TITLE: Nilpotent elements of cohomology ring
QUESTION [6 upvotes]: It is well known that every element of positive degree in the Cech cohomology ring of a compact space is nilpotent. How can ı prove that?

REPLY [13 votes]: The Cech cohomology ring of a space $X$ is the direct limit of the cohomology rings of nerves of open covers of $X$.  If $X$ is compact, then finite open covers are cofinal in all open covers, so every Cech cohomology class on $X$ comes from the nerve of a finite open cover.  But the nerve of a finite open cover of $X$ is just a finite simplicial complex, and in particular has nontrivial cohomology in only finitely many degrees.  Thus every positive-degree cohomology class on the nerve of a finite open cover is nilpotent, and remains nilpotent when mapped into the Cech cohomology of $X$.
(Incidentally, this result is very false for singular cohomology.  By a theorem of Przeździecki (see the references and comment thread here), every topological space of cardinality less than the least measurable cardinal is weak equivalent to a compact Hausdorff space.  In particular, if you take any familiar example of a space $X$ with a non-nilpotent cohomology class in positive degree (e.g., $X=\mathbb{C}P^\infty$), there is a compact Hausdorff space $Y$ which is weak equivalent to $X$ and in particular has an isomorphic singular cohomology ring.)