show/hide this revision's text 2 deleted 3 characters in body

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 $H^\ast(X;\mathbb{Z})$ 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?

show/hide this revision's text 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?