Let $M$ be a simply connected closed Riemannian manifold. How does one find a necessary condition going both ways that may be imposed on $M$ (perhaps on the curvature of $M$ and on torsion) which guarantees that the rational cohomology ring $H^\*(M;\mathbb{Q})$ needs at least two generators? That is, how does one force $M$ not to have rational cohomology that is the quotient of a polynomial ring? 


Cross-posting on MSE: <https://math.stackexchange.com/questions/2857279/show-that-the-rational-cohomology-ring-hm-mathbbq-needs-at-least-two-ge>


Any help would be much appreciated. Thanks in advance!



Show that the rational cohomology ring $H^*(M;\mathbb{Q})$ needs at least two generators
If $M$ is simply-connected and has reducible holonomy, then a [theorem of de Rham implies that $M$ is a product](https://en.wikipedia.org/wiki/Holonomy#Reducible_holonomy_and_the_de_Rham_decomposition), and hence does not have homology generated by one element. 


