Ideal class defined in terms of linear fractional ideal is chern class?

61 Views Asked by At

Let $O$ be a number ring(i.e. dedekind ring). Let $M$ be a finitely generated torsion free and non-zero module over $O$. $M$ can be embedded into a vector space over $Frac(O)=K$. There $M$ is a submodule of some $O$ free module contained in $V$. Fix any free module $F\subset V$. We call $M$ an $O$ lattice if $F$ is of full rank. Define $[F:M]$ as the following.

Consider a place $p$ of $O$. Let $O_p$ be completion of $O$ at $p$ and $K_p$ be the corresponding field completion of $K$ at $p$. Then there is automorphism $l_p\in Aut_{K_p}(V\otimes_KK_p)$ s.t. $l_p(F\otimes_OO_p)=M\otimes_OO_p$. Then define $[F\otimes_OO_p:M\otimes_OO_p]=p^{e}O_p\subset O_p$. Since there are only finitely many places $F$ is distinct from $M$, let $[F:M]=\prod_{i<\infty}p_i^{e_i}O$ where $p_i$ runs through only finitely many places s.t. $[F\otimes_OO_p:M\otimes_OO_p]\neq O_p$. This defines a fractional ideal in $O$.

Let $c(M)=[F:M]$ for a choice of free module $F$ containing $M$ s.t. $F\otimes_OK=V=M\otimes_OK$. First $c(M)$ is independent of choice of $F$. Then one also have $c(M_1\oplus M_2)=c(M_1)c(M_2)$.

$\textbf{Q:}$ It seems that $C(M_1\oplus M_2)=c(M_1)c(M_2)$ defines a chern class. Note that if $M=F$ free module, then $c(F)=1\in Cl(O)$ where $Cl(O)$ is class group of $O$. The only question is naturality here. Do I have a chern class functor defined here?

This is just my thought when I saw the formula on Pg 95 Thm 13 of Taylor Frohlich Algebraic Number Theory.