I am dubious that the Euler characteristic of $X$ can be considered well-defined if $H^\ast(X;\mathbb{Q})$ is finitely-generated but $H^\ast(X;\mathbb{Z})$ is not. If $H^\ast(X;\mathbb{Z})$ H^*(X;\mathbb{Z})$is finitely generated then the Euler characteristic of$H^\ast(X;K)$is constant for all fields$K$. If$X=\mathbb{R}P^\infty$then we have an Euler characteristic of$1$for any field whose characteristic is odd or zero. In characteristic two the Poincare series can be regarded as the rational function$f(t)=1/(1-t)$and by putting$t=-1$you get an Euler characteristic of$1/2$. Perhaps there is some context in which the Euler characteristic can be defined as an adele? 1 This is really a comment rather than an answer. I am dubious that the Euler characteristic of$X$can be considered well-defined if$H^\ast(X;\mathbb{Q})$is finitely-generated but$H^\ast(X;\mathbb{Z})$is not. If$H^\ast(X;\mathbb{Z})$is finitely generated then the Euler characteristic of$H^\ast(X;K)$is constant for all fields$K$. If$X=\mathbb{R}P^\infty$then we have an Euler characteristic of$1$for any field whose characteristic is odd or zero. In characteristic two the Poincare series can be regarded as the rational function$f(t)=1/(1-t)$and by putting$t=-1$you get an Euler characteristic of$1/2\$. Perhaps there is some context in which the Euler characteristic can be defined as an adele?