the stated surjection ${\omega }_{J_i}$. By step 1.4, neither the component ideals nor their induced maps depend on a decomposition choice. [step 1.4, construct]

1.6 The indicated pairs form a set, so the free abelian group exists. The globally lifted component sums form a subset of it; their integer span is a subgroup, and its cosets form an abelian quotient. Steps 1.4–1.5 establish that each relation is unambiguous. [step 1.4, step 1.5, construct] ∎