2
$\begingroup$

I'm sure I am missing something quite simple, but why are divisors on an algebraic variety actually homological cycles? I can see them as chains, but in order to take homology (as one always does), or to integrate along them, I would like to see them as singular cycles ($\partial D=0$). (Not just algebraic cycles, of course, which they are by definition.)

Perhaps I am just a bit confused about definitions. Thank you in advance for any clarifying suggestion.

$\endgroup$
2
  • 1
    $\begingroup$ By algebraic variety you probably mean "complete algebraic variety over ${\mathbb C}$". Otherwise, you have to do more work defining the "correct" homology theory. Compare ac.els-cdn.com/0040938362900940/… $\endgroup$ Mar 6 '18 at 21:08
  • $\begingroup$ Yes, a complete algebraic variety over $\mathbb C$ is the context, thank you for pointing out. $\endgroup$
    – W. Rether
    Mar 6 '18 at 21:16